Rd Sharma Xi 2018 for Class 12 Science Math Chapter 29 - Limits

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Rd Sharma Xi 2018 Solutions for Class 12 Science Math Chapter 29 Limits are provided here with simple step-by-step explanations. These solutions for Limits are extremely popular among Class 12 Science students for Math Limits Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma Xi 2018 Book of Class 12 Science Math Chapter 29 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma Xi 2018 Solutions. All Rd Sharma Xi 2018 Solutions for class Class 12 Science Math are prepared by experts and are 100% accurate.

Page No 29.11:

Question 1:

Show that $\lim_{x \to 0} \frac{x}{|x|}$ does not exist.

Answer:

$\lim_{x \to 0} (\frac{x}{|x|})$

Left hand limit:
$\lim_{x \to 0^{-}} (\frac{x}{|x|}) Let x = 0 - h, where h \to 0 . \Rightarrow \lim_{h \to 0} (\frac{0 - h}{|0 - h|}) = \lim_{h \to 0} (\frac{- h}{h}) = - 1$

Right hand limit:
$\lim_{x \to 0^{+}} \frac{(x)}{|x|} Let x = 0 + h, where h \to 0 . \lim_{h \to 0} (\frac{0 + h}{|0 + h|}) = \lim_{h \to 0} (\frac{h}{h}) = 1$

Left hand limit ≠ Right hand limit
$Thus, \lim_{x \to 0} (\frac{x}{|x|}) does not exist .$

Page No 29.11:

Question 2:

Find k so that $\lim_{x \to 2} f (x)$ may exist, where $f (x) = \{\begin{cases} 2 x + 3, & x \leq 2 \\ x + k, & x > 2 \end{cases} .$

Answer:

$f (x) = \{\begin{cases} 2 x + 3, & x \leq 2 \\ x + k, & x > 2 \end{cases} Left hand limit : \lim_{x \to 2^{-}} f (x) = \lim_{x \to 2^{-}} (2 x + 3) Let x = 2 - h, where h \to 0 . \lim_{h \to 0} [2 (2 - h) + 3] = 2 (2 - 0) + 3 = 7 Right hand limit : \lim_{x \to 2^{+}} f (x) = \lim_{x \to 2^{+}} (x + k) Let x = 2 + h, where h \to 0 . \lim_{h \to 0} (2 + h + k) = 2 + k$

Now, $\lim_{x \to 2} f (x)$ exists if the left hand limit is equal to the right hand limit.

⇒7 = 2 + k
k = 5

Page No 29.11:

Question 3:

Show that $\lim_{x \to 0} \frac{1}{x}$ does not exist.

Answer:

$\lim_{x \to 0} (\frac{1}{x}) Left hand limit : \lim_{x \to 0^{-}} (\frac{1}{x}) Let x = 0 - h, where h \to 0 \lim_{h \to 0} (\frac{1}{0 - h}) = - \infty Right hand limit : \lim_{x \to 0^{+}} (\frac{1}{x}) Let x = 0 + h, where h \to 0 \lim_{h \to 0} (\frac{1}{0 + h}) = \infty \lim_{x \to 0^{-}} (\frac{1}{x}) \neq \lim_{x \to 0^{+}} (\frac{1}{x}) Thus, \lim_{x \to 0} (\frac{1}{x}) does not exist .$

Page No 29.11:

Question 4:

Let f(x) be a function defined by $f (x) = \{\begin{cases} \frac{3 x}{|x| + 2 x}, & x \neq 0 \\ 0, & x = 0 \end{cases} .$
Show that $\lim_{x \to 0} f (x)$ does not exist.

Answer:

$f (x) = \{\begin{cases} \frac{3 x}{|x| + 2 x}, & x \neq 0 \\ 0, & x = 0 \end{cases} Left hand limit : \lim_{x \to 0^{-}} [\frac{3 x}{|x| + 2 x}] Let x = 0 - h, where h \to 0 . \Rightarrow \lim_{h \to 0} [\frac{3 (- h)}{|- h| + 2 (- h)}] = \lim_{h \to 0} [\frac{- 3 h}{h - 2 h}] = \lim_{h \to 0} [\frac{- 3 h}{- h}] = 3 Right hand limit : \lim_{x \to 0^{+}} (\frac{3 x}{|x| + 2 x}) Let x = 0 + h, where h \to 0 . \Rightarrow^{} \lim_{h \to 0} (\frac{3 h}{|h| + 2 h}) = \lim_{h \to 0} (\frac{3 h}{h + 2 h}) = 1 \lim_{x \to 0^{-}} (\frac{3 x}{|x| + 2 x}) \neq \lim_{x \to 0^{+}} (\frac{3 x}{|x| + 2 x}) Thus, \lim_{x \to 0} f (x) does not exist .$

Page No 29.11:

Question 5:

Let $f (x) = \{\begin{array}{l} x + 1, & if x \geq 0 \\ x - 1, & if x < 0 \end{array} .$ Prove that $\lim_{x \to 0} f (x)$ does not exist.

Answer:

$f (x) = \{\begin{cases} x + 1, & x \geq 0 \\ x - 1, & x < 0 \end{cases} RHL : \lim_{x \to 0^{+}} f (x) = \lim_{x \to 0} (x + 1) Let x = 0 + h, where h \to 0 . \lim_{h \to 0} (0 + h + 1) = 1 LHL : \lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{-}} (x - 1) Let x = 0 - h, where h \to 0 . \lim_{h \to 0} (0 - h - 1) = - 1 LHL \neq RHL Thus, \lim_{x \to 0} f (x) does not exist .$

Page No 29.11:

Question 6:

Let $f (x) = \{\begin{cases} x + 5, & if x > 0 \\ x - 4, & if x < 0 \end{cases}$ . Prove that $\lim_{x \to 0} f (x)$ does not exist.

Answer:

We have,

$f (x) = \{\begin{cases} x + 5, & if x > 0 \\ x - 4, & if x < 0 \end{cases}$

LHL of f(x) at x = 0

= $\lim_{x \to 0^{-}} f (x) = \lim_{h \to 0} f (0 - h) = \lim_{h \to 0} (0 - h - 4) = - 4$

RHL of f(x) at x = 0

= $\lim_{x \to 0^{+}} f (x) = \lim_{h \to 0} f (0 + h) = \lim_{h \to 0} (0 + h + 5) = 5$

Clearly, $\lim_{x \to 0^{-}} f (x) \neq \lim_{x \to 0^{+}} f (x)$

Hence, $\lim_{x \to 0} f (x)$ does not exist.

Page No 29.11:

Question 7:

Find $\lim_{x \to 3} f (x)$ , where $f (x) = \{\begin{cases} 4, & if x > 3 \\ x + 1, & if x < 3 \end{cases}$

Answer:

We have,

$f (x) = \{\begin{cases} 4, & if x > 3 \\ x + 1, & if x < 3 \end{cases}$

LHL of f(x) at x = 3

= $\lim_{x \to 3^{-}} f (x) = \lim_{h \to 0} f (3 - h) = \lim_{h \to 0} (3 - h + 1) = 4$

RHL of f(x) at x = 3

= $\lim_{x \to 3^{+}} f (x) = \lim_{h \to 0} f (3 + h) = \lim_{h \to 0} 4 = 4$

Clearly, $\lim_{x \to 3^{-}} f (x) = \lim_{x \to 3^{+}} f (x) = 4$

$∴ \lim_{x \to 3} f (x) = 4$

Page No 29.11:

Question 8:

If $f (x) = \{\begin{array}{l} 2 x + 3, & x \leq 0 \\ 3 (x + 1), & x > 0 \end{array} .$ Find $\lim_{x \to 0} f (x)$ and $\lim_{x \to 1} f (x)$ .

Answer:

$f (x) = \{\begin{cases} 2 x + 3, & x \leq 0 \\ 3 (x + 1), & x > 0 \end{cases} LHL : \lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{-}} (2 x + 3) Let x = 0 - h, where h \to 0 . \lim_{h \to 0} [2 (0 - h) + 3] = \lim_{h \to 0} [- 2 h + 3] = 3 RHL : \lim_{x \to 0^{+}} f (x) \lim_{x \to 0^{+}} 3 (x + 1) Let x = 0 + h, where h \to 0 . \lim_{h \to 0} 3 (0 + h + 1) = \lim_{h \to 0} (3 h + 3) = 3 ∴ LHL = RHL ∴ \lim_{x \to 0} f (x) = 3$

(ii)
$\lim_{x \to 1} f (x) = ? LHL : \lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{-}} 3 (x + 1) Let x = 1 - h, where h \to 0 . \lim_{h \to 0} 3 (1 - h + 1) = 6 RHL : \lim_{x \to 1^{+}} f (x) = \lim_{x \to 1^{+}} 3 (x + 1) Let x = 1 + h, where h \to 0 . \lim_{h \to 0} 3 (1 + h + 1) = 6 ∴ LHL = RHL \lim_{x \to 1} f (x) = 6$

Page No 29.11:

Question 9:

Find $\lim_{x \to 1} f (x)$ , if $f (x) = \{\begin{cases} x^{2} - 1, & x \leq 1 \\ - x^{2} - 1, & x > 1 \end{cases}$

Answer:

$\lim_{x \to 1} f (x) = ? f (x) = \{\begin{cases} x^{2} - 1, & x \leq 1 \\ - x^{2} - 1, & x > 1 \end{cases} LHL : \lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{-}} (x^{2} - 1) Let x = 1 - h, where h \to 0 . \lim_{h \to 0} ({(1 - h)}^{2} - 1) = 0 RHL : \lim_{x \to 1^{+}} f (x) \lim_{x \to 1^{+}} (- x^{2} - 1) Let x = 1 + h, where h \to 0 . \lim_{h \to 0} (- {(1 + h)}^{2} - 1) = - 2 LHL \neq RHL Thus, \lim_{x \to 1} f (x) d oes not exist .$

Page No 29.11:

Question 10:

Evaluate $\lim_{x \to 0} f (x)$ , where $f (x) = \{\begin{cases} \frac{|x|}{x}, & x \neq 0 \\ 0, & x = 0 \end{cases}$

Answer:

$f (x) = \{\begin{cases} \frac{|x|}{x}, & x \neq 0 \\ 0, & x = 0 \end{cases} LHL : \lim_{x \to 0^{-}} (\frac{|x|}{x}) Let x = 0 - h, where h \to 0 . \Rightarrow \lim_{h \to 0} (\frac{|0 - h|}{- h}) = \lim_{h \to 0} (\frac{h}{- h}) = - 1 RHL : \lim_{x \to 0^{+}} f (x) = \lim_{x \to 0^{+}} (\frac{|x|}{x}) Let x = 0 + h, where h \to 0 . \lim_{h \to 0} (\frac{|0 + h|}{0 + h}) = 1 LHL \neq RHL \underset{x \to 0}{Thus, \lim} f (x) does not exist .$

Page No 29.11:

Question 11:

Let a₁, a₂, ..., a_n be fixed real numbers such that
f(x) = (x − a₁) (x − a₂) ... (x − a_n)
What is $\lim_{x \to a_{1}} f (x) ?$ For a ≠ a₁, a₂, ..., a_n. Compute $\lim_{x \to a} f (x) .$

Answer:

$f (x) = (x - a_{1}) (x - a_{2}) . . . (x - a_{n}) \lim_{x \to a_{1}} f (x) = \lim_{x \to a_{1}} [(x - a_{1}) (x - a_{2}) . . . (x - a_{n})] = (a_{1} - a_{1}) (a_{1} - a_{2}) . . . (a_{1} - a_{n}) = 0 \lim_{x \to a} f (x) = \lim_{x \to a} (x - a_{1}) (x - a_{2}) . . . (x - a_{n}) = (a - a_{1}) (a - a_{2}) . . . (a - a_{n})$

Page No 29.11:

Question 12:

Find $\lim_{x \to 1^{+}} (\frac{1}{x - 1}) .$

Answer:

$\lim_{x \to 1^{+}} (\frac{1}{x - 1}) Let x = 1 + h, where h \to 0 . \lim_{h \to 0} (\frac{1}{1 + h - 1}) = \infty$

Page No 29.11:

Question 13:

Evaluate the following one sided limits:

(i)
$\lim_{x \to 2^{+}} \frac{x - 3}{x^{2} - 4}$

(ii)
$\lim_{x \to 2^{-}} \frac{x - 3}{x^{2} - 4}$

(iii)
$\lim_{x \to 0^{+}} \frac{1}{3 x}$

(iv)
$\lim_{x \to - 8^{+}} \frac{2 x}{x + 8}$

(v)
$\lim_{x \to 0^{+}} \frac{2}{x^{1 / 5}}$

(vi)
$\lim_{x \to \frac{π}{2}} \tan x$

(vii)
$\lim_{x \to - π / 2^{+}} \sec x$

(viii)
$\lim_{x \to 0^{-}} \frac{x^{2} - 3 x + 2}{x^{3} - 2 x^{2}}$

(ix)
$\lim_{x \to - 2^{+}} \frac{x^{2} - 1}{2 x + 4}$

(x)
$\lim_{x \to 0^{-}} 2 - \cot x$

(xi)
$\lim_{x \to 0^{-}} 1 + cosec x$

Answer:

(i)
$\lim_{x \to 2^{+}} (\frac{x - 3}{x^{2} - 4}) Let x = 2 + h, where h \to 0 . \Rightarrow \lim_{h \to 0} [\frac{(2 + h) - 3}{{(2 + h)}^{2} - 2^{2}}] = \lim_{h \to 0} [\frac{- 1 + h}{(2 + h - 2) (2 + h + 2)}] = - \infty$

(ii)
$\lim_{x \to 2^{-}} (\frac{x - 3}{x^{2} - 4}) Let x = 2 - h, where h \to 0 . \Rightarrow \lim_{h \to 0} [\frac{2 - h - 3}{{(2 - h)}^{2} - 2^{2}}] = \lim_{h \to 0} [\frac{- h - 1}{(2 - h - 2) (2 - h + 2)}] = \lim_{h \to 0} [\frac{- h - 1}{(- h) (4 - h)}] = \lim_{h \to 0} [\frac{1 + h}{h (4 - h)}] = \infty$

(iii)
$\lim_{x \to 0^{+}} (\frac{1}{3 x}) Let x = 0 + h, where h \to 0 . \Rightarrow \lim_{h \to 0} (\frac{1}{3 h}) = \infty$

(iv)
$\lim_{x \to - 8^{+}} (\frac{2 x}{x + 8}) Let x = - 8 + h, where h \to 0 . \Rightarrow \lim_{h \to 0} (\frac{2 (- 8 + h)}{- 8 + h + 8}) = \lim_{h \to 0} (\frac{- 16 + 2 h}{h}) = - \infty$

(v)
$\lim_{x \to 0^{+}} \frac{2}{x^{\frac{1}{5}}} Let x = 0 + h, where h \to 0 . \Rightarrow \lim_{h \to 0} [\frac{2}{h^{\frac{1}{5}}}] = \infty$

(vi)
$\lim_{x \to {\frac{π}{2}}^{-}} (\tan x) Let x = \frac{π}{2} - h, where h \to 0 . \Rightarrow \lim_{h \to 0} [\tan (\frac{π}{2} - h)] = \lim_{h \to 0} (\cot h) = \infty$

(vii)
$\lim_{x \to - {\frac{π}{2}}^{+}} \sec x Let x = - \frac{π}{2} + h, where h \to 0 . \lim_{h \to 0} \sec (- \frac{π}{2} + h) = \lim_{h \to 0} \sec (\frac{π}{2} - h) [\sec (- θ) = \sec θ] = \lim_{h \to 0} cosec h = \infty$

(viii)
$\lim_{x \to 0^{-}} [\frac{x^{2} - 3 x + 2}{x^{3} - 2 x^{2}}] = \lim_{x \to 0^{-}} [\frac{x^{2} - 2 x - x + 2}{x^{2} (x - 1)}] = \lim_{x \to 0^{-}} [\frac{x (x - 2) - 1 (x - 2)}{x^{2} (x - 2)}] = \lim_{x \to 0^{-}} [\frac{(x - 1) (x - 2)}{x^{2} (x - 2)}] Let x = 0 - h, where h \to 0 . \Rightarrow \lim_{h \to 0} [\frac{(- h - 1)}{{(- h)}^{2}}] = - \infty$

(ix)
$\lim_{x \to - 2^{+}} (\frac{x^{2} - 1}{2 x + 4}) Let x = - 2 + h, where h \to 0 . \lim_{h \to 0} [\frac{{(- 2 + h)}^{2} - 1}{2 (- 2 + h) + 4}] = \frac{4 - 1}{- 4 + 4} = \frac{3}{0} = \infty$

(x)
$\lim_{x \to 0^{-}} (2 - \cot x) Let x = 0 - h, where h \to 0 . \lim_{h \to 0} [2 - \cot (- h)] = \lim_{h \to 0} (2 + \cot h) = 2 + \infty = \infty$

(xi)
$\lim_{x \to 0^{-}} [1 + \cos e c x] Let x = 0 - h, where h \to 0 . \lim_{h \to 0} [1 + cosec (- h)] = \lim_{h \to 0} [1 - cosec h] [cosec (- θ) = - cosec θ] = 1 - \infty = - \infty$

Page No 29.12:

Question 14:

Show that $\lim_{x \to 0} e^{- 1 / x}$ does not exist.

Answer:

$\lim_{x \to 0} e^{- \frac{1}{x}} Left hand limit : \lim_{x \to 0^{-}} (e^{- \frac{1}{x}}) Let x = 0 - h, where h \to 0 . \Rightarrow \lim_{h \to 0} e^{(\frac{- 1}{0 - h})} = \lim_{h \to 0} (e^{\frac{1}{h}}) = \lim_{h \to 0} (e^{\frac{1}{h}}) [When h \to 0, then \frac{1}{h} \to \infty .] = e^{\infty} = \infty Right hand limit : \lim_{x \to 0^{+}} (e^{- \frac{1}{x}}) Let x = 0 + h, where h \to 0 . \lim_{h \to 0} (e^{\frac{- 1}{0 + h}}) = \lim_{h \to 0} (\frac{1}{e^{\frac{1}{h}}}) = \frac{1}{\infty} = 0 \lim_{x \to 0^{-}} e^{- \frac{1}{x}} \neq \lim_{x \to 0^{+}} e^{\frac{1}{x}} Thus, \underset{x \to 0}{limit} e^{- \frac{1}{x}} does not exist .$

Page No 29.12:

Question 15:

Find:

$(i) \lim_{x \to 2} [x] (ii) \lim_{x \to \frac{5}{2}} [x] (iii) \lim_{x \to 1} [x]$

Answer:

(i)

$\lim_{x \to 2} [x] LHL : \lim_{x \to 2^{-}} [x] Let x = 2 - h, where h \to 0 . \lim_{h \to 0} [2 - h] = 1 RHL : \lim_{x \to 2^{+}} [x] Let x = 2 + h, where h \to 0 . \lim_{h \to 0} [2 + h] = 2 LHL \neq RHL Thus, \lim_{x \to 2} [x] does not exist .$

(ii)
$\lim_{x \to \frac{5}{2}} [x] LHL \lim_{x \to {\frac{5}{2}}^{-}} [x] Let x = \frac{5}{2} - h, where h \to 0 . \lim_{h \to 0} [\frac{5}{2} - h] = 2 RHL : \lim_{x \to {\frac{5}{2}}^{+}} [x] Let x = \frac{5}{2} + h, where h \to 0 . \Rightarrow \lim_{h \to 0} [\frac{5}{2} + h] ∴ \lim_{x \to \frac{5}{2}} [x] = 2 = 2 ∴ \lim_{x \to \frac{5}{2}} [x] = 2$

(iii)
$\lim_{x \to 1} [x] LHL : \lim_{x \to 1^{-}} [x] Let x = 1 - h, where h \to 0 . \lim_{h \to 0} [1 - h] = 0 RHL : \lim_{x \to 1^{+}} [x] Let x = 1 + h, where h \to 0 . \lim_{h \to 0} [1 + h] = 1 LHL \neq RHL Thus, \lim_{x \to 1} [x] does not exist .$

Page No 29.12:

Question 16:

Prove that $\lim_{x \to a^{+}} [x] = [a]$ for all a ∈ R. Also, prove that $\lim_{x \to 1^{-}} [x] = 0 .$

Answer:

$\lim_{x \to a^{+}} [x] Let x = a + h, where h \to 0 . \lim_{h \to 0} [a + h] = a \lim_{x \to 1^{-}} [x] Let x = 1 - h, where h \to 0 . \lim_{h \to 0} [1 - h] = 0$

Page No 29.12:

Question 17:

Show that $\lim_{x \to 2^{-}} \frac{x}{[x]} \neq \lim_{x \to 2^{+}} \frac{x}{[x]} .$

Answer:

$\lim_{x \to 2^{-}} \frac{x}{[x]} Let x = 2 - h, where h \to 0 . \lim_{h \to 0} \frac{(2 - h)}{[2 - h]} = \frac{2}{1} \lim_{x \to 2^{+}} \frac{x}{[x]} Let x = 2 + h, where h \to 0 . \lim_{h \to 0} \frac{(2 + h)}{[2 + h]} = \frac{2}{2} = 1 ∴ \lim_{x \to 2^{-}} \frac{x}{[x]} \neq \lim_{x \to 2^{+}} \frac{x}{[x]}$

Page No 29.12:

Question 18:

Find $\lim_{x \to 3^{+}} \frac{x}{[x]} .$ Is it equal to $\lim_{x \to 3^{-}} \frac{x}{[x]} .$

Answer:

$\lim_{x \to 3^{+}} \frac{x}{[x]} Let x = 3 + h, where h \to 0 . \Rightarrow \lim_{h \to 0} (\frac{3 + h}{[3 + h]}) = \frac{3}{3} = 1 Also, \lim_{x \to 3^{-}} (\frac{x}{[x]}) Let x = 3 - h, where h \to 0 . \lim_{h \to 0} (\frac{3 - h}{[3 - h]}) = \frac{3}{2} ∴ \lim_{x \to 3^{-}} \frac{x}{[x]} \neq \lim_{x \to 3^{+}} \frac{x}{[x]}$

Page No 29.12:

Question 19:

Find $\lim_{x \to 5 / 2} [x] .$

Answer:

$\lim_{x \to \frac{5}{2}} [x] LHL : \lim_{x \to {\frac{5}{2}}^{-}} [x] Let x = \frac{5}{2} - h, where h \to 0 . \lim_{h \to 0} [\frac{5}{2} - h] = 2 RHL : \lim_{x \to {\frac{5}{2}}^{+}} [x] Let x = \frac{5}{2} + h, where h \to 0 . \lim_{h \to 0} [\frac{5}{2} + h] = 2$

We know:
$\lim_{x \to \frac{5}{2}} [x] = \lim_{x \to {\frac{5}{2}}^{-}} [x] = \lim_{x \to {\frac{5}{2}}^{+}} [x]$

$∴ \lim_{x \to \frac{5}{2}} [x] = 2$

Page No 29.12:

Question 20:

Evaluate $\lim_{x \to 2} f (x)$ (if it exists), where $f (x) = \{\begin{array}{l} x - [x], & x < 2 \\ 4, & x = 2 \\ 3 x - 5, & x > 2 \end{array} .$

Answer:

$f (x) = \{\begin{matrix} x - [x], & x < 2 \\ 4, & x = 2 \\ 3 x - 5, & x > 2 \end{matrix} LHL : \lim_{x \to 2^{-}} f (x) = \lim_{x \to 2^{-}} \{x - [x]\} Let x = 2 - h, where h \to 0 . \Rightarrow \lim_{h \to 0} [(2 - h) - [2 - h]] = 2 - (1) = 1 RHL : \lim_{x \to 2^{+}} f (x) = \lim_{x \to 2^{+}} (3 x - 5) Let x = 2 + h, where h \to 0 . \Rightarrow \lim_{h \to 0} [3 (2 + h) - 5] = 6 - 5 = 1 Here, LHL = RHL = 1 ∴ \lim_{x \to 2} f (x) = 1$

Page No 29.12:

Question 21:

Show that $\lim_{x \to 0} \sin \frac{1}{x}$ does not exist.

Answer:

$\lim_{x \to 0} \sin (\frac{1}{x}) L . H . L \lim_{x \to 0^{-}} f (x) \lim_{x \to 0^{-}} \sin (\frac{1}{x}) Let x = 0 - h where h \to 0 = \lim_{h \to 0} \sin (- \frac{1}{h}) = - \lim_{h \to 0} \sin (\frac{1}{h})$

$\lim_{x \to 0^{-}} f (x)$ = – (An oscillating number that oscillates between –1 and 1)

$Thus, \lim_{x \to 0^{-}} f (x) does not exist . Similarly, \lim_{x \to 0^{+}} f (x) does not exist .$

Page No 29.12:

Question 22:

Let $f (x) = \{\begin{cases} \frac{k \cos x}{π - 2 x}, & where x \neq \frac{π}{2} \\ 3, & where x = \frac{π}{2} \end{cases}$ and if $\lim_{x \to \frac{π}{2}} f (x) = f (\frac{π}{2})$ , find the value of k.

Answer:

We have,

$f (x) = \{\begin{cases} \frac{k \cos x}{π - 2 x}, & where x \neq \frac{π}{2} \\ 3, & where x = \frac{π}{2} \end{cases}$

It is given that,

$\lim_{x \to \frac{π}{2}} f (x) = f (\frac{π}{2}) \Rightarrow \lim_{x \to \frac{π}{2}} \frac{k \cos x}{π - 2 x} = 3 \Rightarrow \frac{k}{2} \times \lim_{x - \frac{π}{2} \to 0} \frac{\sin (\frac{π}{2} - x)}{\frac{π}{2} - x} = 3 \Rightarrow \frac{k}{2} \times \lim_{h \to 0} \frac{\sin (- h)}{- h} = 3 (Put x - \frac{π}{2} = h)$
$\Rightarrow \frac{k}{2} \times \lim_{h \to 0} \frac{\sin h}{h} = 3 [\sin (- θ) = - \sin θ] \Rightarrow \frac{k}{2} = 3 (\lim_{x \to 0} \frac{\sin x}{x} = 1) \Rightarrow k = 6$

Hence, the value of k is 6.

Page No 29.18:

Question 1:

$\lim_{x \to 1} \frac{x^{2} + 1}{x + 1}$

Answer:

$\lim_{x \to 1} (\frac{x^{2} + 1}{x + 1}) = \frac{1^{2} + 1}{1 + 1} = 1$

Page No 29.18:

Question 2:

$\lim_{x \to 0} \frac{2 x^{2} + 3 x + 4}{x^{2} + 3 x + 2}$

Answer:

$\lim_{x \to 0} (\frac{2 x^{2} + 3 x + 4}{x^{2} + 3 x + 2}) = \frac{2 \times 0 + 3 \times 0 + 4}{0 + 3 \times 0 + 2} = \frac{4}{2} = 2$

Page No 29.18:

Question 3:

$\lim_{x \to 3} \frac{\sqrt{2 x + 3}}{x + 3}$

Answer:

$\lim_{x \to 3} [\frac{\sqrt{2 x + 3}}{x + 3}] = \frac{\sqrt{2 \times 3 + 3}}{3 + 3} = \frac{3}{6} = \frac{1}{2}$

Page No 29.18:

Question 4:

$\lim_{x \to 1} \frac{\sqrt{x + 8}}{\sqrt{x}}$

Answer:

$\lim_{x \to 1} (\frac{\sqrt{x + 8}}{\sqrt{x}}) = \frac{\sqrt{1 + 8}}{1} = 3$

Page No 29.18:

Question 5:

$\lim_{x \to a} \frac{\sqrt{x} + \sqrt{a}}{x + a}$

Answer:

$\lim_{x \to a} (\frac{\sqrt{x} + \sqrt{a}}{x + a}) = \frac{\sqrt{a} + \sqrt{a}}{a + a} = \frac{2 \sqrt{a}}{2 a} = \frac{1}{\sqrt{a}}$

Page No 29.18:

Question 6:

$\lim_{x \to 1} \frac{1 + {(x - 1)}^{2}}{1 + x^{2}}$

Answer:

$\lim_{x \to 1} [\frac{1 + {(x - 1)}^{2}}{1 + x^{2}}] = \frac{1 + {(1 - 1)}^{2}}{1 + 1^{2}} = \frac{1}{2}$

Page No 29.18:

Question 7:

$\lim_{x \to 0} \frac{x^{2 / 3} - 9}{x - 27}$

Answer:

$\lim_{x \to 0} [\frac{x^{2 / 3} - 9}{x - 27}] = \frac{0 - 9}{0 - 27} = \frac{1}{3}$

Page No 29.18:

Question 8:

$\lim_{x \to 0} 9$

Answer:

$\lim_{x \to 0} (9) = 9$

f(x) = 9 is a constant function.
Its value does not depend on x.

Page No 29.18:

Question 9:

$\lim_{x \to 2} (3 - x)$

Answer:

$\lim_{x \to 2} (3 - x) = 3 - 2 = 1$

Page No 29.18:

Question 10:

$\lim_{x \to - 1} (4 x^{2} + 2)$

Answer:

$\lim_{x \to - 1} (4 x^{2} + 2) = 4 {(- 1)}^{2} + 2 = 4 + 2 = 6$

Page No 29.18:

Question 11:

$\lim_{x \to - 1} \frac{x^{3} - 3 x + 1}{x - 1}$

Answer:

$\lim_{x \to - 1} (\frac{x^{3} - 3 x + 1}{x - 1}) = \frac{{(- 1)}^{3} - 3 (- 1) + 1}{- 1 - 1} = \frac{- 1 + 3 + 1}{- 2} = \frac{- 3}{2}$

Page No 29.18:

Question 12:

$\lim_{x \to 0} \frac{3 x + 1}{x + 3}$

Answer:

$\lim_{x \to 0} (\frac{3 x + 1}{x + 3}) = \frac{3 \times 0 + 1}{0 + 3} = \frac{1}{3}$

Page No 29.18:

Question 13:

$\lim_{x \to 3} \frac{x^{2} - 9}{x + 2}$

Answer:

$\lim_{x \to 3} (\frac{x^{2} - 9}{x + 2}) = \frac{3^{2} - 9}{3 + 2} = \frac{9 - 9}{5} = 0$

Page No 29.18:

Question 14:

$\lim_{x \to 0} \frac{a x + b}{c x + d}, d \neq 0$

Answer:

$\lim_{x \to 0} (\frac{a x + b}{c x + d}) = \frac{a \times 0 + b}{c \times 0 + d} = \frac{b}{d}$

Page No 29.23:

Question 1:

$\lim_{x \to - 5} \frac{2 x^{2} + 9 x - 5}{x + 5}$

Answer:

$\lim_{x \to - 5} [\frac{2 x^{2} + 9 x - 5}{x + 5}] It is of the form \frac{0}{0} . \lim_{x \to - 5} [\frac{2 x^{2} + 10 x - x - 5}{x + 5}] = \lim_{x \to - 5} [\frac{2 x (x + 5) - 1 (x + 5)}{x + 5}] = \lim_{x \to - 5} [\frac{(2 x - 1) (x + 5)}{x + 5}] = \lim_{x \to - 5} (2 x - 1) = 2 (- 5) - 1 = - 11$

Page No 29.23:

Question 2:

$\lim_{x \to 3} \frac{x^{2} - 4 x + 3}{x^{2} - 2 x - 3}$

Answer:

$\lim_{x \to 3} [\frac{x^{2} - 4 x + 3}{x^{2} - 2 x - 3}] It is of the form \frac{0}{0} . \lim_{x \to 3} [\frac{x^{2} - x - 3 x + 3}{x^{2} - 3 x + x - 3}] = \lim_{x \to 3} [\frac{(x - 3) (x - 1)}{(x + 1) (x - 3)}] = \lim_{x \to 3} (\frac{x - 1}{x + 1}) = \frac{3 - 1}{3 + 1} = \frac{1}{2}$

Page No 29.23:

Question 3:

$\lim_{x \to 3} \frac{x^{4} - 81}{x^{2} - 9}$

Answer:

$\lim_{x \to 3} [\frac{x^{4} - 81}{x^{2} - 9}] It is of the form \frac{0}{0} . \lim_{x \to 3} [\frac{{(x^{2})}^{2} - 9^{2}}{x^{2} - 9}] = \lim_{x \to 3} [\frac{(x^{2} - 9) (x^{2} + 9)}{x^{2} - 9}] = \lim_{x \to 3} (x^{2} + 9) = 3^{2} + 9 = 18$

Page No 29.23:

Question 4:

$\lim_{x \to 2} \frac{x^{3} - 8}{x^{2} - 4}$

Answer:

$\lim_{x \to 2} [\frac{x^{3} - 8}{x^{2} - 4}] It is of the form \frac{0}{0} . \lim_{x \to 2} [\frac{(x - 2) (x^{2} + 2 x + 4)}{(x - 2) [x + 2]}] [\begin{matrix} ∵ A^{3} - B^{3} = (A - B) (A^{2} + A B + B^{2}) \\ ∵ A^{2} - B^{2} = (A - B) (A + B) \end{matrix}] = \frac{2^{2} + 2 \times 2 + 4}{2 + 2} = \frac{12}{4} = 3$

Page No 29.23:

Question 5:

$\lim_{x \to - 1 / 2} \frac{8 x^{3} + 1}{2 x + 1}$

Answer:

$\lim_{x \to - 1 / 2} [\frac{8 x^{3} + 1}{2 x + 1}] It is of the form \frac{0}{0} . \lim_{x \to - 1 / 2} [\frac{{(2 x)}^{3} + 1}{2 x + 1}] = \lim_{x \to - 1 / 2} [\frac{(2 x + 1) \{{(2 x)}^{2} - 2 x \times 1 + 1^{2}\}}{(2 x + 1)}] [∵ A^{3} + B^{3} = (A + B) (A^{2} - A B + B^{2})] = \lim_{x \to - 1 / 2} [{(2 x)}^{2} - 2 x + 1] = {(2 \times \frac{- 1}{2})}^{2} - 2 \times \frac{- 1}{2} + 1 = 1 + 1 + 1 = 3$

Page No 29.23:

Question 6:

$\lim_{x \to 4} \frac{x^{2} - 7 x + 12}{x^{2} - 3 x - 4}$

Answer:

$\lim_{x \to 4} [\frac{x^{2} - 7 x + 12}{x^{2} - 3 x - 4}] It is of the form \frac{0}{0} . \lim_{x \to 4} [\frac{x^{2} - 3 x - 4 x + 12}{x^{2} - 4 x + x - 4}] = \lim_{x \to 4} [\frac{x (x - 3) - 4 (x - 3)}{x (x - 4) + 1 (x - 4)}] = \lim_{x \to 4} [\frac{(x - 4) (x - 3)}{(x - 4) (x + 1)}] = \frac{4 - 3}{4 + 1} = \frac{1}{5}$

Page No 29.23:

Question 7:

$\lim_{x \to 2} \frac{x^{4} - 16}{x - 2}$

Answer:

$\lim_{x \to 2} [\frac{x^{4} - 16}{x - 2}] It is of the form \frac{0}{0} . \lim_{x \to 2} [\frac{{(x^{2})}^{2} - {(4)}^{2}}{x - 2}] = \lim_{x \to 2} [\frac{(x^{2} - 4) (x^{2} + 4)}{x - 2}] = \lim_{x \to 2} [\frac{(x - 2) (x + 2) (x^{2} + 4)}{x - 2}] = (2 + 2) (2^{2} + 4) = 32$

Page No 29.23:

Question 8:

\lim_{x \to 5} \frac{x^{2} - 9 x + 20}{x^{2} - 6 x + 5}

Answer:

$\lim_{x \to 5} [\frac{x^{2} - 9 x + 20}{x^{2} - 6 x + 5}] It is of the form \frac{0}{0} . \lim_{x \to 5} [\frac{x^{2} - 4 x - 5 x + 20}{x^{2} - x - 5 x + 5}] = \lim_{x \to 5} [\frac{x (x - 4) - 5 (x - 4)}{x (x - 1) - 5 (x - 1)}] = \lim_{x \to 5} [\frac{(x - 5) (x - 4)}{(x - 1) (x - 5)}] = \frac{5 - 4}{5 - 1} = \frac{1}{4}$

Page No 29.23:

Question 9:

$\lim_{x \to - 1} \frac{x^{3} + 1}{x + 1}$

Answer:

$\lim_{x \to - 1} [\frac{x^{3} + 1}{x + 1}] It is of the form \frac{0}{0} . \lim_{x \to - 1} [\frac{x^{3} + 1^{3}}{x + 1}] = \lim_{x \to - 1} [\frac{(x + 1) (x^{2} - x + 1)}{(x + 1)}] = {(- 1)}^{2} - (- 1) + 1 = 1 + 1 + 1 = 3$

Page No 29.23:

Question 10:

$\lim_{x \to 5} \frac{x^{3} - 125}{x^{2} - 7 x + 10}$

Answer:

$\lim_{x \to 5} [\frac{x^{3} - 125}{x^{2} - 7 x + 10}] It is of the form \frac{0}{0} . \lim_{x \to 5} [\frac{x^{3} - 5^{3}}{x^{2} - 2 x - 5 x + 10}] = \lim_{x \to 5} [\frac{(x - 5) (x^{2} + 5 x + 5^{2})}{x (x - 2) - 5 (x - 2)}] [∵ A^{3} - B^{3} = (A - B) (A^{2} + A B + B^{2})] = \lim_{x \to 5} [\frac{(x - 5) (x^{2} + 5 x + 25)}{(x - 2) (x - 5)}] = \frac{5^{2} + 5 \times 5 + 25}{5 - 2} = \frac{75}{3} = 25$

Page No 29.23:

Question 11:

$\lim_{x \to \sqrt{2}} \frac{x^{2} - 2}{x^{2} + \sqrt{2} x - 4}$

Answer:

$\lim_{x \to \sqrt{2}} [\frac{x^{2} - 2}{x^{2} + \sqrt{2} x - 4}] It is of the form \frac{0}{0} . \lim_{x \to \sqrt{2}} [\frac{x^{2} - {(\sqrt{2})}^{2}}{x^{2} + 2 \sqrt{2} x - \sqrt{2} x - 4}] = \lim_{x \to \sqrt{2}} [\frac{(x - \sqrt{2}) (x + \sqrt{2})}{x (x + 2 \sqrt{2}) - \sqrt{2} (x + 2 \sqrt{2})}] = \lim_{x \to \sqrt{2}} [\frac{(x - \sqrt{2}) (x + \sqrt{2})}{(x - \sqrt{2}) (x + 2 \sqrt{2})}] = \frac{\sqrt{2} + \sqrt{2}}{\sqrt{2} + 2 \sqrt{2}} = \frac{2 \sqrt{2}}{3 \sqrt{2}} = \frac{2}{3}$

Page No 29.23:

Question 12:

$\lim_{x \to \sqrt{3}} \frac{x^{2} - 3}{x^{2} + 3 \sqrt{3} x - 12}$

Answer:

$\lim_{x \to \sqrt{3}} [\frac{x^{2} - 3}{x^{2} + 3 \sqrt{3} x - 12}] It is of the form \frac{0}{0} . \lim_{x \to \sqrt{3}} [\frac{x^{2} - {(\sqrt{3})}^{2}}{x^{2} + 4 \sqrt{3} x - \sqrt{3} x - 12}] = \lim_{x \to \sqrt{3}} [\frac{(x - \sqrt{3}) (x + \sqrt{3})}{x (x + 4 \sqrt{3}) - \sqrt{3} (x + 4 \sqrt{3})}] = \lim_{x \to \sqrt{3}} [\frac{(x - \sqrt{3}) (x + \sqrt{3})}{(x - \sqrt{3}) (x + 4 \sqrt{3})}] = \frac{\sqrt{3} + \sqrt{3}}{\sqrt{3} + 4 \sqrt{3}} = \frac{2}{5}$

Page No 29.23:

Question 13:

$\lim_{x \to \sqrt{3}} \frac{x^{4} - 9}{x^{2} + 4 \sqrt{3} x - 15}$

Answer:

$\lim_{x \to \sqrt{3}} [\frac{x^{4} - 9}{x^{2} + 4 \sqrt{3} x - 15}] It is of the form \frac{0}{0} . \lim_{x \to \sqrt{3}} [\frac{{(x^{2})}^{2} - {(3)}^{2}}{x^{2} + 5 \sqrt{3} x - \sqrt{3} x - 15}] = \lim_{x \to \sqrt{3}} [\frac{(x^{2} - 3) (x^{2} + 3)}{x (x + 5 \sqrt{3}) - \sqrt{3} (x + 5 \sqrt{3})}] = \lim_{x \to \sqrt{3}} [\frac{\{x^{2} - {(\sqrt{3})}^{2}\} (x^{2} + 3)}{(x - \sqrt{3}) (x + 5 \sqrt{3})}] = \lim_{x \to \sqrt{3}} [\frac{(x - \sqrt{3}) (x + \sqrt{3}) (x^{2} + 3)}{(x - \sqrt{3}) (x + 5 \sqrt{3})}] = \frac{(\sqrt{3} + \sqrt{3}) (3 + 3)}{\sqrt{3} + 5 \sqrt{3}} = \frac{2 \sqrt{3} \times 6}{6 \sqrt{3}} = 2$

Page No 29.23:

Question 14:

$\lim_{x \to 2} (\frac{x}{x - 2} - \frac{4}{x^{2} - 2 x})$

Answer:

$\lim_{x \to 2} [\frac{x}{x - 2} - \frac{4}{x^{2} - 2 x}] = \lim_{x \to 2} [\frac{x}{x - 2} - \frac{4}{x (x - 2)}] = \lim_{x \to 2} [\frac{x^{2} - 4}{x (x - 2)}] = \lim_{x \to 2} [\frac{(x - 2) (x + 2)}{x (x - 2)}] = \lim_{x \to 2} [\frac{(x + 2)}{x}] = \frac{2 + 2}{2} = 2$

Page No 29.23:

Question 15:

$\lim_{x \to 1} (\frac{1}{x^{2} + x - 2} - \frac{x}{x^{3} - 1})$

Answer:

$\lim_{x \to 1} [\frac{(x^{3} - 1) - x (x^{2} + x - 2)}{(x^{2} + x - 2) (x^{3} - 1)}] = \lim_{x \to 1} [\frac{(x^{3} - 1) - x^{3} - x^{2} + 2 x}{(x^{2} + x - 2) (x^{3} - 1)}] = \lim_{x \to 1} [\frac{- x^{2} + 2 x - 1}{(x^{2} + x - 2) (x - 1) (x^{2} + x + 1)}] = \lim_{x \to 1} [\frac{- (x^{2} - 2 x + 1)}{(x^{2} + x - 2) (x - 1) (x^{2} + x + 1)}] = \lim_{x \to 1} [\frac{- {(x - 1)}^{2}}{(x^{2} + x - 2) (x - 1) (x^{2} + x + 1)}] = \lim_{x \to 1} [\frac{- (x - 1)}{(x^{2} + 2 x - x - 2) (x^{2} + x + 1)}] = \lim_{x \to 1} [\frac{- (x - 1)}{\{x (x + 2) - 1 (x + 2)\} (x^{2} + x + 1)}] = \lim_{x \to 1} [\frac{- (x - 1)}{(x - 1) (x + 2) (x^{2} + x + 1)}] = \frac{- 1}{(1 + 2) (1 + 1 + 1)} = \frac{- 1}{9}$

Page No 29.23:

Question 16:

$\lim_{x \to \sqrt{3}} \frac{x^{4} - 9}{x^{2} + 4 \sqrt{3} x - 15}$

Answer:

$\lim_{x \to 3} [\frac{1}{x - 3} - \frac{2}{x^{2} - 4 x + 3}] = \lim_{x \to 3} [\frac{1}{x - 3} - \frac{2}{x^{2} - 3 x - x + 3}] = \lim_{x \to 3} [\frac{1}{x - 3} - \frac{2}{x (x - 3) - 1 (x - 3)}] = \lim_{x \to 3} [\frac{1}{x - 3} - \frac{2}{(x - 1) (x - 3)}] = \lim_{x \to 3} [\frac{x - 1 - 2}{(x - 3) (x - 1)}] = \lim_{x \to 3} [\frac{1}{x - 1}] = \frac{1}{3 - 1} = \frac{1}{2}$

Page No 29.23:

Question 17:

$\lim_{x \to 2} (\frac{1}{x - 2} - \frac{2}{x^{2} - 2 x})$

Answer:

$\lim_{x \to 2} [\frac{1}{x - 2} - \frac{2}{x^{2} - 2 x}] = \lim_{x \to 2} [\frac{1}{x - 2} - \frac{2}{x (x - 2)}] = \lim_{x \to 2} [\frac{x - 2}{x (x - 2)}] = \lim_{x \to 2} [\frac{1}{x}] = \frac{1}{2}$

Page No 29.23:

Question 18:

$\lim_{x \to 1 / 4} \frac{4 x - 1}{2 \sqrt{x} - 1}$

Answer:

$\lim_{x \to 1 / 4} [\frac{4 x - 1}{2 \sqrt{x} - 1}] It is of the form \frac{0}{0} . \lim_{x \to 1 / 4} [\frac{{(2 \sqrt{x})}^{2} - 1^{2}}{2 \sqrt{x} - 1}] = \lim_{x \to 1 / 4} [\frac{(2 \sqrt{x} - 1) (2 \sqrt{x} + 1)}{(2 \sqrt{x} - 1)}] = 2 \sqrt{\frac{1}{4}} + 1 = 2 \times \frac{1}{2} + 1 = 2$

Page No 29.23:

Question 19:

$\lim_{x \to 4} \frac{x^{2} - 16}{\sqrt{x} - 2}$

Answer:

$\lim_{x \to 4} [\frac{x^{2} - 16}{\sqrt{x} - 2}] It is of the form \frac{0}{0} . \lim_{x \to 4} [\frac{x^{2} - 4^{2}}{\sqrt{x} - 2}] = \lim_{x \to 4} [\frac{(x - 4) (x + 4)}{(\sqrt{x} - 2)}] = \lim_{x \to 4} [\frac{\{{(\sqrt{x})}^{2} - 2^{2}\} (x + 4)}{(\sqrt{x} - 2)}] = \lim_{x \to 4} [\frac{(\sqrt{x} - 2) (\sqrt{x} + 2) (x + 4)}{(\sqrt{x} - 2)}] = (2 + 2) (4 + 4) = 32$

Page No 29.23:

Question 20:

$\lim_{x \to 0} \frac{{(a + x)}^{2} - a^{2}}{x}$

Answer:

$\lim_{x \to 0} [\frac{{(a + x)}^{2} - a^{2}}{x}] It is of the form \frac{0}{0} . \lim_{x \to 0} [\frac{a^{2} + x^{2} + 2 a x - a^{2}}{x}] = \lim_{x \to 0} [\frac{x (x + 2 a)}{x}] = \lim_{x \to 0} [x + 2 a] = 0 + 2 a = 2 a$

Page No 29.23:

Question 21:

$\lim_{x \to 2} (\frac{1}{x - 2} - \frac{4}{x^{3} - 2 x^{2}})$

Answer:

$\lim_{x \to 2} [\frac{1}{x - 2} - \frac{4}{x^{3} - 2 x^{2}}] = \lim_{x \to 2} [\frac{1}{x - 2} - \frac{4}{x^{2} (x - 2)}] = \lim_{x \to 2} [\frac{x^{2} - 4}{x^{2} (x - 2)}] = \lim_{x \to 2} [\frac{(x - 2) (x + 2)}{x^{2} (x - 2)}] = \lim_{x \to 2} [\frac{x + 2}{x^{2}}] = \frac{2 + 2}{2^{2}} = 1$

Page No 29.23:

Question 22:

$\lim_{x \to 3} (\frac{1}{x - 3} - \frac{3}{x^{2} - 3 x})$

Answer:

$\lim_{x \to 3} [\frac{1}{x - 3} - \frac{3}{x^{2} - 3 x}] = \lim_{x \to 3} [\frac{1}{x - 3} - \frac{3}{x (x - 3)}] = \lim_{x \to 3} [\frac{x - 3}{x (x - 3)}] = \lim_{x \to 3} [\frac{1}{x}] = \frac{1}{3}$

Page No 29.23:

Question 23:

$\lim_{x \to 1} (\frac{1}{x - 1} - \frac{2}{x^{2} - 1})$

Answer:

$\lim_{x \to 1} [\frac{1}{x - 1} - \frac{2}{x^{2} - 1}] = \lim_{x \to 1} [\frac{1}{x - 1} - \frac{2}{(x - 1) (x + 1)}] = \lim_{x \to 1} [\frac{x + 1 - 2}{(x - 1) (x + 1)}] = \lim_{x \to 1} [\frac{(x - 1)}{(x - 1) (x + 1)}] = \lim_{x \to 1} [\frac{1}{x + 1}] = \frac{1}{1 + 1} = \frac{1}{2}$

Page No 29.23:

Question 24:

$\lim_{x \to 3} (x^{2} - 9) [\frac{1}{x + 3} + \frac{1}{x - 3}]$

Answer:

$\lim_{x \to 3} [(x^{2} - 9) \{\frac{1}{x + 3} + \frac{1}{x - 3}\}] = \lim_{x \to 3} [(x^{2} - 9) \{\frac{x - 3 + x + 3}{(x + 3) (x - 3)}\}] = \lim_{x \to 3} [(x^{2} - 9) (\frac{2 x}{x^{2} - 9})] = \lim_{x \to 3} (2 x) = 2 \times 3 = 6$

Page No 29.23:

Question 25:

$\lim_{x \to 1} \frac{x^{4} - 3 x^{3} + 2}{x^{3} - 5 x^{2} + 3 x + 1}$

Answer:

p(x) = x⁴ $-$ 3x³ + 2
p(1) = 1 $-$ 3 + 2
= 0
Now, $(x - 1)$ is a factor of p(x).

$x^{3} - 2 x^{2} - 2 x - 2 x - 1 x^{4} - 3 x^{3} + 2 x^{4} - x^{3} - + - 2 x^{3} + 2 - 2 x^{3} + 2 x^{2} + - - 2 x^{2} + 2 - 2 x^{2} + 2 x + - - 2 x + 2 - 2 x + 2 + -___________0$

q(x) = x³ $-$ 5x² + 3x + 1
q(1) = 1 $-$ 5 + 3 + 1
= 0
$Now, (x - 1)$ is a factor of q(x).

$x^{2} - 4 x - 1 x - 1 x^{3} - 5 x^{2} + 3 x + 1 x^{3} - x^{2} - + 4 x^{2} + 3 x + 1 4 x^{2} + 4 x - - - x + 1 - x + 1 + - 0$

$\Rightarrow \lim_{x \to 1} [\frac{x^{4} - 3 x^{3} + 2}{x^{3} - 5 x^{2} + 3 x + 1}] = \lim_{x \to 1} [\frac{(x - 1) (x^{3} - 2 x^{2} - 2 x - 2)}{(x - 1) (x^{2} - 4 x - 1)}] = \frac{(1)^{3} - 2 {(1)}^{2} - 2 (1) - 2}{{(1)}^{2} - 4 \times 1 - 1} = \frac{1 - 2 - 2 - 2}{1 - 4 - 1} = \frac{- 5}{- 4} = \frac{5}{4}$

Page No 29.23:

Question 26:

$\lim_{x \to 2} \frac{x^{3} + 3 x^{2} - 9 x - 2}{x^{3} - x - 6}$

Answer:

$\lim_{x \to 2} [\frac{x^{3} + 3 x^{2} - 9 x - 2}{x^{3} - x - 6}]$

It is of the form $\frac{0}{0} .$

Let p(x) = x³ + 3x² $-$ 9x $-$ 2
p(2) = 8 + 12 $-$ 18 $-$ 2
= 0
Now, $(x - 2)$ is a factor of p(x).

$x^{2} + 5 x + 1 x - 2 x^{3} + 3 x^{2} - 9 x - 2 x^{3} - 2 x^{2} - + 5 x^{2} - 9 x - 2 5 x^{2} - 10 x - + x - 2 x - 2 - + 0$

Let q(x) = x³ – x – 6
q(2) = 8 – 2 – 6
= 0
$Now, (x - 2)$ is a factor of q(x).

$x^{2} + 2 x + 3 x - 2 x^{3} - x - 6 x^{3} - 2 x^{2} - + 2 x^{2} - x - 6 2 x^{2} - 4 x - + 3 x - 6 3 x - 6 - + \times \times \times$

$\Rightarrow \lim_{x \to 2} [\frac{x^{3} + 3 x^{2} - 9 x - 2}{x^{3} - x - 6}] = \lim_{x \to 2} [\frac{(x - 2) (x^{2} + 5 x + 1)}{(x - 2) (x^{2} + 2 x + 3)}] = \lim_{x \to 2} [\frac{x^{2} + 5 x + 1}{x^{2} + 2 x + 3}] = \frac{(2)^{2} + 5 \times 2 + 1}{{(2)}^{2} + 2 \times 2 + 3} = \frac{4 + 10 + 1}{4 + 4 + 3} = \frac{15}{11}$

Page No 29.23:

Question 27:

$\lim_{x \to 1} \frac{1 - x^{- 1 / 3}}{1 - x^{- 2 / 3}}$

Answer:

$\lim_{x \to 1} [\frac{1 - x^{- 1 / 3}}{1 - x^{- 2 / 3}}] It is of the form \frac{0}{0} . \lim_{x \to 1} [\frac{(1 - x^{- 1 / 3})}{{(1)}^{2} - {(x^{- 1 / 3})}^{2}}] = \lim_{x \to 1} [\frac{(1 - x^{- 1 / 3})}{(1 - x^{- 13}) (1 + x^{- 1 / 3})}] = \lim_{x \to 1} [\frac{1}{1 + x^{- 1 / 3}}] = \frac{1}{1 + 1} = \frac{1}{2}$

Page No 29.23:

Question 28:

$\lim_{x \to 3} \frac{x^{2} - x - 6}{x^{3} - 3 x^{2} + x - 3}$

Answer:

$\lim_{x \to 3} [\frac{x^{2} - x - 6}{x^{3} - 3 x^{2} + x - 3}] It is of the form \frac{0}{0} . \lim_{x \to 3} [\frac{x^{2} - 3 x + 2 x - 6}{x^{2} (x - 3) + 1 (x - 3)}] = \lim_{x \to 3} [\frac{x (x - 3) + 2 (x - 3)}{(x^{2} + 1) (x - 3)}] = \lim_{x \to 3} [\frac{(x + 2) (x - 3)}{(x^{2} + 1) (x - 3)}] = \frac{3 + 2}{3^{2} + 1} = \frac{5}{10} = \frac{1}{2}$

Page No 29.23:

Question 29:

$\lim_{x \to - 2} \frac{x^{3} + x^{2} + 4 x + 12}{x^{3} - 3 x + 2}$

Answer:

Let p(x) = x³ + x² + 4x + 12
p(–2) = 0
Thus, x = –2 is the root of p(x).
Now, $(x + 2)$ is a factor of p(x).

$x^{2} - x + 6 x + 2 x^{3} + x^{2} + 4 x + 12 x^{3} + 2 x^{2} - - - x^{2} + 4 x + 12 - x^{2} - 2 x + + 6 x + 12 6 x + 12 - - 0$

p(x) = x³ + x² + 4x + 12
= (x + 2)(x² – x + 6)

Let q(x) = x³ – 3x + 2
q $(- 2)$ = $-$ 8 + 6 + 2
= 0
Thus, x = $-$ 2 is the root of q(x).
Now, $(x + 2)$ is a factor of q(x).

$x^{2} - 2 x + 1 x + 2 x^{3} - 3 x + 2 x^{3} + 2 x^{2} - - - 2 x^{2} - 3 x + 2 - 2 x^{2} - 4 x + + x + 2 x + 2 - - \times \times \times$

q(x) = (x + 2)(x² – 2x + 1)

$\Rightarrow \lim_{x \to - 2} [\frac{x^{3} + x^{2} + 4 x + 12}{x^{3} - 3 x + 2}] = \lim_{x \to - 2} [\frac{(x + 2) (x^{2} - x + 6)}{(x + 2) (x^{2} - 2 x + 1)}] = \frac{(- 2)^{2} - (- 2) + 6}{{(- 2)}^{2} - 2 (- 2) + 1} = \frac{4 + 2 + 6}{4 + 4 + 1} = \frac{12}{9} = \frac{4}{3}$

Page No 29.23:

Question 30:

$\lim_{x \to 1} \frac{x^{3} + 3 x^{2} - 6 x + 2}{x^{3} + 3 x^{2} - 3 x - 1}$

Answer:

Let p(x) = x³ + 3x² $-$ 6x + 2
p(1) = 1 + 3 $-$ 6 + 2
= 0
Now, $(x + 2)$ is a factor of p(x).

$x^{2} + 4 x - 2 x - 1 x^{3} + 3 x^{2} - 6 x + 2 x^{3} - x^{2} - + - 4 x^{2} - 6 x + 2 - 4 x^{2} - 4 x + + - 2 x + 2 - 2 x + 2 + - 0$

p(x) = (x $-$ 1)(x² + 4x $-$ 2)

q(x) = x³ + 3x² $-$ 3x + 2
q(1) = 1 + 3 $-$ 3 $-$ 1
= 0
$Now, (x + 2)$ is a factor of p(x).

$x^{2} + 4 x + 1 x - 1 x^{3} + 3 x^{2} - 3 x - 1 x^{3} - x^{2} - + 4 x^{2} - 3 x - 1 4 x^{2} - 4 x - + x - 1 x - 1 - + 0$

$\Rightarrow \lim_{x \to 1} [\frac{x^{3} + 3 x^{2} - 6 x + 2}{x^{3} + 3 x^{2} - 3 x - 1}] = \lim_{x \to 1} [\frac{(x - 1) (x^{2} + 4 x - 2)}{(x - 1) (x^{2} + 4 x + 1)}] = \frac{(1)^{2} + 4 \times 1 - 2}{{(1)}^{2} + 4 \times 1 + 1} = \frac{1 + 4 - 2}{1 + 4 + 1} = \frac{3}{6} = \frac{1}{2}$

Page No 29.24:

Question 31:

$\lim_{x \to 2} [\frac{1}{x - 2} - \frac{2 (2 x - 3)}{x^{3} - 3 x^{2} + 2 x}]$

Answer:

$\lim_{x \to 2} [\frac{1}{x - 2} - \frac{2 (2 x - 3)}{x^{3} - 3 x^{2} + 2 x}] = \lim_{x \to 2} [\frac{1}{x - 2} - \frac{2 (2 x - 3)}{x (x^{2} - 3 x + 2)}] = \lim_{x \to 2} [\frac{1}{x - 2} - \frac{- 2 (2 x - 3)}{x (x^{2} - 2 x - x + 2)}] = \lim_{x \to 2} [\frac{1}{x - 2} - \frac{2 (2 x - 3)}{x \{x (x - 2) - 1 (x - 2)\}}] = \lim_{x \to 2} [\frac{1}{x - 2} - \frac{2 (2 x - 3)}{x (x - 1) (x - 2)}] = \lim_{x \to 2} [\frac{x (x - 1) - 2 (2 x - 3)}{x (x - 1) (x + 2)}] = \lim_{x \to 2} [\frac{x^{2} - x - 4 x + 6}{x (x - 1) (x + 2)}] = \lim_{x \to 2} [\frac{x^{2} - 5 x + 6}{x (x - 1) (x + 2)}] = \lim_{x \to 2} [\frac{x^{2} - 2 x - 3 x + 6}{x (x - 1) (x - 2)}] = \lim_{x \to 2} [\frac{x (x - 2) - 3 (x - 2)}{x (x - 1) (x - 2)}] = \lim_{x \to 2} [\frac{(x - 3) (x - 2)}{x (x - 1) (x - 2)}] = \frac{2 - 3}{2 (2 - 1)} = - \frac{1}{2}$

Page No 29.24:

Question 32:

$\lim_{x \to 1} \frac{\sqrt{x^{2} - 1} + \sqrt{x - 1}}{\sqrt{x^{2} - 1}}, x > 1$

Answer:

$\lim_{x \to 1} [\frac{\sqrt{x^{2} - 1} + \sqrt{x - 1}}{\sqrt{x^{2} - 1}}] It is of the form \frac{0}{0} . \lim_{x \to 1} [\frac{\sqrt{x^{2} - 1}}{\sqrt{x^{2} - 1}} + \frac{\sqrt{x - 1}}{\sqrt{x^{2} - 1}}] = \lim_{x \to 1} [1 + \frac{\sqrt{x - 1}}{(\sqrt{x - 1}) (\sqrt{x + 1})}] = 1 + \frac{1}{\sqrt{1 + 1}} = 1 + \frac{1}{\sqrt{2}} = \frac{\sqrt{2} + 1}{\sqrt{2}}$

Page No 29.24:

Question 33:

$\lim_{x \to 1} \{\frac{x - 2}{x^{2} - x} - \frac{1}{x^{3} - 3 x^{2} + 2 x}\}$

Answer:

$\lim_{x \to 1} [\frac{x - 2}{x^{2} - x} - \frac{1}{x^{3} - 3 x^{2} + 2 x}] = \lim_{x \to 1} [\frac{x - 2}{x (x - 1)} - \frac{1}{x (x^{2} - 3 x + 2)}] = \lim_{x \to 1} [\frac{x - 2}{x (x - 1)} - \frac{1}{x (x^{2} - 2 x - x + 2)}] = \lim_{x \to 1} [\frac{x - 2}{x (x - 1)} - \frac{1}{x \{x (x - 2) - 1 (x - 2)\}}]$
$= \lim_{x \to 1} [\frac{x - 2}{x (x - 1)} - \frac{1}{x (x - 1) (x - 2)}] = \lim_{x \to 1} [\frac{{(x - 2)}^{2} - 1}{x (x - 1) (x - 2)}]$

$= \lim_{x \to 1} [\frac{(x - 1) (x - 3)}{x (x - 1) (x - 2)}] = \frac{(1 - 3)}{1 (1 - 2)} = \frac{- 2}{- 1} = 2$

Page No 29.24:

Question 34:

Evaluate the following limits:

$\lim_{x \to 1} \frac{x^{7} - 2 x^{5} + 1}{x^{3} - 3 x^{2} + 2}$

Answer:

When x = 1, the expression $\frac{x^{7} - 2 x^{5} + 1}{x^{3} - 3 x^{2} + 2}$ assumes the form $\frac{0}{0}$ . So, (x − 1) is a factor of numerator and denominator.

Using long division method, we get

$x^{7} - 2 x^{5} + 1 = (x - 1) (x^{6} + x^{5} - x^{4} - x^{3} - x^{2} - x - 1)$

and $x^{3} - 3 x^{2} + 2 = (x - 1) (x^{2} - 2 x - 2)$

$∴ \lim_{x \to 1} \frac{x^{7} - 2 x^{5} + 1}{x^{3} - 3 x^{2} + 2} = \lim_{x \to 1} \frac{(x - 1) (x^{6} + x^{5} - x^{4} - x^{3} - x^{2} - x - 1)}{(x - 1) (x^{2} - 2 x - 2)} = \lim_{x \to 1} \frac{x^{6} + x^{5} - x^{4} - x^{3} - x^{2} - x - 1}{x^{2} - 2 x - 2} = \frac{1 + 1 - 1 - 1 - 1 - 1 - 1}{1 - 2 - 2} = \frac{- 3}{- 3} = 1$

Page No 29.28:

Question 1:

$\lim_{x \to 0} \frac{\sqrt{1 + x + x^{2}} - 1}{x}$

Answer:

$\lim_{x \to 0} [\frac{\sqrt{1 + x + x^{2}} - 1}{x}]$

When x = 0, the expression $\frac{\sqrt{1 + x + x^{2}} - 1}{x}$ takes the form $\frac{0}{0}$ .
Rationalising the numerator:

$\lim_{x \to 0} [\frac{(\sqrt{1 + x + x^{2}} - 1) (\sqrt{1 + x + x^{2}} + 1)}{x (\sqrt{1 + x + x^{2}} + 1)}] = \lim_{x \to 0} [\frac{1 + x + x^{2} - 1}{x (\sqrt{1 + x + x^{2}} + 1)}] = \lim_{x \to 0} [\frac{x (1 + x)}{x (\sqrt{1 + x + x^{2}} + 1)}] = \frac{1 + 0}{\sqrt{1 + 0 + 0} + 1} = \frac{1}{2}$

Page No 29.28:

Question 2:

$\lim_{x \to 0} \frac{2 x}{\sqrt{a + x} - \sqrt{a - x}}$

Answer:

$\lim_{x \to 0} [\frac{2 x}{\sqrt{a + x} - \sqrt{a - x}}]$

When x = 0, then the expression $\frac{2 x}{\sqrt{a + x} - \sqrt{a - x}}$ becomes $\frac{0}{0}$ .

Rationalising the denominator:

$\lim_{x \to 0} [\frac{2 x}{(\sqrt{a + x} - \sqrt{a - x})} \times \frac{(\sqrt{a + x} + \sqrt{a - x})}{(\sqrt{a + x} + \sqrt{a - x})}]$

= $\lim_{x \to 0} [\frac{(2 x) (\sqrt{a + x} + \sqrt{a - x})}{(a + x) - (a - x)}]$

= $\lim_{x \to 0} [\frac{2 x (\sqrt{a + x} + \sqrt{a - x})}{2 x}]$

= $\sqrt{a} + \sqrt{a}$
= $2 \sqrt{a}$

Page No 29.28:

Question 3:

$\lim_{x \to 0} \frac{\sqrt{a^{2} + x^{2}} - a}{x^{2}}$

Answer:

$\lim_{x \to 0} [\frac{\sqrt{a^{2} + x^{2}} - a}{x^{2}}]$

On putting x = 0 in the expression $\sqrt{a^{2} + x^{2}} - a$ , it becomes $\frac{0}{0} .$
Rationalising the numerator:

$\lim_{x \to 0} [\frac{(\sqrt{a^{2} + x^{2}} - a) (\sqrt{a^{2} + x^{2}} + a)}{x^{2} (\sqrt{a^{2} + x^{2}} + a)}]$

= $\lim_{x \to 0} [\frac{a^{2} + x^{2} - a^{2}}{x^{2} (\sqrt{a^{2} + x^{2}} + a)}]$

= $\frac{1}{\sqrt{a^{2}} + a}$

= $\frac{1}{2 a}$

Page No 29.28:

Question 4:

$\lim_{x \to 0} \frac{\sqrt{1 + x} - \sqrt{1 - x}}{2 x}$

Answer:

$\lim_{x \to 0} [\frac{\sqrt{1 + x} - \sqrt{1 - x}}{2 x}]$

It is of the form $\frac{0}{0} .$

Rationalising the numerator:

$\lim_{x \to 0} [\frac{(\sqrt{1 + x} - \sqrt{1 - x}) (\sqrt{1 + x} + \sqrt{1 - x})}{2 x (\sqrt{1 + x} + \sqrt{1 - x})}]$

= $\lim_{x \to 0} [\frac{(1 + x) - (1 - x)}{2 x (\sqrt{1 + x} + \sqrt{1 - x})}]$

= $\lim_{x \to 0} [\frac{2 x}{2 x (\sqrt{1 + x} + \sqrt{1 - x})}]$

= $\frac{1}{\sqrt{1 + 0} + \sqrt{1 - 0}}$
= $\frac{1}{2}$

Page No 29.28:

Question 5:

$\lim_{x \to 2} \frac{\sqrt{3 - x} - 1}{2 - x}$

Answer:

$\lim_{x \to 2} [\frac{\sqrt{3 - x} - 1}{2 - x}]$

It is of the form $\frac{0}{0} .$
Rationalising the numerator:

$\lim_{x \to 2} [\frac{(\sqrt{3 - x} - 1) (\sqrt{3 - x} + 1)}{(2 - x) (\sqrt{3 - x} + 1)}]$

$= \lim_{x \to 2} [\frac{3 - x - 1}{(2 - x) (\sqrt{3 - x} + 1)}]$

$= \lim_{x \to 2} [\frac{(2 - x)}{(2 - x) (\sqrt{3 - x} + 1)}]$

$= \frac{1}{\sqrt{3 - 2} + 1} = \frac{1}{1 + 1} = \frac{1}{2}$

Page No 29.28:

Question 6:

$\lim_{x \to 3} \frac{x - 3}{\sqrt{x - 2} - \sqrt{4 - x}}$

Answer:

$\lim_{x \to 3} [\frac{x - 3}{\sqrt{x - 2} - \sqrt{4 - x}}]$

It is of form $\frac{0}{0} .$

Rationalising the denominator:

$\lim_{x \to 3} [\frac{(x - 3) (\sqrt{x - 2} + \sqrt{4 - x})}{(\sqrt{x - 2} - \sqrt{4 - x}) (\sqrt{x - 2} + \sqrt{4 - x})}]$

= $\lim_{x \to 3} [\frac{(x - 3) (\sqrt{x - 2} + \sqrt{4 - x})}{(x - 2) - (4 - x)}]$

= $\lim_{x \to 3} [\frac{(x - 3) (\sqrt{x - 2} + \sqrt{4 - x})}{2 x - 6}]$

= $\lim_{x \to 3} [\frac{(x - 3) (\sqrt{x - 2} + \sqrt{4 - x})}{2 (x - 3)}]$

= $\frac{\sqrt{3 - 2} + \sqrt{4 - 3}}{2}$

= $\frac{\sqrt{1} + \sqrt{1}}{2}$
= $\frac{2}{2} = 1$

Page No 29.28:

Question 7:

$\lim_{x \to 0} \frac{x}{\sqrt{1 + x} - \sqrt{1 - x}}$

Answer:

$\lim_{x \to 0} [\frac{x}{\sqrt{1 + x} - \sqrt{1 - x}}]$

It is of the form $\frac{0}{0}$ .

Rationalising the denominator:

$\lim_{x \to 0} [\frac{x}{(\sqrt{1 + x} - \sqrt{1 - x})} \times \frac{(\sqrt{1 + x} + \sqrt{1 - x})}{(\sqrt{1 + x} + \sqrt{1 - x})}]$

= $\lim_{x \to 0} [\frac{x (\sqrt{1 + x} + \sqrt{1 - x})}{(1 + x) - (1 - x)}]$

= $\lim_{x \to 0} [\frac{x (\sqrt{1 + x} + \sqrt{1 - x})}{2 x}]$

= $\frac{\sqrt{1} + \sqrt{1}}{2}$

= $\frac{2}{2}$
= 1

Page No 29.28:

Question 8:

$\lim_{x \to 1} \frac{\sqrt{5 x - 4} - \sqrt{x}}{x - 1}$

Answer:

$\lim_{x \to 1} [\frac{\sqrt{5 x - 4} - \sqrt{x}}{x - 1}]$

It is of the form $\frac{0}{0} .$
Rationalising the numerator:

$\lim_{x \to 1} [(\frac{\sqrt{5 x - 4} - \sqrt{x}}{x - 1}) (\frac{\sqrt{5 x - 4} + \sqrt{x}}{\sqrt{5 x - 4} + \sqrt{x}})]$

= $\lim_{x \to 1} [\frac{5 x - 4 - x}{(x - 1) (\sqrt{5 x - 4} + \sqrt{x})}]$

= $\lim_{x \to 1} [\frac{4 (x - 1)}{(x - 1) (\sqrt{5 x - 4} + \sqrt{x})}]$

= $\frac{4}{\sqrt{5 - 4} + \sqrt{1}}$

= $\frac{4}{2}$

= 2

Page No 29.28:

Question 9:

$\lim_{x \to 1} \frac{x - 1}{\sqrt{x^{2} + 3 - 2}}$

Answer:

$\lim_{x \to 1} [\frac{x - 1}{\sqrt{x^{2} + 3} - 2}]$

It is of the form $\frac{0}{0}$ .
Rationalising the denominator:

$\lim_{x \to 1} [\frac{(x - 1) (\sqrt{x^{2} + 3} + 2)}{(\sqrt{x^{2} + 3} - 2) (\sqrt{x^{2} + 3} + 2)}]$

= $\lim_{x \to 1} [\frac{(x - 1) (\sqrt{x^{2} + 3} + 2)}{x^{2} + 3 - 4}]$

= $\lim_{x \to 1} [\frac{(x - 1) (\sqrt{x^{2} + 3} + 2)}{(x^{2} - 1)}]$

= $\lim_{x \to 1} [\frac{(x - 1) (\sqrt{x^{2} + 3} + 2)}{(x - 1) (x + 1)}]$

= $\frac{\sqrt{1 + 3} + 2}{1 + 1}$

= $\frac{4}{2}$

= 2

Page No 29.28:

Question 10:

$\lim_{x \to 3} \frac{\sqrt{x + 3} - \sqrt{6}}{x^{2} - 9}$

Answer:

$\lim_{x \to 3} [\frac{\sqrt{x + 3} - \sqrt{16}}{x^{2} - 9}]$

It is of the form $\frac{0}{0}$ .
Rationalising the numerator:

$\lim_{x \to 3} [\frac{(\sqrt{x + 3} - \sqrt{6}) (\sqrt{x + 3} + \sqrt{6})}{(x^{2} - 9) (\sqrt{x + 3} + \sqrt{6})}]$

= $\lim_{x \to 3} [\frac{(x + 3 - 6)}{(x - 3) (x + 3) (\sqrt{x + 3} + \sqrt{6})}]$

= $\frac{1}{6 \times 2 \sqrt{6}}$

= $\frac{1}{12 \sqrt{6}}$

Page No 29.28:

Question 11:

$\lim_{x \to 1} \frac{\sqrt{5 x - 4} - \sqrt{x}}{x^{2} - 1}$

Answer:

$\lim_{x \to 1} [\frac{\sqrt{5 x - 4} - \sqrt{x}}{x^{2} - 1}]$

It is of the form $\frac{0}{0}$ .
Rationalising the numerator:

$\lim_{x \to 1} [\frac{(\sqrt{5 x - 4} - \sqrt{x}) (\sqrt{5 x - 4} + \sqrt{x})}{(\sqrt{5 x - 4} + \sqrt{x}) (x^{2} - 1)}]$

= $\lim_{x \to 1} [\frac{5 x - 4 - x}{(\sqrt{5 x - 4} + \sqrt{x}) (x - 1) (x + 1)}]$

= $\lim_{x \to 1} [\frac{4 (x - 1)}{(\sqrt{5 x - 4} + \sqrt{x}) (x - 1) (x + 1)}]$

= $\frac{4}{(\sqrt{5 - 4} + \sqrt{1}) (1 + 1)}$

= $\frac{4}{2 \times 2}$

= 1

Page No 29.28:

Question 12:

$\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{x}$

Answer:

$\lim_{x \to 0} [\frac{\sqrt{1 + x} - 1}{x}]$

It is of the form $\frac{0}{0}$ .
Rationalising the numerator:

$\lim_{x \to 0} [\frac{(\sqrt{1 + x} - 1) (\sqrt{1 + x} + 1)}{x (\sqrt{1 + x} + 1)}]$

= $\lim_{x \to 0} [\frac{1 + x - 1}{x (\sqrt{1 + x} + 1)}]$

= $\frac{1}{\sqrt{1 + 0} + 1}$

= $\frac{1}{2}$

Page No 29.28:

Question 13:

$\lim_{x \to 2} \frac{\sqrt{x^{2} + 1} - \sqrt{5}}{x - 2}$

Answer:

$\lim_{x \to 2} [\frac{\sqrt{x^{2} + 1} - \sqrt{5}}{x - 2}]$

It is of the form $\frac{0}{0}$ .
Rationalising the numerator:

$\lim_{x \to 2} [\frac{(\sqrt{x^{2} + 1} - \sqrt{5}) (\sqrt{x^{2} + 1} + \sqrt{5})}{(x - 2) (\sqrt{x^{2} + 1} + \sqrt{5})}]$

= $\lim_{x \to 2} [\frac{x^{2} + 1 - 5}{(x - 2) (\sqrt{x^{2} + 1} + \sqrt{5})}]$

= $\lim_{x \to 2} [\frac{x^{2} - 4}{(x - 2) (\sqrt{x^{2} + 1} + \sqrt{5})}]$

= $\lim_{x \to 2} [\frac{(x - 2) (x + 2)}{(x - 2) (\sqrt{x^{2} + 1} + \sqrt{5})}]$

= $\frac{4}{2 \sqrt{5}}$

= $\frac{2}{\sqrt{5}}$

Page No 29.28:

Question 14:

$\lim_{x \to 2} \frac{x - 2}{\sqrt{x} - \sqrt{2}}$

Answer:

$\lim_{x \to 2} [\frac{x - 2}{\sqrt{x} - \sqrt{2}}]$

It is of the form $\frac{0}{0}$ .

⇒ $\lim_{x \to 2} [\frac{{(\sqrt{x})}^{2} - {(\sqrt{2})}^{2}}{x - \sqrt{2}}]$

= $\lim_{x \to 2} [\frac{(\sqrt{x} - \sqrt{2}) (\sqrt{x} + \sqrt{2})}{(x - \sqrt{2})}]$

= $\sqrt{2} + \sqrt{2}$

= $2 \sqrt{2}$

Page No 29.28:

Question 15:

$\lim_{x \to 7} \frac{4 - \sqrt{9 + x}}{1 - \sqrt{8 - x}}$

Answer:

$\lim_{x \to 7} [\frac{4 - \sqrt{9 + x}}{1 - \sqrt{8 - x}}]$

It is of the form $\frac{0}{0}$ .
Rationalising the numerator and the denominator:

$\lim_{x \to 7} [\frac{(4 - \sqrt{9 + x})}{1} \times \frac{(4 + \sqrt{9 + x})}{(4 + \sqrt{9 + x})} \times \frac{1}{(1 - \sqrt{8 - x})} \times \frac{(1 + \sqrt{8 - x})}{(1 + \sqrt{8 - x})}]$

= $\lim_{x \to 7} [\frac{16 - (9 + x)}{(4 + \sqrt{9 + x})} \times \frac{(1 + \sqrt{8 - x})}{1 - (8 - x)}]$

= $\lim_{x \to 7} [\frac{- 1 (- 7 + x) (1 + \sqrt{8 - x})}{(4 + \sqrt{9 + x}) (- 7 + x)}]$

= $\lim_{x \to 7} [\frac{- (1 + \sqrt{8 - x})}{4 + \sqrt{9 + x}}]$

= $- (\frac{1 + \sqrt{8 - 7}}{4 + \sqrt{9 + 7}})$

= $\frac{- 2}{4 + 4}$

= $\frac{- 1}{4}$

Page No 29.28:

Question 16:

$\lim_{x \to 0} \frac{\sqrt{a + x} - \sqrt{a}}{x \sqrt{a^{2} + a x}}$

Answer:

$\lim_{x \to 0} [\frac{\sqrt{a + x} - \sqrt{a}}{x \sqrt{a^{2} + a x}}]$

It is of the form $\frac{0}{0}$ .
Rationalising the numerator:

$\lim_{x \to 0} [\frac{(\sqrt{a + x} - \sqrt{a}) \times (\sqrt{a + x} + \sqrt{a})}{(x \sqrt{a^{2} + a x}) \times (\sqrt{a + x} + \sqrt{a})}]$

= $\lim_{x \to 0} [\frac{a + x - a}{x (\sqrt{a^{2} + a x}) (\sqrt{a + x} + \sqrt{a})}]$

= $\frac{1}{(\sqrt{a^{2}}) (\sqrt{a} + \sqrt{a})}$

= $\frac{1}{2 a \sqrt{a}}$

Page No 29.28:

Question 17:

$\lim_{x \to 5} \frac{x - 5}{\sqrt{6 x - 5} - \sqrt{4 x + 5}}$

Answer:

$\lim_{x \to 5} [\frac{x - 5}{\sqrt{6 x - 5} - \sqrt{4 x + 5}}]$

It is of the form $\frac{0}{0}$ .
Rationalising the denominator:

$\lim_{x \to 5} [\frac{(x - 5) (\sqrt{6 x - 5} + \sqrt{4 x + 5})}{(\sqrt{6 x - 5} - \sqrt{4 x + 5}) (\sqrt{6 x - 5} + \sqrt{4 x + 5})}]$

= $\lim_{x \to 5} [\frac{(x - 5) (\sqrt{6 x - 5} + \sqrt{4 x + 5})}{(6 x - 5) - (4 x + 5)}]$

= $\lim_{x \to 5} [\frac{(x - 5) (\sqrt{6 x - 5} + \sqrt{4 x + 5})}{2 x - 10}]$

= $\lim_{x \to 5} [\frac{(x - 5) (\sqrt{6 x - 5} + \sqrt{4 x + 5})}{2 (x - 5)}]$

= $\frac{\sqrt{6 \times 5 - 5} + \sqrt{4 \times 5 + 5}}{2}$

= $\frac{5 + 5}{2}$

= 5

Page No 29.28:

Question 18:

$\lim_{x \to 1} \frac{\sqrt{5 x - 4} - \sqrt{x}}{x^{3} - 1}$

Answer:

$\lim_{x \to 1} [\frac{\sqrt{5 x - 4} - \sqrt{x}}{x^{3} - 1}]$

It is of the form $\frac{0}{0}$ .
Rationalising the numerator:

$\lim_{x \to 1} [\frac{(\sqrt{5 x - 4} - \sqrt{x}) (\sqrt{5 x - 4} + \sqrt{x})}{(x^{3} - 1) (\sqrt{5 x - 4} + \sqrt{x})}]$

= $\lim_{x \to 1} [\frac{5 x - 4 - x}{(x - 1) (x^{2} + x + 1) (\sqrt{5 x - 4} + \sqrt{x})}]$

= $\lim_{x \to 1} [\frac{4 (x - 1)}{(x - 1) (x^{2} + x + 1) (\sqrt{5 x - 4} + \sqrt{x})}]$

= $\frac{4}{3 (1 + 1)}$

= $\frac{2}{3}$

Page No 29.28:

Question 19:

$\lim_{x \to 2} \frac{\sqrt{1 + 4 x} - \sqrt{5 + 2 x}}{x - 2}$

Answer:

$\lim_{x \to 2} [\frac{\sqrt{1 + 4 x} - \sqrt{5 + 2 x}}{x - 2}]$

It is of the from $\frac{0}{0}$ .
Rationalising the numerator:

$\lim_{x \to 2} [\frac{(\sqrt{1 + 4 x} - \sqrt{5 + 2 x}) (\sqrt{1 + 4 x} + \sqrt{5 + 2 x})}{(x - 2) (\sqrt{1 + 4 x} + \sqrt{5 + 2 x})}]$

= $\lim_{x \to 2} [\frac{(1 + 4 x) - (5 + 2 x)}{(x - 2) (\sqrt{1 + 4 x} + \sqrt{5 + 2 x})}]$

= $\lim_{x \to 2} [\frac{2 (x - 2)}{(x - 2) (\sqrt{1 + 4 x} + \sqrt{5 + 2 x})}]$

= $\frac{2}{(\sqrt{1 + 4 \times 2} + \sqrt{5 + 2 \times 2})}$

= $\frac{2}{3 + 3}$

= $\frac{1}{3}$

Page No 29.28:

Question 20:

$\lim_{x \to 1} \frac{\sqrt{3 + x} - \sqrt{5 - x}}{x^{2} - 1}$

Answer:

$\lim_{x \to 1} [\frac{\sqrt{3 + x} - \sqrt{5 - x}}{x^{2} - 1}]$

It is of the form $\frac{0}{0}$ .
Rationalising the numerator:

$\lim_{x \to 1} [\frac{(\sqrt{3 + x} - \sqrt{5 - x}) (\sqrt{3 + x} + \sqrt{5 - x})}{(x - 1) (x + 1) (\sqrt{3 + x} + \sqrt{5 - x})}]$

= $\lim_{x \to 1} [\frac{(3 + x) - (5 - x)}{(x - 1) (x + 1) \{\sqrt{3 + x} + \sqrt{5 - x}\}}]$

= $\lim_{x \to 1} [\frac{2 x - 2}{(x - 1) (x + 1) \{\sqrt{3 + x} + \sqrt{5 - x}\}}]$

= $\lim_{x \to 1} [\frac{2 (x - 1)}{(x - 1) (x + 1) \{\sqrt{3 + x} + \sqrt{5 - x}\}}]$

= $\frac{2}{2 (\sqrt{4} + \sqrt{4})}$

= $\frac{1}{4}$

Page No 29.29:

Question 21:

$\lim_{x \to 0} \frac{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}{x}$

Answer:

$\lim_{x \to 0} [\frac{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}{x}]$

It is of the form $\frac{0}{0}$ .
Rationalising the numerator:

$\lim_{x \to 0} [\frac{(\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}) (\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}})}{(\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}) x}]$

= $\lim_{x \to 0} [\frac{(1 + x^{2}) - (1 - x^{2})}{x \{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}\}}]$

= $\lim_{x \to 0} [\frac{2 x^{2}}{x \{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}\}}]$

= $\frac{2 \times 0}{\sqrt{1 + 0} + \sqrt{1 - 0}}$

= 0

Page No 29.29:

Question 22:

$\lim_{x \to 0} \frac{\sqrt{1 + x + x^{2}} - \sqrt{x + 1}}{2 x^{2}}$

Answer:

$\lim_{x \to 0} [\frac{\sqrt{1 + x + x^{2}} - \sqrt{x + 1}}{2 x^{2}}]$

It is of the form $\frac{0}{0}$ .
Rationalising the numerator:

$\lim_{x \to 0} [\frac{(\sqrt{1 + x + x^{2}} - \sqrt{x + 1}) (\sqrt{1 + x + x^{2}} + \sqrt{x + 1})}{(\sqrt{1 + x + x^{2}} + \sqrt{x + 1}) 2 x^{2}}]$

= $\lim_{x \to 0} [\frac{(1 + x + x^{2}) - (x + 1)}{(\sqrt{1 + x + x^{2}} + \sqrt{x + 1}) 2 x^{2}}]$

= $\lim_{x \to 0} [\frac{x^{2}}{(\sqrt{1 + x + x^{2}} + \sqrt{x + 1}) (2 x^{2})}]$

= $\frac{1}{(\sqrt{1 + 0 + 0} + \sqrt{0 + 1}) \times 2}$

= $\frac{1}{2} \times \frac{1}{2}$

= $\frac{1}{4}$

Page No 29.29:

Question 23:

$\lim_{x \to 4} \frac{2 - \sqrt{x}}{4 - x}$

Answer:

$\lim_{x \to 4} [\frac{2 - \sqrt{x}}{4 - x}]$

= $\lim_{x \to 4} [\frac{2 - \sqrt{x}}{2 - {(\sqrt{x})}^{2}}]$

= $\lim_{x \to 4} [\frac{(2 - \sqrt{x})}{(2 - \sqrt{x}) (2 + \sqrt{x})}]$

= $\frac{1}{2 + \sqrt{4}}$

= $\frac{1}{2 + 2}$

= $\frac{1}{4}$

Page No 29.29:

Question 24:

$\lim_{x \to a} \frac{x - a}{\sqrt{x} - \sqrt{a}}$

Answer:

$\lim_{x \to a} [\frac{x - a}{\sqrt{x} - \sqrt{a}}]$

= $\lim_{x \to a} [\frac{{(\sqrt{x})}^{2} - a^{2}}{\sqrt{x} - \sqrt{a}}]$

= $\lim_{x \to a} [\frac{(\sqrt{x} - \sqrt{a}) (\sqrt{x} + \sqrt{a})}{(\sqrt{x} - \sqrt{a})}]$

= $\sqrt{a} + \sqrt{a}$

= $2 \sqrt{a}$

Page No 29.29:

Question 25:

$\lim_{x \to 0} \frac{\sqrt{1 + 3 x} - \sqrt{1 - 3 x}}{x}$

Answer:

$\lim_{x \to 0} [\frac{\sqrt{1 + 3 x} - \sqrt{1 - 3 x}}{x}]$

It is of the form $\frac{0}{0}$ .

Rationalising the numerator:

$\lim_{x \to 0} [\frac{(\sqrt{1 + 3 x} - \sqrt{1 - 3 x}) (\sqrt{1 + 3 x} + \sqrt{1 - 3 x})}{x (\sqrt{1 + 3 x} + \sqrt{1 - 3 x})}] = \lim_{x \to 0} [\frac{(1 + 3 x) - (1 - 3 x)}{x (\sqrt{1 + 3 x} + \sqrt{1 - 3 x})}] = \lim_{x \to 0} [\frac{6 x}{x (\sqrt{1 + 3 x} + \sqrt{1 - 3 x})}] = \frac{6}{\sqrt{1} + \sqrt{1}} = \frac{6}{2} = 3$

Page No 29.29:

Question 26:

$\lim_{x \to 0} \frac{\sqrt{2 - x} - \sqrt{2 + x}}{x}$

Answer:

$\lim_{x \to 0} [\frac{\sqrt{2 - x} - \sqrt{2 + x}}{x}]$

It is of the form $\frac{0}{0}$ .
Rationalising the numerator:

$\lim_{x \to 0} [\frac{(\sqrt{2 - x} - \sqrt{2 + x}) (\sqrt{2 - x} + \sqrt{2 + x})}{x (\sqrt{2 - x} + \sqrt{2 + x})}]$

= $\lim_{x \to 0} [\frac{(2 - x) - (2 + x)}{x (\sqrt{2 - x} + \sqrt{2 + x})}]$

= $\lim_{x \to 0} [\frac{- 2 x}{x (\sqrt{2 - x} + \sqrt{2 + x})}]$

= $\frac{- 2}{2 \sqrt{2}}$

= $- \frac{1}{\sqrt{2}}$

Page No 29.29:

Question 27:

$\lim_{x \to 1} \frac{\sqrt{3 + x} - \sqrt{5 - x}}{x^{2} - 1}$

Answer:

$\lim_{x \to 1} [\frac{\sqrt{3 + x} - \sqrt{5 - x}}{x^{2} - 1}]$

It is of the form $\frac{0}{0}$ .

Rationalising the numerator:

$\lim_{x \to 1} [\frac{(\sqrt{3 + x} - \sqrt{5 - x}) (\sqrt{3 + x} + \sqrt{5 - x})}{(x - 1) (x + 1) (\sqrt{3 + x} + \sqrt{5 - x})}] = \lim_{x \to 1} [\frac{(3 + x) - (5 - x)}{(x - 1) (x + 1) (\sqrt{3 + x} + \sqrt{5 - x})}] = \lim_{x \to 1} \frac{2 (x - 1)}{(x - 1) (x + 1) (\sqrt{3 + x} + \sqrt{5 - x})} = \frac{2}{(1 + 1) (\sqrt{3 + 1} + \sqrt{5 - 1})} = \frac{2}{2 \times (2 + 2)} = \frac{1}{4}$

Page No 29.29:

Question 28:

$\lim_{x \to 1} \frac{(2 x - 3) (\sqrt{x} - 1)}{3 x^{2} + 3 x - 6}$

Answer:

$\lim_{x \to 1} [\frac{(2 x - 3) (\sqrt{x} - 1)}{3 x^{2} + 3 x - 6}]$

It is of the form $\frac{0}{0}$ .

⇒ $\lim_{x \to 1} [\frac{(2 x - 3) (\sqrt{x} - 1)}{3 (x^{2} + x - 2)}]$

= $\lim_{x \to 1} [\frac{(2 x - 3) (\sqrt{x} - 1)}{3 (x^{2} + 2 x - x - 2)}]$

= $\lim_{x \to 1} [\frac{(2 x - 3) (\sqrt{x} - 1)}{3 (x (x + 2) - 1 (x + 2))}]$

= $\lim_{x \to 1} [\frac{(2 x - 3) (\sqrt{x} - 1)}{3 (x - 1) (x + 2)}]$

= $\lim_{x \to 1} [\frac{(2 x - 3) (\sqrt{x} - 1)}{3 ({(\sqrt{x})}^{2} - 1^{2}) (x + 2)}]$

= $\lim_{x \to 1} [\frac{(2 x - 3) (\sqrt{x} - 1)}{3 (\sqrt{x} + 1) (\sqrt{x} - 1) (x + 2)}]$

= $\frac{- 1}{3 (2) \times 3}$

= $\frac{- 1}{18}$

Page No 29.29:

Question 29:

$\lim_{x \to 0} \frac{\sqrt{1 + x^{2}} - \sqrt{1 + x}}{\sqrt{1 + x^{3}} - \sqrt{1 + x}}$

Answer:

$\lim_{x \to 0} [\frac{\sqrt{1 - x^{2}} - \sqrt{1 + x}}{\sqrt{1 + x^{3}} - \sqrt{1 + x}}]$

It is of the form $\frac{0}{0}$ .
Rationalising the numerator and the denominator:

$\lim_{x \to 0} [\frac{(\sqrt{1 + x^{2}} - \sqrt{1 + x})}{1} \times \frac{(\sqrt{1 + x^{2}} + \sqrt{1 + x})}{(\sqrt{1 + x^{2}} + \sqrt{1 + x})} \times \frac{1}{(\sqrt{1 + x^{3}} - \sqrt{1 + x})} \times \frac{(\sqrt{1 + x^{3}} - \sqrt{1 + x})}{(\sqrt{1 + x^{3}} + \sqrt{1 + x})}] = \lim_{x \to 0} [\frac{[(1 + x^{2}) - (1 + x)]}{[(1 + x^{3}) - (1 + x)]} \times \frac{(\sqrt{1 + x^{3}} + \sqrt{1 + x})}{(\sqrt{1 + x^{2}} + \sqrt{1 + x})}] = \lim_{x \to 0} [(\frac{x^{2} - x}{x^{3} - x}) \times \frac{(\sqrt{1 + x^{3}} + \sqrt{1 + x})}{(\sqrt{1 + x^{2}} + \sqrt{1 + x})}] = \lim_{x \to 0} [\frac{x (x - 1)}{x (x^{2} - 1)} \frac{(\sqrt{1 + x^{3}} + \sqrt{1 + x})}{(\sqrt{1 + x^{2}} + \sqrt{1 + x})}] = \lim_{x \to 0} [\frac{(x - 1) (\sqrt{1 + x^{3}} + \sqrt{1 + x})}{(x - 1) (x + 1) (\sqrt{1 + x^{2}} + \sqrt{1 + x})}] = \frac{(\sqrt{1 + 0} + \sqrt{1 + 0})}{(0 + 1) (\sqrt{1 + 0} + \sqrt{1 + 0})} = \frac{2}{2} = 1$

Page No 29.29:

Question 30:

$\lim_{x \to 1} \frac{x^{2} - \sqrt{x}}{\sqrt{x} - 1}$

Answer:

$\lim_{x \to 1} [\frac{x^{2} - \sqrt{x}}{\sqrt{x} - 1}]$

It is of the form $\frac{0}{0}$ .

$\lim_{x \to 1} [\frac{\sqrt{x} (x \sqrt{x} - 1)}{\sqrt{x} - 1}]$

= $\lim_{x \to 1} [\frac{\sqrt{x} (x^{3 / 2} - 1)}{x^{\frac{1}{2}} - 1}]$

= $\lim_{x \to 1} [\frac{\sqrt{x} ({(\sqrt{x})}^{3} - 1^{3})}{(\sqrt{x} - 1)}] [A^{3} - B^{3} = (A - B) (A^{2} + A B + B^{2})]$

= $\lim_{x \to 1} [\frac{(\sqrt{x}) (\sqrt{x} - 1) (x + \sqrt{x} + 1)}{(\sqrt{x} - 1)}]$

= 1 (1 + 1 + 1)

= 3

Page No 29.29:

Question 31:

$\lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h}, x \neq 0$

Answer:

$\lim_{h \to 0} [\frac{\sqrt{x + h} - \sqrt{x}}{h}]$

It is of the form $\frac{0}{0}$ .
Rationalising the numerator:
= $\lim_{h \to 0} [\frac{(\sqrt{x + h} - \sqrt{x})}{h} \times \frac{(\sqrt{x + h} + \sqrt{x})}{(\sqrt{x + h} + \sqrt{x})}]$

= $\lim_{h \to 0} [\frac{(x + h) - x}{h (\sqrt{x + h} + \sqrt{x})}]$

= $\frac{1}{\sqrt{x} + \sqrt{x}}$

= $\frac{1}{2 \sqrt{x}}$

Page No 29.29:

Question 32:

$\lim_{x \to \sqrt{10}} \frac{\sqrt{7 + 2 x} - (\sqrt{5} + \sqrt{2})}{x^{2} - 10}$

Answer:

$\lim_{x \to \sqrt{10}} [\frac{\sqrt{7 + 2 x} - (\sqrt{5} + \sqrt{2})}{x^{2} - {(\sqrt{10})}^{2}}]$

= $\lim_{x \to \sqrt{10}} [\frac{\sqrt{7 + 2 x} - \sqrt{{(\sqrt{5} + \sqrt{2})}^{2}}}{(x - \sqrt{10}) (x + \sqrt{10})}]$

= $\lim_{x \to \sqrt{10}} [\frac{\sqrt{7 + 2 x} - \sqrt{5 + 2 + 2 \sqrt{5} \sqrt{2}}}{(x - \sqrt{10}) (x + \sqrt{10})}]$

= $\lim_{x \to \sqrt{10}} [\frac{\sqrt{7 + 2 x} - \sqrt{7 + 2 \sqrt{10}}}{(x - \sqrt{10}) (x + \sqrt{10})}]$

Rationalising the numerator:

$\lim_{x \to \sqrt{10}} [\frac{(\sqrt{7 + 2 x} - \sqrt{7 + 2 \sqrt{10}}) (\sqrt{7 + 2 x} + \sqrt{7 + 2 \sqrt{10}})}{(x - \sqrt{10}) (x + \sqrt{10}) (\sqrt{7 + 2 x} + \sqrt{7 + 2 \sqrt{10}})}]$

= $\lim_{x \to \sqrt{10}} [\frac{(7 + 2 x) - (7 + 2 \sqrt{10})}{(x - \sqrt{10}) (x + \sqrt{10}) (\sqrt{7 + 2 x} + \sqrt{7 + 2 \sqrt{10}})}]$

= $\lim_{x \to \sqrt{10}} [\frac{2 (x - \sqrt{10})}{(x - \sqrt{10}) (x + \sqrt{10}) (\sqrt{7 + 2 x} + \sqrt{7 + 2 \sqrt{10}})}]$

= $\frac{2}{(\sqrt{10} + \sqrt{10}) (2 \sqrt{7 + 2 \sqrt{10}})}$

$= \frac{2}{(2 \sqrt{10}) \times 2 \sqrt{7 + 2 \sqrt{10}}} = \frac{1}{2 \sqrt{10} \sqrt{{(\sqrt{5} + \sqrt{2})}^{2}}}$

= $\frac{1}{2 \sqrt{10} (\sqrt{5} + \sqrt{2})} \times \frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} - \sqrt{2}}$

$= \frac{1}{2 \sqrt{10}} [\frac{\sqrt{5} - \sqrt{2}}{{(\sqrt{5})}^{2} - {(\sqrt{2})}^{2}}]$

= $\frac{\sqrt{5} - \sqrt{2}}{6 \sqrt{10}}$

Page No 29.29:

Question 33:

$\lim_{x \to \sqrt{6}} \frac{\sqrt{5 + 2 x} - (\sqrt{3} + \sqrt{2})}{x^{2} - 6}$

Answer:

$\lim_{x \to \sqrt{6}} [\frac{\sqrt{5 + 2 x} - (\sqrt{3} + \sqrt{2})}{x^{2} - 6}]$

= $\lim_{x \to \sqrt{6}} [\frac{\sqrt{5 + 2 x} - \sqrt{{(\sqrt{3} + \sqrt{2})}^{2}}}{x^{2} - {(\sqrt{6})}^{2}}]$

= $\lim_{x \to \sqrt{6}} [\frac{\sqrt{5 + 2 x} - \sqrt{3 + 2 + 2 \sqrt{6}}}{(x - \sqrt{6}) (x + \sqrt{6})}]$

= $\lim_{x \to \sqrt{6}} [\frac{\sqrt{5 + 2 x} - \sqrt{5 + 2 \sqrt{6}}}{(x - \sqrt{6}) (x + \sqrt{6})}]$

Rationalising the numerator:

$\lim_{x \to \sqrt{6}} [\frac{(\sqrt{5 + 2 x} - \sqrt{5 + 2 \sqrt{6}}) (\sqrt{5 + 2 x} + \sqrt{5 + 2 \sqrt{6}})}{(x - \sqrt{6}) (x + \sqrt{6}) (\sqrt{5 + 2 x} + \sqrt{5 + 2 \sqrt{6}})}]$

= $\lim_{x \to \sqrt{6}} [\frac{(5 + 2 x) - (5 + 2 \sqrt{6})}{(x - \sqrt{6}) (x + \sqrt{6}) (\sqrt{5 + 2 x} + \sqrt{5 + 2 \sqrt{6}})}]$

= $\lim_{x \to \sqrt{6}} [\frac{2 (x - \sqrt{6})}{(x - \sqrt{6}) (x + \sqrt{6}) (\sqrt{5 + 2 x} + \sqrt{5 + 2 \sqrt{6}})}]$

= $\frac{2}{(\sqrt{6} + \sqrt{6}) (\sqrt{5 + 2 \sqrt{6}} + \sqrt{5 + 2 \sqrt{6}})}$

= $\frac{1}{2 \sqrt{6} (\sqrt{{(\sqrt{3} + \sqrt{2})}^{2}})}$

= $\frac{1}{2 \sqrt{6} (\sqrt{3} + \sqrt{2})}$

= $\frac{1}{2 \sqrt{6} (\sqrt{3} + \sqrt{2})} \times \frac{(\sqrt{3} - \sqrt{2})}{(\sqrt{3} - \sqrt{2})}$

= $\frac{\sqrt{3} - \sqrt{2}}{2 \sqrt{6} (3 - 2)}$

= $\frac{\sqrt{3} - \sqrt{2}}{2 \sqrt{6}}$

Page No 29.29:

Question 34:

$\lim_{x \to \sqrt{2}} \frac{\sqrt{3 + 2 x} - (\sqrt{2} + 1)}{x^{2} - 2}$

Answer:

$\lim_{x \to \sqrt{2}} [\frac{\sqrt{3 + 2 x} - (\sqrt{2} + 1)}{x^{2} - 2}]$

= $\lim_{x \to \sqrt{2}} [\frac{\sqrt{3 + 2 x} - \sqrt{{(\sqrt{2} + 1)}^{2}}}{(x - \sqrt{2}) (x + \sqrt{2})}]$

= $\lim_{x \to \sqrt{2}} [\frac{\sqrt{3 + 2 x} - \sqrt{2 + 1 + 2 \sqrt{2}}}{(x - \sqrt{2}) (x + \sqrt{2})}]$

= $\lim_{x \to \sqrt{2}} [\frac{(\sqrt{3 + 2 x} - \sqrt{3 + 2 \sqrt{2}})}{(x - \sqrt{2}) (x + \sqrt{2})}]$

Rationalising the numerator:

$\lim_{x \to \sqrt{2}} [\frac{(\sqrt{3 + 2 x} - \sqrt{3 + 2 \sqrt{2}}) (\sqrt{3 + 2 x} + \sqrt{3 + 2 \sqrt{2}})}{(x - \sqrt{2}) (x + \sqrt{2}) (\sqrt{3 + 2 x} + \sqrt{3 + 2 \sqrt{2}})}]$

= $\lim_{x \to \sqrt{2}} [\frac{(3 + 2 x) - (3 + 2 \sqrt{2})}{(x - \sqrt{2}) (x + \sqrt{2}) (\sqrt{3 + 2 x} + \sqrt{3 + 2 \sqrt{2}})}]$

= $\lim_{x \to \sqrt{2}} [\frac{2 (x - \sqrt{2})}{(x - \sqrt{2}) (x + \sqrt{2}) (\sqrt{3 + 2 x} + \sqrt{3 + 2 \sqrt{2}})}]$

= $\frac{2}{(\sqrt{2} + \sqrt{2}) (\sqrt{3 + 2 \sqrt{2}} + \sqrt{3 + 2 \sqrt{2}})}$

= $\frac{2}{(2 \sqrt{2}) (2 \sqrt{3 + 2 \sqrt{2}})}$

= $\frac{1}{2 \sqrt{2} (\sqrt{3 + 2 \sqrt{2}})}$

= $\frac{1}{2 \sqrt{2} \sqrt{{(\sqrt{2} + 1)}^{2}}}$

= $\frac{1}{2 \sqrt{2} (\sqrt{2} + 1)} \times \frac{\sqrt{2} - 1}{\sqrt{2} - 1}$

= $\frac{\sqrt{2} - 1}{2 \sqrt{2} (2 - 1)}$

= $\frac{\sqrt{2} - 1}{2 \sqrt{2}}$

Page No 29.33:

Question 1:

$\lim_{x \to a} \frac{{(x + 2)}^{5 / 2} - {(a + 2)}^{5 / 2}}{x - a}$

Answer:

$\lim_{x \to a} [\frac{{(x + 2)}^{\frac{5}{2}} - {(a + 2)}^{\frac{5}{2}}}{x - a}] = \lim_{x \to a} [\frac{{(x + 2)}^{\frac{5}{2}} - {(a + 2)}^{\frac{5}{2}}}{(x + 2) - (a + 2)}]$

Let y = x + 2 and b = a + 2.

When x → a, then x + 2 → a + 2.
⇒ y → b

$\lim_{y \to b} [\frac{y^{\frac{5}{2}} - b^{\frac{5}{2}}}{y - b}] = \frac{5}{2} {(b)}^{\frac{5}{2} - 1} = \frac{5}{2} b^{\frac{3}{2}} = \frac{5}{2} {(a + 2)}^{\frac{3}{2}}$

Page No 29.33:

Question 2:

$\lim_{x \to a} \frac{{(x + 2)}^{3 / 2} - {(a + 2)}^{3 / 2}}{x - a}$

Answer:

$\lim_{x \to a} = [\frac{{(x + 2)}^{\frac{3}{2}} - {(a + 2)}^{\frac{3}{2}}}{x - a}] = \lim_{x \to a} [\frac{{(x + 2)}^{\frac{3}{2}} - {(a + 2)}^{\frac{3}{2}}}{(x + 2) - (a + 2)}]$

Let y = x + 2 and b = a + 2.

When x → a and x + 2 → a + 2.
$\Rightarrow$ y → b

$\lim_{y \to b} [\frac{y^{\frac{3}{2}} - b^{\frac{3}{2}}}{y - b}] = \frac{3}{2} {(b)}^{\frac{3}{2} - 1} = \frac{3}{2} b^{\frac{1}{2}} = \frac{3}{2} {(a + 2)}^{\frac{1}{2}}$

Page No 29.33:

Question 3:

$\lim_{x \to 0} \frac{{(1 + x)}^{6} - 1}{{(1 + x)}^{2} - 1}$

Answer:

$\lim_{x \to 0} [\frac{{(1 + x)}^{6} - 1}{{(1 + x)}^{2} - 1}] = \lim_{x \to 0} [\frac{{(1 + x)}^{6} - 1}{x} \times \frac{x}{{(1 + x)}^{2} - 1}] = \lim_{x \to 0} [\frac{{(1 + x)}^{6} - 1^{6}}{(1 + x) - 1} \times \frac{(1 + x) - 1}{{(1 + x)}^{2} - 1}]$

Let y = 1 + x

When x → 0, then 1 + x → 1.
$\Rightarrow$ y → 1

$\lim_{y \to 1} [(\frac{y^{6} - 1^{6}}{y - 1}) \times \frac{(y - 1)}{y^{2} - 1^{2}}] = \frac{6 \times {(1)}^{6 - 1}}{2 \times {(1)}^{2 - 1}} = \frac{6}{2} = 3$

Page No 29.33:

Question 4:

$\lim_{x \to a} \frac{x^{2 / 7} - a^{2 / 7}}{x - a}$

Answer:

$\lim_{x \to a} [\frac{x^{\frac{2}{7}} - a^{\frac{2}{7}}}{x - a}] = \frac{2}{7} a^{\frac{2}{7} - 1} [∵ \lim_{x \to a} [\frac{x^{n} - a^{n}}{x - a}] = n a^{n - 1}] = \frac{2}{7} a^{- \frac{5}{7}}$

Page No 29.33:

Question 5:

$\lim_{x \to a} \frac{x^{5 / 7} - a^{5 / 7}}{x^{2 / 7} - a^{2 / 7}}$

Answer:

$\lim_{x \to a} [\frac{x^{\frac{5}{7}} - a^{\frac{5}{7}}}{x^{\frac{2}{7}} - a^{\frac{2}{7}}}] = \lim_{x \to a} [(\frac{x^{\frac{5}{7}} - a^{\frac{5}{7}}}{x - a}) \times (\frac{x - a}{x^{\frac{2}{7}} - a^{\frac{2}{7}}})] = \frac{5}{7} a^{\frac{5}{7} - 1} \times \frac{1}{\frac{2}{7} a^{\frac{2}{7} - 1}} = \frac{5}{2} a^{- \frac{2}{7}} a^{\frac{5}{7}} = \frac{5}{2} a^{- \frac{2}{7} + \frac{5}{7}} = \frac{5}{2} a^{\frac{3}{7}}$

Page No 29.33:

Question 6:

$\lim_{x \to - 1 / 2} \frac{8 x^{3} + 1}{2 x + 1}$

Answer:

$\lim_{x \to - \frac{1}{2}} [\frac{8 x^{3} + 1}{2 x + 1}] = \lim_{x \to - \frac{1}{2}} [\frac{{(2 x)}^{3} - (- 1)}{2 x - (- 1)}] When x \to - \frac{1}{2}, then 2x → –1 .$
Let y = 2x

$\lim_{y \to - 1} [\frac{y^{3} - {(- 1)}^{3}}{y - (- 1)}] = 3 {(- 1)}^{3 - 1} = 3 \times {(- 1)}^{2} = 3$

Page No 29.33:

Question 7:

$\lim_{x \to 27} \frac{(x^{1 / 3} + 3) (x^{1 / 3} - 3)}{x - 27}$

Answer:

$\lim_{x \to 27} \frac{[x^{\frac{1}{3}} + 3] [x^{\frac{1}{3}} - 3]}{x - 27} = \lim_{x \to 27} [\frac{(x^{\frac{1}{3}} + 3) (x^{\frac{1}{3}} - 3)}{{(x^{\frac{1}{3}})}^{3} - 3^{3}}] x \to 27 ∴ x^{\frac{1}{3}} \to 3 Let y = x^{\frac{1}{3}} \lim_{y \to 3} [\frac{(y + 3) (y - 3)}{y^{3} - 3^{3}}] = \frac{(3 + 3)}{3 \times 3^{3 - 1}} = \frac{6}{3 \times 9} = \frac{2}{9}$

Page No 29.33:

Question 8:

$\lim_{x \to 4} \frac{x^{3} - 64}{x^{2} - 16}$

Answer:

$\lim_{x \to 4} [\frac{x^{3} - 64}{x^{2} - 16}] = \lim_{x \to 4} [\frac{x^{3} - 64}{x - 4} \times \frac{x - 4}{x^{2} - 16}] = \lim_{x \to 4} [\frac{x^{3} - 4^{3}}{x - 4} \times \frac{x - 4}{x^{2} - 4^{2}}] = 3 {(4)}^{3 - 1} \times \frac{1}{2 {(4)}^{2 - 1}} = \frac{3 \times 16}{2 \times 4} = 6$

Page No 29.33:

Question 9:

$\lim_{x \to 1} \frac{x^{15} - 1}{x^{10} - 1}$

Answer:

$\lim_{x \to 1} [\frac{x^{15} - 1}{x^{10} - 1}] = \lim_{x \to 1} [\frac{x^{15} - {(1)}^{15}}{x - 1} \times \frac{x - 1}{x^{10} - {(1)}^{10}}] = \frac{15 {(1)}^{15 - 1}}{10 {(1)}^{10 - 1}} = \frac{3}{2}$

Page No 29.33:

Question 10:

$\lim_{x \to - 1} \frac{x^{3} + 1}{x + 1}$

Answer:

$\lim_{x \to - 1} [\frac{x^{3} + 1}{x + 1}] = \lim_{x \to - 1} [\frac{x^{3} - (- 1)}{x - (- 1)}] = \lim_{x \to - 1} [\frac{x^{3} - {(- 1)}^{3}}{x - (- 1)}] = 3 {(- 1)}^{3 - 1} = 3$

Page No 29.33:

Question 11:

$\lim_{x \to a} \frac{x^{2 / 3} - a^{2 / 3}}{x^{3 / 4} - a^{3 / 4}}$

Answer:

$\lim_{x \to a} [\frac{x^{\frac{2}{3}} - a^{\frac{2}{3}}}{x^{\frac{3}{4}} - a^{\frac{3}{4}}}] = \lim_{a \to a} [(\frac{x^{\frac{2}{3}} - a^{\frac{2}{3}}}{x - a}) \times (\frac{x - a}{x^{\frac{3}{4}} - a^{\frac{3}{4}}})] = \frac{2}{3} {(a)}^{\frac{2}{3} - 1} \times \frac{1}{\frac{3}{4} a^{\frac{3}{4} - 1}} = \frac{8}{9} \frac{a^{- \frac{1}{3}}}{a^{- \frac{1}{4}}} = \frac{8}{9} a^{- \frac{1}{3} + \frac{1}{4}} = \frac{8}{9} a^{- \frac{1}{12}}$

Page No 29.33:

Question 12:

If $\lim_{x \to 3} \frac{x^{n} - 3^{n}}{x - 3} = 108,$ find the value of n.

Answer:

$\lim_{x \to 3} [\frac{x^{n} - 3^{n}}{x - 3}] = 108$

⇒ x(3)^{n – 1} = 108
⇒ x(3)^{n – 1} = 4 × 3³

On comparing LHS and RHS, we observe that x is equal to 4.

$\begin{matrix} 2 & 108 \\ 2 & 54 \\ 3 & 27 \\ 3 & 9 \\ 3 & 3 \\ 1 \end{matrix}$

Page No 29.33:

Question 13:

If $\lim_{x \to a} \frac{x^{9} - a^{9}}{x - a} = 9,$ find all possible values of a.

Answer:

$\lim_{x \to a} [\frac{x^{9} - a^{9}}{x - a}] = 9 \Rightarrow 9 a^{8} = 9 \Rightarrow a^{8} = 1 \Rightarrow a = \pm 1$

Page No 29.33:

Question 14:

If $\lim_{x \to a} \frac{x^{5} - a^{5}}{x - a} = 405,$ find all possible values of a.

Answer:

$\lim_{x \to a} [\frac{x^{5} - a^{5}}{x - a}] = 405 \Rightarrow 5 a^{4} = 405 \Rightarrow a^{4} = 81 \Rightarrow a^{2} = 9 \Rightarrow a = \pm 3$

Page No 29.33:

Question 15:

If $\lim_{x \to a} \frac{x^{9} - a^{9}}{x - a} = \lim_{x \to 5} (4 + x),$ find all possible values of a.

Answer:

$\lim_{x \to a} [\frac{x^{9} - a^{9}}{x - a}] = \lim_{x \to 5} (4 + x) \Rightarrow 9 a^{9 + 1} = 9 \Rightarrow a^{8} = 1 \Rightarrow a = \pm 1$

Page No 29.33:

Question 16:

If $\lim_{x \to a} \frac{x^{3} - a^{3}}{x - a} = \lim_{x \to 1} \frac{x^{4} - 1}{x - 1},$ find all possible values of a.

Answer:

$\lim_{x \to a} [\frac{x^{3} - a^{3}}{x - a}] = \lim_{x \to 1} [\frac{x^{4} - 1^{4}}{x - 1}] \Rightarrow 3 a^{3 - 1} = 4 {(1)}^{4 - 1} \Rightarrow 3 a^{2} = 4 \Rightarrow a^{2} = \frac{4}{3} \Rightarrow a = \pm \frac{2}{\sqrt{3}}$

Page No 29.38:

Question 1:

$\lim_{x \to \infty} \frac{(3 x - 1) (4 x - 2)}{(x + 8) (x - 1)}$

Answer:

$\lim_{x \to \infty} \frac{(3 x - 1) (4 x - 2)}{(x + 8) (x - 1)} Dividing the numerator and the denominator by x^{2} : \lim_{x \to \infty} \frac{(\frac{3 x - 1}{x}) (\frac{4 x - 2}{x})}{(\frac{x + 8}{x}) (\frac{x - 1}{x})} = \lim_{x \to \infty} [\frac{(3 - \frac{1}{x}) (4 - \frac{2}{x})}{(1 + \frac{8}{x}) (1 - \frac{1}{x})}] When x \to \infty, then \frac{1}{x} \to 0 . \frac{3 \times 4}{1} = 12$

Page No 29.38:

Question 2:

$\lim_{x \to \infty} \frac{3 x^{3} - 4 x^{2} + 6 x - 1}{2 x^{3} + x^{2} - 5 x + 7}$

Answer:

$\lim_{x \to \infty} [\frac{3 x^{3} - 4 x^{2} + 6 x - 1}{2 x^{3} + x^{2} - 5 x + 7}] Dividing the numerator and the denominator by x^{3} : \lim_{x \to \infty} [\frac{3 - \frac{4}{x} + \frac{6}{x^{2}} - \frac{1}{x^{3}}}{2 + \frac{1}{x} - \frac{5}{x^{2}} + \frac{7}{x^{3}}}] When x \to \infty, then \frac{1}{x}, \frac{1}{x^{2}}, \frac{1}{x^{3}} \to 0 . \Rightarrow \frac{3}{2}$

Page No 29.38:

Question 3:

$\lim_{x \to \infty} \frac{5 x^{3} - 6}{\sqrt{9 + 4 x^{6}}}$

Answer:

$\lim_{x \to \infty} [\frac{5 x^{3} - 6}{\sqrt{9 + 4 x^{6}}}] Dividing the numerator and the denominator by x : \lim_{x \to \infty} [\frac{\frac{5 x^{3} - 6}{x^{3}}}{\frac{\sqrt{9 + 4 x^{6}}}{x^{3}}}] = \lim_{x \to \infty} [\frac{5 - \frac{6}{x^{3}}}{\sqrt{\frac{9}{x^{6}} + 4}}] = \frac{5}{\sqrt{4}} = \frac{5}{2}$

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Question 4:

$\lim_{x \to \infty} \sqrt{x^{2} + c x - x}$

Answer:

$\lim_{x \to \infty} (\sqrt{x^{2} + c x - x})$

It is of the form ∞ - ∞.

Rationalising the numerator:

$\lim_{x \to \infty} (\sqrt{x^{2} + c x} - x) \frac{(\sqrt{x^{2} + c x} + x)}{(\sqrt{x^{2} + c x} + x)} = \lim_{x \to \infty} (\frac{x^{2} + c x - x^{2}}{\sqrt{x^{2} + c x} + x}) Dividing the numerator and the denomiator by x : \lim_{x \to \infty} [\frac{\frac{c x}{x}}{\frac{\sqrt{x^{2} + c x} + x}{x}}] = \lim_{x \to \infty} [\frac{c}{\sqrt{1 + \frac{c}{x}} + 1}] When x \to \infty, then \frac{1}{x} \to 0 . \Rightarrow \frac{c}{\sqrt{1} + 1} = \frac{c}{2}$

Page No 29.38:

Question 5:

$\lim_{x \to \infty} \sqrt{x + 1} - \sqrt{x}$

Answer:

$\lim_{x \to \infty} (\sqrt{x + 1} - \sqrt{x})$

It is of the form ∞–∞.

On rationalising, we get:
$\lim_{x \to \infty} (\sqrt{x + 1} - \sqrt{x}) \times (\frac{\sqrt{x + 1} + \sqrt{x}}{\sqrt{x + 1} + \sqrt{x}}) = \lim_{x \to \infty} (\frac{x + 1 - x}{(\sqrt{x + 1} + \sqrt{x})}) = \frac{1}{\infty} = 0$

Page No 29.38:

Question 6:

$\lim_{x \to \infty} \sqrt{x^{2} + 7 x - x}$

Answer:

$\lim_{x \to \infty} [\sqrt{x^{2} + 7 x} - x] Rationalising the numerator : \lim_{x \to \infty} [(\sqrt{x^{2} + 7 x} - x) \frac{(\sqrt{x^{2} + 7 x} + x)}{(\sqrt{x^{2} + 7 x} + x)}] = \lim_{x \to \infty} [\frac{x^{2} + 7 x - x^{2}}{(\sqrt{x^{2} + 7 x} + x)}] Dividing the numerator and the denominator by x : \lim_{x \to \infty} [\frac{7}{\frac{\sqrt{x^{2} + 7 x}}{x} + 1}] = \lim_{x \to \infty} [\frac{7}{\frac{\sqrt{x^{2} + 7 x}}{x} + 1}] When x \to \infty, then \frac{1}{x} \to 0 . \Rightarrow \frac{7}{\sqrt{1} + 1} = \frac{7}{2}$

Page No 29.38:

Question 7:

$\lim_{x \to \infty} \frac{x}{\sqrt{4 x^{2} + 1} - 1}$

Answer:

$\lim_{x \to \infty} [\frac{x}{\sqrt{4 x^{2} + 1} - 1}] Rationalising the denominator : \lim_{x \to \infty} [\frac{x}{(\sqrt{4 x^{2} + 1} - 1)} \frac{(\sqrt{4 x^{2} + 1} + 1)}{(\sqrt{4 x^{2} + 1} + 1)}] = \lim_{x \to \infty} [\frac{x (\sqrt{4 x^{2} + 1} + 1)}{4 x^{2} + 1 - 1}] = \lim_{x \to \infty} [\frac{\sqrt{4 x^{2} + 1} + 1}{4 x}] Dividing the numerator and the denominator by x : \lim_{x \to \infty} [\frac{\frac{\sqrt{4 x^{2} + 1}}{x} + \frac{1}{x}}{4}] = \lim_{x \to \infty} [\frac{\sqrt{\frac{4 x^{2} + 1}{x^{2}}} + \frac{1}{x}}{4}] = \lim_{x \to \infty} [\frac{\sqrt{4 + \frac{1}{x^{2}}} + \frac{1}{x}}{4}] x \to \infty ∴ \frac{1}{x}, \frac{1}{x^{2}} \to 0 = \frac{\sqrt{4}}{4} = \frac{2}{4} = \frac{1}{2}$

Page No 29.38:

Question 8:

$\lim_{n \to \infty} \frac{n^{2}}{1 + 2 + 3 + . . . + n}$

Answer:

$\lim_{n \to \infty} [\frac{n^{2}}{1 + 2 + 3 . . . . . n}] It is of the form \frac{\infty}{\infty} . \Rightarrow \lim_{n \to \infty} [\frac{n^{2}}{n \frac{(n + 1)}{2}}] = \lim_{n \to \infty} [\frac{2 n}{n + 1}] Dividing the numerator and the denominator by n: \lim_{n \to \infty} \frac{2}{1 + \frac{1}{n}} = 2$

Page No 29.39:

Question 9:

$\lim_{x \to \infty} \frac{3 x^{- 1} + 4 x^{- 2}}{5 x^{- 1} + 6 x^{- 2}}$

Answer:

$\lim_{x \to \infty} [\frac{3 x^{- 1} + 4 x^{- 2}}{5 x^{- 1} + 6 x^{- 2}}] = \lim_{x \to \infty} [\frac{\frac{3}{x} + \frac{4}{x^{2}}}{\frac{5}{x} + \frac{6}{x^{2}}}] = \lim_{x \to \infty} [\frac{3 x + 4}{5 x + 6}] Dividing the numerator and the denominator by x: \lim_{x \to \infty} [\frac{3 + \frac{4}{x}}{5 + \frac{6}{x}}] = \frac{3}{5}$

Page No 29.39:

Question 10:

$\lim_{x \to \infty} \frac{\sqrt{x^{2} + a^{2}} - \sqrt{x^{2} + b^{2}}}{\sqrt{x^{2} + c^{2}} - \sqrt{x^{2} + d^{2}}}$

Answer:

$\lim_{x \to \infty} [\frac{\sqrt{x^{2} + a^{2}} - \sqrt{x^{2} + b^{2}}}{\sqrt{x^{2} + c^{2}} - \sqrt{x^{2} + d^{2}}}] Rationalising the numerator and the denominator: \lim_{x \to \infty} [\frac{(\sqrt{x^{2} + a^{2}} - \sqrt{x^{2} + b^{2}})}{(\sqrt{x^{2} + c^{2}} - \sqrt{x^{2} + d^{2}})} \times \frac{(\sqrt{x^{2} + c^{2}} + \sqrt{x^{2} + d^{2}})}{(\sqrt{x^{2} + c^{2}} + \sqrt{x^{2} + d^{2}})} \times \frac{(\sqrt{x^{2} + a^{2}} + \sqrt{x^{2} + b^{2}})}{(\sqrt{x^{2} + a^{2}} + \sqrt{x^{2} + b^{2}})}] = \lim_{x \to \infty} [\frac{(\sqrt{x^{2} + a^{2}} - \sqrt{x^{2} + b^{2}}) (\sqrt{x^{2} + a^{2}} + \sqrt{x^{2} + b^{2}}) (\sqrt{x^{2} + c^{2}} + \sqrt{x^{2} + d^{2}})}{(\sqrt{x^{2} + c^{2}} - \sqrt{x^{2} + d^{2}}) (\sqrt{x^{2} + c^{2}} + \sqrt{x^{2} + d^{2}}) (\sqrt{x^{2} + a^{2}} + \sqrt{x^{2} + b^{2}})}] = \lim_{x \to \infty} \frac{(x^{2} + a^{2}) - (x^{2} + b^{2})}{(x^{2} + c^{2}) - (x^{2} + d^{2})} \times (\frac{\sqrt{x^{2} + c^{2}} + \sqrt{x^{2} + d^{2}}}{\sqrt{x^{2} + a^{2}} + \sqrt{x^{2} + b^{2}}}) \underset{x \to \infty}{= \lim} (\frac{a^{2} - b^{2}}{c^{2} - d^{2}}) (\frac{\sqrt{x^{2} + c^{2}} + \sqrt{x^{2} + d^{2}}}{\sqrt{x^{2} + a^{2}} + \sqrt{x^{2} + b^{2}}}) Dividing the numerator and the denominator by x : \lim_{x \to \infty} (\frac{a^{2} - b^{2}}{c^{2} - d^{2}}) (\frac{\sqrt{1 + \frac{c^{2}}{x^{2}}} + \sqrt{1 + \frac{d^{2}}{x^{2}}}}{\sqrt{1 + \frac{1}{x^{2}}} + \sqrt{1 + \frac{b^{2}}{x^{2}}}}) As x \to \infty, \frac{1}{x}, \frac{1}{x^{2}} \to 0 = (\frac{a^{2} - b^{2}}{c^{2} - d^{2}}) (\frac{\sqrt{1} + \sqrt{1}}{\sqrt{1} + \sqrt{1}}) = \frac{a^{2} - b^{2}}{c^{2} - d^{2}}$

Page No 29.39:

Question 11:

$\lim_{n \to \infty} [\frac{(n + 2)! + (n + 1)!}{(n + 2)! - (n + 1)!}]$

Answer:

$\lim_{n \to \infty} [\frac{(n + 2)! + (n + 1)!}{(n + 2)! - (n + 1)!}] = \lim_{n \to \infty} [\frac{(n + 2) (n + 1)! + (n + 1)!}{(n + 2) (n + 1)! - (n + 1)!}] = \lim_{n \to \infty} [\frac{(n + 1)!}{(n + 1)!} \times \frac{(n + 2 + 1)}{(n + 2 - 1)}] = \lim_{n \to \infty} [\frac{n + 3}{n + 1}]$

Dividing the numerator and the denominator by n:

$\lim_{n \to \infty} [\frac{1 + \frac{3}{n}}{1 + \frac{1}{n}}] When n \to \infty, then \frac{1}{n} \to 0 . \Rightarrow \frac{1}{1} = 1$

Page No 29.39:

Question 12:

$\lim_{x \to \infty} [x \{\sqrt{x^{2} + 1} - \sqrt{x^{2} - 1}\}]$

Answer:

$\lim_{x \to \infty} [x \{\sqrt{x^{2} + 1} - \sqrt{x^{2} - 1}\}] Rationalising the numerator : \lim_{x \to \infty} [x \{\frac{(\sqrt{x^{2} + 1} - \sqrt{x^{2} - 1}) (\sqrt{x^{2} + 1} + \sqrt{x^{2} - 1})}{(\sqrt{x^{2} + 1} + \sqrt{x^{2} - 1})}\}] = \lim_{x \to \infty} [x \{\frac{(x^{2} + 1) - (x^{2} - 1)}{(\sqrt{x^{2} + 1} + \sqrt{x^{2} - 1})}\}] = \lim_{x \to \infty} [\frac{x \times 2}{(\sqrt{x^{2} + 1} + \sqrt{x^{2} - 1})}] Dividing the numerator and the denominator by x : \lim_{x \to \infty} [\frac{2}{\sqrt{\frac{x^{2} + 1}{x^{2}}} + \sqrt{\frac{x^{2} - 1}{x^{2}}}}] = \lim_{x \to \infty} [\frac{2}{\sqrt{\frac{x^{2} + 1}{x^{2}}} + \sqrt{\frac{x^{2} - 1}{x^{2}}}}] = \lim_{x \to \infty} [\frac{2}{\sqrt{1 + \frac{1}{x^{2}}} + \sqrt{1 - \frac{1}{x^{2}}}}] = \frac{2}{\sqrt{1} + \sqrt{1}} = 2$

Page No 29.39:

Question 13:

$\lim_{x \to \infty} [\{\sqrt{x + 1} - \sqrt{x}\} \sqrt{x + 2}]$

Answer:

$\lim_{x \to \infty} [\{\sqrt{x + 1} - \sqrt{x}\} \sqrt{x + 2}] Rationalising the numerator : \lim_{x \to \infty} [\frac{(\sqrt{x + 2}) \{\sqrt{x + 1} - \sqrt{x}\} \{\sqrt{x + 1} + \sqrt{x}\}}{(\sqrt{x + 1} + \sqrt{x})}] \Rightarrow \lim_{x \to \infty} [\frac{(\sqrt{x + 2}) (x + 1 - x)}{(\sqrt{x + 1} + \sqrt{x})}] Dividing the numerator and the denominator by \sqrt{x} : \lim_{x \to \infty} [\frac{\frac{\sqrt{x + 2}}{\sqrt{x}}}{\frac{\sqrt{x + 1} + \sqrt{x}}{\sqrt{x}}}] = \lim_{x \to \infty} [\frac{\sqrt{1 + \frac{2}{x}}}{\sqrt{1 + \frac{1}{x}} + 1}] = \frac{1}{\sqrt{1} + 1} = \frac{1}{2}$

Page No 29.39:

Question 14:

$\lim_{n \to \infty} [\frac{1^{2} + 2^{2} + . . . + n^{2}}{n^{3}}]$

Answer:

$= \lim_{n \to \infty} [\frac{1^{2} + 2^{2} + . . . + n^{2}}{n^{3}}] = \lim_{n \to \infty} [\frac{n (n + 1) (2 n + 1)}{6 n^{3}}] = \lim_{n \to \infty} [\frac{(n + 1) (2 n + 1)}{6 n^{2}}] = \lim_{n \to \infty} [(\frac{n + 1}{n}) (\frac{2 n + 1}{n}) \times \frac{1}{6}] = \lim_{n \to \infty} [(1 + \frac{1}{n}) (2 + \frac{1}{n}) \times \frac{1}{6}] = \frac{2}{6} = \frac{1}{3}$

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Question 15:

$\lim_{n \to \infty} [\frac{1 + 2 + 3 . . . . . . n - 1}{n^{2}}]$

Answer:

$\Rightarrow \lim_{n \to \infty} (\frac{1 + 2 + 3 + . . . n - 1}{n^{2}}) \Rightarrow \lim_{n \to \infty} [\frac{n (n - 1)}{2 n^{2}}] \Rightarrow \lim_{n \to \infty} [(1 - \frac{1}{n}) \times \frac{1}{2}] When n \to \infty, then \frac{1}{n} \to 0 . = \frac{1}{2}$

Page No 29.39:

Question 16:

$\lim_{n \to \infty} [\frac{1^{3} + 2^{3} + . . . . n^{3}}{n^{4}}]$

Answer:

$\lim_{n \to \infty} [\frac{1^{3} + 2^{3} + . . . + n^{3}}{n^{4}}] = \lim_{n \to \infty} [\frac{{[\frac{n (n + 1)}{2}]}^{2}}{n^{4}}] = \lim_{n \to \infty} [\frac{n^{2} {(n + 1)}^{2}}{4 n^{4}}] = \lim_{n \to \infty} [\frac{n^{2} {(n + 1)}^{2}}{4 n^{4}}] = \lim_{n \to \infty} [\frac{{(n + 1)}^{2}}{4 n^{2}}] = \lim_{n \to \infty} [{(\frac{n + 1}{n})}^{2} \times \frac{1}{4}] = \lim_{n \to \infty} [{(1 + \frac{1}{n})}^{2} \times \frac{1}{4}] When n \to \infty, then \frac{1}{n} \to 0 . \Rightarrow \frac{1}{4}$

Page No 29.39:

Question 17:

$\lim_{n \to \infty} [\frac{1^{3} + 2^{3} + . . . n^{3}}{{(n - 1)}^{4}}]$

Answer:

$\lim_{n \to \infty} [\frac{1^{3} + 2^{3} + . . . + n^{3}}{{(n - 1)}^{4}}] = \lim_{n \to \infty} [\frac{{[\frac{n (n + 1)}{2}]}^{2}}{{(n - 1)}^{4}}] = \lim_{n \to \infty} [\frac{n^{2} {(n + 1)}^{2}}{4 {(n - 1)}^{4}}]$

Dividing the numerator and the denominator by n⁴:

$\lim_{n \to \infty} [\frac{\frac{n^{2} {(n + 1)}^{2}}{n^{4}}}{4 \frac{{(n - 1)}^{4}}{n^{4}}}] = \lim_{n \to \infty} [\frac{\frac{{(n + 1)}^{2}}{n^{2}}}{4 {(\frac{n - 1}{n})}^{4}}] = \lim_{n \to \infty} [\frac{{(1 + \frac{1}{n})}^{2}}{4 {(1 - \frac{1}{n})}^{4}}] When n \to \infty, then \frac{1}{n} \to 0 . \frac{{(1 + 0)}^{2}}{4 {(1 - 0)}^{4}} = \frac{1}{4}$

Page No 29.39:

Question 18:

$\lim_{x \to \infty} [\sqrt{x} \{\sqrt{x + 1} - \sqrt{x}\}]$

Answer:

$\lim_{x \to \infty} [\sqrt{x} \{\sqrt{x + 1} - \sqrt{x}\}] = \lim_{x \to \infty} [\sqrt{x} \{(\sqrt{x + 1} - \sqrt{x}) \frac{\sqrt{x + 1} + \sqrt{x}}{(\sqrt{x + 1} + \sqrt{x})}\}] = \lim_{x \to \infty} [\frac{\sqrt{x} \{(x + 1) - x\}}{(\sqrt{x + 1} + \sqrt{x})}]$

Dividing the numerator and the denominator by $\sqrt{x}$ :

$\lim_{x \to \infty} [\frac{1}{\frac{\sqrt{x + 1} + \sqrt{x}}{\sqrt{x}}}] = \lim_{x \to \infty} [\frac{1}{\sqrt{1 + \frac{1}{x}} + 1}] As x \to \infty, \frac{1}{x} \to 0 = \frac{1}{\sqrt{1 + 0} + 1} = \frac{1}{2}$

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Question 19:

$\lim_{n \to \infty} [\frac{1}{3} + \frac{1}{3^{2}} + \frac{1}{3^{3}} + . . . + \frac{1}{3^{n}}]$

Answer:

$\lim_{n \to \infty} [\frac{1}{3} + \frac{1}{3^{2}} + \frac{1}{3^{3}} + . . . \frac{1}{3^{n}}] = \lim_{n \to \infty} [\frac{1}{3} (1 + \frac{1}{3} + \frac{1}{3^{2}} + . . . . . \frac{1}{3^{n - 1}})] = \lim_{n \to \infty} \frac{[\frac{1}{3} (1 - \frac{1}{3^{n}})]}{1 - \frac{1}{3}} [a + a r + a r^{2} + . . . . . . a r^{n - 1} = a (\frac{1 - r^{n}}{1 - r})] = \lim_{n \to \infty} [\frac{\frac{1}{3} (1 - \frac{1}{3^{n}})}{\frac{2}{3}}] = \lim_{n \to \infty} \frac{1}{2} [(1 - \frac{1}{3^{n}})] As n \to \infty, 3^{n} \to \infty, \frac{1}{3^{n}} \to 0 = \frac{1}{2} (1 - 0) = \frac{1}{2}$

Page No 29.39:

Question 20:

$\lim_{x \to \infty} [\frac{x^{4} + 7 x^{3} + 46 x + a}{x^{4} + 6}]$ , where a is a non-zero real number.

Answer:

$\lim_{x \to \infty} [\frac{x^{4} + 7 x^{3} + 46 x + a}{x^{4} + 6}]$

Dividing the numerator and the denominator by x⁴:

$\lim_{x \to \infty} [\frac{1 + \frac{7}{x} + \frac{46}{x^{2}} + \frac{9}{x^{4}}}{1 + \frac{6}{x^{4}}}] As x \to \infty, \frac{1}{x}, \frac{1}{x^{2}}, \frac{1}{x^{3}}, \frac{1}{x^{4}} \to 0 = \frac{1}{1}$

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Question 21:

$f (x) = \frac{a x^{2} + b}{x^{2} + 1}, \lim_{x \to 0} f (x) = 1$ and $\lim_{x \to \infty} f (x) = 1,$ then prove that f(−2) = f(2) = 1

Answer:

$f (x) = \frac{a x^{2} + b}{x^{2} + 1} \lim_{x \to 0} f (x) = 1 \Rightarrow \lim_{x \to 0} [\frac{a x^{2} + b}{x^{2} + 1}] = 1 \Rightarrow \frac{a \times 0 + b}{a + 1} = 1 \Rightarrow b = 1 Also, \lim_{x \to \infty} f (x) = 1 \lim_{x \to \infty} [\frac{a x^{2} + b}{x^{2} + 1}] = 1$

Dividing the numerator and the denominator by x²:

$\Rightarrow \lim_{x \to \infty} [\frac{a + \frac{b}{x^{2}}}{1 + \frac{1}{x^{2}}}] = 1 As x \to \infty, \frac{1}{x}, \frac{1}{x^{2}} \to 0 \Rightarrow \frac{a + 0}{1 + 0} = 1 \Rightarrow a = 1 ∴ a = 1, b = 1 \Rightarrow f (x) = \frac{x^{2} + 1}{x^{2} + 1} = 1 f (- 2) = 1 [Since f (x) is a constant function, its value does not depend on x .] f (2) = 1$

Page No 29.39:

Question 22:

Show that $\lim_{x \to \infty} (\sqrt{x^{2} + x + 1} - x) \neq \lim_{x \to \infty} (\sqrt{x^{2} + 1} - x)$

Answer:

$\lim_{x \to \infty} (\sqrt{x^{2} + x + 1} - x) \neq \lim_{x \to \infty} (\sqrt{x^{2} + 1} - x) LHS : \lim_{x \to \infty} ((\sqrt{x^{2} + x + 1} - x)) Rationalising the numerator : \lim_{x \to \infty} [\frac{(\sqrt{x^{2} + x + 1} - x) (\sqrt{x^{2} + x + 1} + x)}{(\sqrt{x^{2} + x + 1} + x)}] = \lim_{x \to \infty} [\frac{(x^{2} + x + 1) - x^{2}}{(\sqrt{x^{2} + x + 1} + x)}] = \lim_{x \to \infty} [\frac{x + 1}{(\sqrt{x^{2} + x + 1} + x)}] Dividing the numerator and the denominator by x : \lim_{x \to \infty} [\frac{1 + \frac{1}{x}}{\frac{\sqrt{x^{2} + x + 1}}{x} + 1}] = \lim_{x \to \infty} [\frac{1 + \frac{1}{x}}{\sqrt{\frac{x^{2} + x + 1}{x^{2}}} + 1}] = \lim_{x \to \infty} [\frac{1 + \frac{1}{x}}{\sqrt{1 + \frac{1}{x} + \frac{1}{x^{2}}} + 1}] When x \to \infty, then \frac{1}{x} \to 0 . \frac{1}{\sqrt{1} + 1} = \frac{1}{2} RHS : \lim_{x \to \infty} (\sqrt{x^{2} + 1} - x) [from \infty - \infty]$

Rationalising the numerator:

$\lim_{x \to \infty} [\frac{(\sqrt{x^{2} + 1} - x) (\sqrt{x^{2} + 1} + x)}{(\sqrt{x^{2} + 1} + x)}] = \lim_{x \to \infty} [\frac{x^{2} + 1 - x^{2}}{(\sqrt{x^{2} + 1} + x)}] = \frac{1}{\infty} = 0 ∴ \lim_{x \to \infty} [\sqrt{x^{2} + x + 1} - x] \neq \lim_{x \to \infty} (\sqrt{x^{2} + 1} - x)$

Page No 29.39:

Question 23:

$\lim_{x \to - \infty} (\sqrt{4 x^{2} - 7 x} + 2 x)$

Answer:

$\lim_{x \to - \infty} (\sqrt{4 x^{2} - 7 x} + 2 x)$

Let x = $-$ m
When n → – ∞, then m → ∞.

$\Rightarrow \lim_{m \to \infty} [\sqrt{4 m^{2} + 7 m} - 2 m] = \lim_{m \to \infty} [(\sqrt{4 m^{2} = 7 m} - 2 m) \times \frac{(\sqrt{4 m^{2} + 7 m} + 2 m)}{(\sqrt{4 m^{2} + 7 m} + 2 m)}] = \lim_{m \to \infty} [\frac{(4 m^{2} + 7 m) - {(2 m)}^{2}}{\sqrt{4 m^{2} + 7 m} + 2 m}] = \lim_{m \to \infty} [\frac{4 m^{2} + 7 m - 4 m^{2}}{\sqrt{4 m^{2} + 7 m} + 2 m}]$

Dividing the numerator and the denominator by m:

$\lim_{m \to \infty} [\frac{7}{\sqrt{\frac{4 m^{2} + 7 m}{m^{2}}} + \frac{2 m}{m}}] = \lim_{m \to \infty} [\frac{7}{\sqrt{\frac{4 m^{2}}{m^{2}} + \frac{7 m}{m^{2}}} + 2}] = \lim_{m \to \infty} [\frac{7}{\sqrt{4 + \frac{7}{m}} + 2}] As m \to \infty, \frac{1}{m} \to 0 = \frac{7}{\sqrt{4} + 2} = \frac{7}{4}$

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Question 24:

$\lim_{x \to - \infty} (\sqrt{x^{2} - 8 x} + x)$

Answer:

$\lim_{x \to - \infty} (\sqrt{x^{2} - 8 x} + x)$

Let x = –m

When x → –∞, then m → ∞.

$\lim_{m \to \infty} (\sqrt{m^{2} + 8 m} - m) = \lim_{m \to \infty} (\frac{(\sqrt{m^{2} + 8 m} - m) (\sqrt{m^{2} + 8 m} + m)}{(\sqrt{m^{2} + 8 m} + m)}) = \lim_{m \to \infty} [\frac{m^{2} + 8 m - m^{2}}{\sqrt{m^{2} + 8 m} + m}]$

Dividing the numerator and the denominator by m:

$\lim_{m \to \infty} [\frac{8}{\frac{\sqrt{m^{2} + 8 m} + 1}{m}}] = \lim_{m \to \infty} [\frac{8}{\sqrt{\frac{m^{2}}{m^{2}} + \frac{8 m}{m^{2}}} + 1}] = \lim_{m \to \infty} [\frac{8}{\sqrt{1 + \frac{8}{m}} + 1}] As m \to \infty, \frac{1}{m} \to 0 = \frac{8}{\sqrt{1 + 0} + 1} = 4$

Page No 29.39:

Question 25:

Evaluate: $\lim_{n \to \infty} \frac{1^{4} + 2^{4} + 3^{4} + . . . + n^{4}}{n^{5}} - \lim_{n \to \infty} \frac{1^{3} + 2^{3} + . . . + n^{3}}{n^{5}}$

Answer:

Consider the identity

${(k + 1)}^{5} - k^{5} = 5 k^{4} + 10 k^{3} + 10 k^{2} + 5 k + 1$ .....(1)

Putting k = 1, 2, 3,..., n in (1) and then adding the equations, we have

${(n + 1)}^{5} - 1 = 5 \sum_{k = 1}^{n} k^{4} + 10 \sum_{k = 1}^{n} k^{3} + 10 \sum_{k = 1}^{n} k^{2} + 5 \sum_{k = 1}^{n} k + \sum_{k = 1}^{n} 1 \Rightarrow n^{5} + 5 n^{4} + 10 n^{3} + 10 n^{2} + 5 n = 5 \sum_{k = 1}^{n} k^{4} + \frac{10 n^{2} {(n + 1)}^{2}}{4} + \frac{10 n (n + 1) (2 n + 1)}{6} + \frac{5 n (n + 1)}{2} + n \Rightarrow 5 \sum_{k = 1}^{n} k^{4} = n^{5} + 5 n^{4} + 10 n^{3} + 10 n^{2} + 4 n - \frac{5 n^{2} {(n + 1)}^{2}}{2} - \frac{5 n (n + 1) (2 n + 1)}{3} - \frac{5 n (n + 1)}{2} \Rightarrow 5 \sum_{k = 1}^{n} k^{4} = n^{5} + \frac{5 n^{4}}{2} + \frac{5 n^{3}}{3} - \frac{n}{6}$

This expression on further simplification gives

$\sum_{k = 1}^{n} k^{4} = \frac{n (n + 1) (2 n + 1) (3 n^{2} + 3 n - 1)}{30}$

$∴ \lim_{n \to \infty} \frac{1^{4} + 2^{4} + 3^{4} + . . . + n^{4}}{n^{5}} - \lim_{n \to \infty} \frac{1^{3} + 2^{3} + . . . + n^{3}}{n^{5}} = \lim_{n \to \infty} \frac{n (n + 1) (2 n + 1) (3 n^{2} + 3 n - 1)}{30 n^{5}} - \lim_{n \to \infty} \frac{n^{2} {(n + 1)}^{2}}{4 n^{5}} = \frac{1}{30} \lim_{n \to \infty} (1 + \frac{1}{n}) (2 + \frac{1}{n}) (3 + \frac{3}{n} - \frac{1}{n^{2}}) - \frac{1}{4} \lim_{n \to \infty} \frac{1}{n} {(1 + \frac{1}{n})}^{2} = \frac{1}{30} \times (1 + 0) \times (2 + 0) \times (3 + 0 - 0) - \frac{1}{4} \times 0 (\lim_{n \to \infty} \frac{1}{n} = \lim_{n \to \infty} \frac{1}{n^{2}} = . . . = 0)$
$= \frac{1}{30} \times 6 - 0 = \frac{1}{5}$

Page No 29.39:

Question 26:

Evaluate: $\lim_{n \to \infty} \frac{1.2 + 2.3 + 3.4 + . . . + n (n + 1)}{n^{3}}$

Answer:

$\lim_{n \to \infty} \frac{1.2 + 2.3 + 3.4 + . . . + n (n + 1)}{n^{3}} = \lim_{n \to \infty} \frac{\sum_{k = 1}^{n} k (k + 1)}{n^{3}} = \lim_{n \to \infty} \frac{\sum_{k = 1}^{n} k^{2} + \sum_{k = 1}^{n} k}{n^{3}} = \lim_{n \to \infty} \frac{\frac{n (n + 1) (2 n + 1)}{6} + \frac{n (n + 1)}{2}}{n^{3}}$
$= \lim_{n \to \infty} \frac{n (n + 1) (2 n + 1 + 3)}{6 n^{3}} = \lim_{n \to \infty} \frac{2 n (n + 1) (n + 2)}{6 n^{3}} = \frac{1}{3} \lim_{n \to \infty} (1 + \frac{1}{n}) (1 + \frac{2}{n}) = \frac{1}{3} \times (1 + 0) \times (1 + 0) (\lim_{n \to \infty} \frac{1}{n} = 0) = \frac{1}{3}$

Page No 29.49:

Question 1:

$\lim_{x \to 0} \frac{\sin 3 x}{5 x}$

Answer:

$\lim_{x \to 0} [\frac{\sin 3 x}{5 x}]$

= $\frac{1}{5} \lim_{x \to 0} [\frac{\sin 3 x}{3 x} \times 3]$ $[∵ \lim_{x \to 0} (\frac{\sin 3 x}{3 x}) = 1]$

= $\frac{1}{5} \times 1 \times 3$

= $\frac{3}{5}$

Page No 29.49:

Question 2:

$\lim_{x \to 0} \frac{\sin x^{0}}{x}$

Answer:

We know that $x ° = \frac{π}{180} x$ .

$∴ \lim_{x \to 0} \frac{\sin x^{0}}{x} = \lim_{x \to 0} \frac{\sin \frac{π}{180} x}{x} = \lim_{x \to 0} \frac{\sin (\frac{π}{180} x)}{(\frac{π}{180} x)} \times \frac{π}{180} = \frac{π}{180} \times 1 = \frac{π}{180}$

Page No 29.49:

Question 3:

$\lim_{x \to 0} \frac{x^{2}}{\sin x^{2}}$

Answer:

$\lim_{x \to 0} [\frac{x^{2}}{\sin x^{2}}]$

$When x \to 0, then x^{2} \to 0 .$

Let $θ = x^{2}$

⇒ $\lim_{θ \to 0} (\frac{θ}{\sin θ})$

= 1

Page No 29.49:

Question 4:

$\lim_{x \to 0} \frac{\sin x \cos x}{3 x}$

Answer:

$\lim_{x \to 0} (\frac{\sin x \cos x}{3 x})$

= $\frac{1}{3} \lim_{x \to 0} (\frac{\sin x}{x}) \times \cos x$

= $\frac{1}{3} \times 1 \times \cos 0$

= $\frac{1}{3} \times 1$

= $\frac{1}{3}$

Page No 29.50:

Question 5:

$\lim_{x \to 0} \frac{3 \sin x - 4 \sin^{3} x}{x}$

Answer:

$\lim_{x \to 0} [\frac{3 \sin x - 4 \sin^{3} x}{x}]$

= $\lim_{x \to 0} [\frac{\sin 3 x}{x}]$ $[∵ \sin^{3} A = 3 sinA - 4 \sin^{3} A]$

= $\lim_{x \to 0} [\frac{\sin 3 x}{3 x} \times 3]$

= 1 × 3

= 3

Page No 29.50:

Question 6:

$\lim_{x \to 0} \frac{\tan 8 x}{\sin 2 x}$

Answer:

$\lim_{x \to 0} [\frac{\tan 8 x}{\sin 2 x}]$

$\Rightarrow \lim_{x \to 0} [\frac{\tan 8 x}{8 x} \times \frac{8 x}{\frac{\sin 2 x}{2 x} \times 2 x}] [∵ \lim_{x \to 0} \frac{\tan 8 x}{8 x} = 1, \lim_{x \to 0} \frac{\sin 2 x}{2 x} = 1] \Rightarrow 4$

Page No 29.50:

Question 7:

$\lim_{x \to 0} \frac{\tan m x}{\tan n x}$

Answer:

$\lim_{x \to 0} [\frac{\tan m x}{\tan n x}]$

$\Rightarrow \lim_{x \to 0} (\frac{\tan m x}{m x} \times \frac{m x}{n x} \times \frac{n x}{\tan n x}) \Rightarrow \lim_{x \to 0} (\frac{\tan m x}{m x} \times \frac{m x}{\frac{\tan n x}{n x}} \times n x) \Rightarrow \frac{m}{n} [∵ \lim_{x \to 0} \frac{\tan x}{x} = 1]$

Page No 29.50:

Question 8:

$\lim_{x \to 0} \frac{\sin 5 x}{\tan 3 x}$

Answer:

$\lim_{x \to 0} [\frac{\sin 5 x}{\tan 3 x}]$

$\Rightarrow \lim_{x \to 0} [\lim_{x \to 0} \frac{\sin 5 x}{5 x} \times \frac{5 x}{\frac{\tan 3 x}{3 x} \times 3 x}] \Rightarrow \frac{5}{3} [∵ \lim_{x \to 0} \frac{\sin x}{x} = 1, \lim_{x \to 0} \frac{\tan x}{x} = 1]$

Page No 29.50:

Question 9:

$\lim_{x \to 0} \frac{\sin x^{n}}{x^{n}}$

Answer:

$\lim_{x \to 0} \frac{\sin x^{n}}{x^{n}}$

It is of the form $(\frac{0}{0})$ .

$Let y = x^{n} \Rightarrow \lim_{x \to 0} = \lim_{y \to 0} \Rightarrow \lim_{y \to 0} \frac{\sin y}{y} \Rightarrow 1 \{∵ \lim_{x \to 0} \frac{\sin x}{x} = 1\}$

Page No 29.50:

Question 10:

$\lim_{x \to 0} \frac{7 x \cos x - 3 \sin x}{4 x + \tan x}$

Answer:

$\lim_{x \to 0} [\frac{7 x \cos x - 3 \sin x}{4 x + \tan x}]$

It is of the form $(\frac{0}{0})$ .
Dividing the numerator and the denominator by x:

$\Rightarrow \lim_{x \to 0} \frac{7 \cos x - 3 (\frac{\sin x}{x})}{4 + (\frac{\tan x}{x})} \Rightarrow \frac{7 \lim_{x \to 0} (\cos x) - 3 \lim_{x \to 0} (\frac{\sin x}{x})}{4 + \lim_{x \to 0} (\frac{\tan x}{x})} \Rightarrow \frac{7.1 - 3.1}{4 + 1} \Rightarrow \frac{4}{5}$

Page No 29.50:

Question 11:

$\lim_{x \to 0} \frac{\cos a x - \cos b x}{\cos c x - \cos d x}$

Answer:

$\lim_{x \to 0} [\frac{\cos a x - \cos b x}{\cos c x - \cos d x}]$

It is of the form $(\frac{0}{0})$ .

$\Rightarrow \lim_{x \to 0} [\frac{- 2 \sin (\frac{a x + b x}{2}) \sin (\frac{a x - b x}{2})}{- 2 \sin (\frac{c x + d x}{a 2}) \sin (\frac{c x - d x}{2})}] = \lim_{x \to 0} [\frac{\frac{\sin (\frac{a x + b x}{2})}{(\frac{a x + b}{x})} \times (\frac{a x + b x}{2}) \times \frac{\sin (\frac{a x - b x}{2})}{(\frac{a x - b x}{2})} \times (\frac{a x - b x}{2})}{\frac{\sin (\frac{c x + d x}{2})}{(\frac{c x + d x}{2})} \times (\frac{c x + d x}{2}) \times \frac{\sin (\frac{c x - d x}{2})}{(\frac{c x - d x}{2})} \times (\frac{c x - d x}{2})}] = 1 \times \lim_{x \to 0} [\frac{(\frac{a x + b x}{2}) (\frac{a x - b x}{2})}{(\frac{c x + d x}{2}) (\frac{c x - d x}{2})}] = \lim_{x \to 0} \frac{x^{2} (a + b) (a - b)}{x^{2} (c + d) (c - d)} = \frac{a^{2} - b^{2}}{c^{2} - d^{2}}$

Page No 29.50:

Question 12:

$\lim_{x \to 0} \frac{\tan^{2} 3 x}{x^{2}}$

Answer:

$\lim_{x \to 0} [\frac{\tan^{2} 3 x}{x^{2}}]$

$= \lim_{x \to 0} [\frac{\tan 3 x}{3 x} \times \frac{\tan 3 x}{3 x}] \times 9 = 1 \times 1 \times 9 [∵ \lim_{x \to 0} \frac{\tan x}{x} = 1] = 9$

Page No 29.50:

Question 13:

$\lim_{x \to 0} \frac{1 - \cos m x}{x^{2}}$

Answer:

$\lim_{x \to 0} [\frac{1 - \cos (m x)}{x^{2}}]$

$= \lim_{x \to 0} [\frac{2 \sin^{2} (\frac{m x}{2})}{x^{2}}] [∵ 1 - \cos A = 2 \sin^{2} (\frac{A}{2})] = 2 \lim_{x \to 0} [\frac{\sin (\frac{m x}{2})}{\frac{m x}{2}} \times \frac{\sin (\frac{m x}{2})}{\frac{m x}{2}}] \times \frac{m}{2} \times \frac{m}{2} = 2 \times \frac{m}{2} \times \frac{m}{2} [∵ \lim_{x \to 0} \frac{\sin x}{x} = 1] = \frac{m^{2}}{2}$

Page No 29.50:

Question 14:

$\lim_{x \to 0} \frac{3 \sin 2 x + 2 x}{3 x + 2 \tan 3 x}$

Answer:

$\lim_{x \to 0} [\frac{3 \sin (2 x) + 2 x}{3 x + 2 \tan (3 x)}]$

$= \lim_{x \to 0} [\frac{\frac{3 \sin 2 x}{x} + 2}{3 + \frac{2 \tan 3 x}{x}}] = \frac{3 \lim_{x \to 0} (\frac{\sin 2 x}{2 x}) \times 2 + 2}{3 + 2 \lim_{x \to 0} (\frac{\tan 3 x}{x}) \times 3} = \frac{6 + 2}{3 + 6} = \frac{8}{9}$

Page No 29.50:

Question 15:

$\lim_{x \to 0} \frac{\cos 3 x - \cos 7 x}{x^{2}}$

Answer:

$\lim_{x \to 0} [\frac{\cos 3 x - \cos 7 x}{x^{2}}]$
$= \lim_{x \to 0} [\frac{- 2 \sin (\frac{3 x + 7 x}{2}) \sin \frac{(3 x - 7 x)}{2}}{x^{2}}] [\cos C - \cos D = - 2 \sin (\frac{C + D}{2}) \sin (\frac{C - D}{2})] = \lim_{x \to 0} [\frac{- 2 \sin 5 x \sin (- 2 x)}{x^{2}}] = \lim_{x \to 0} [\frac{2 \sin 5 x \sin 2 x}{x^{2}}] [∵ \sin (- θ) = - \sin θ] = 2 \lim_{x \to 0} [\frac{\sin 5 x}{5 x} \times \frac{\sin 2 x}{2 x}] \times 5 \times 2 = 2 \times 5 \times 2 = 20$

Page No 29.50:

Question 16:

$\lim_{θ \to 0} \frac{\sin 3 θ}{\tan 2 θ}$

Answer:

$\lim_{θ \to 0} [\frac{\sin 3 θ}{\tan 2 θ}]$

$= \lim_{θ \to 0} [\frac{\sin 3 θ}{3 θ} \times \frac{3 θ}{\frac{\tan 2 θ}{2 θ} \times 2 θ}] = \frac{3}{2}$

Page No 29.50:

Question 17:

$\lim_{x \to 0} \frac{\sin x^{2} (1 - \cos x^{2})}{x^{6}}$

Answer:

$\lim_{x \to 0} [\frac{\sin x^{2} (1 - \cos x^{2})}{x^{6}}]$
$= \lim_{x \to 0} [\frac{\sin x^{2} \times 2 \sin^{2} (\frac{x^{2}}{2})}{x^{6}}] [∵ 1 - \cos A = 2 \sin^{2} (\frac{A}{2})] = 2 \lim_{x \to 0} [\frac{\sin x^{2}}{x^{2}} \times \frac{\sin (\frac{x^{2}}{2})}{2 \times \frac{x^{2}}{2}} \times \frac{\sin (\frac{x^{2}}{2})}{2 \times \frac{x^{2}}{2}}] = \frac{2}{2 \times 2} = \frac{1}{2}$

Page No 29.50:

Question 18:

$\lim_{x \to 0} \frac{\sin^{2} 4 x^{2}}{x^{4}}$

Answer:

$\lim_{x \to 0} [\frac{\sin^{2} 4 x^{2}}{x^{4}}]$

$= \lim_{x \to 0} [\frac{\sin (4 x^{2})}{x^{2}} \times \frac{\sin (4 x^{2})}{x^{2}}] = \lim_{x \to 0} [\frac{\sin (4 x^{2})}{4 x^{2}} \times 4 \times \frac{\sin (4 x^{2})}{4 x^{2}} \times 4] = 4 \times 4 [∵ \lim_{x \to 0} \frac{\sin x}{x} = 1] = 16$

Page No 29.50:

Question 19:

$\lim_{x \to 0} \frac{x \cos x + 2 \sin x}{x^{2} + \tan x}$

Answer:

$\lim_{x \to 0} [\frac{x \cos x + 2 \sin x}{x^{2} + \tan x}]$

Dividing the numerator and the denominator by x, we get:

$\lim_{x \to 0} [\frac{\cos x + \frac{2 \sin x}{x}}{x + \frac{\tan x}{x}}] = \frac{\cos 0 + 2 \times 1}{0 + 1} [∵ \lim_{x \to 0} (\frac{\sin x}{x}) = 1, \lim_{x \to 0} (\frac{\tan x}{x}) = 1] = \frac{3}{1}$

Page No 29.50:

Question 20:

$\lim_{x \to 0} \frac{2 x - \sin x}{\tan x + x}$

Answer:

$\lim_{x \to 0} [\frac{2 x - \sin x}{\tan x + x}]$

Dividing the numerator and the denominator by x, we get:

$\lim_{x \to 0} [\frac{2 - \frac{\sin x}{x}}{\frac{\tan x}{x} + 1}] = \frac{2 - 1}{1 + 1} [∵ \lim_{x \to 0} \frac{\tan x}{x} = 1, \lim_{x \to 0} \frac{\sin x}{x} = 1] = \frac{1}{2}$

Page No 29.50:

Question 21:

$\lim_{x \to 0} \frac{5 x \cos x + 3 \sin x}{3 x^{2} + \tan x}$

Answer:

$\lim_{x \to 0} [\frac{5 x \cos x + 3 \sin x}{3 x^{2} + \tan x}]$

Dividing the numerator and the denominator by x:

$\lim_{x \to 0} [\frac{5 \cos x + 3 (\frac{\sin x}{x})}{3 x + (\frac{\tan x}{x})}] = [\frac{5 \cos 0 + 3}{3 \times 0 + 1}] [∵ \lim_{x \to 0} \frac{\sin x}{x} = 1, \lim_{x \to 0} \frac{\tan x}{x} = 1] = \frac{5 + 3}{0 + 1} = 8$

Page No 29.50:

Question 22:

$\lim_{x \to 0} \frac{\sin 3 x - \sin x}{\sin x}$

Answer:

$\lim_{x \to 0} [\frac{\sin 3 x - \sin x}{\sin x}]$

$= \lim_{x \to 0} [\frac{2 \cos (\frac{3 x + x}{2}) \sin (\frac{3 x - x}{2})}{\sin x}] [∵ \sin C - \sin D = 2 \cos (\frac{C + D}{2}) \sin (\frac{C - D}{2})] = \lim_{x \to 0} [\frac{2 \cos 2 x . \sin x}{\sin x}] = 2 \cos 0 = 2$

Page No 29.50:

Question 23:

$\lim_{x \to 0} \frac{\sin 5 x - \sin 3 x}{\sin x}$

Answer:

$\lim_{x \to 0} [\frac{\sin 5 x - \sin 3 x}{\sin x}]$

$= \lim_{x \to 0} [\frac{2 \cos (\frac{5 x + 3 x}{2}) \sin (\frac{5 x - 3 x}{2})}{\sin x}] [∵ sinC - sinD = 2 \cos (\frac{C + D}{2}) \sin (\frac{C - D}{2})] = \lim_{x \to 0} [\frac{2 \cos 4 x . \sin x}{\sin x}] = 2 \cos 0 = 2$

Page No 29.50:

Question 24:

$\lim_{x \to 0} \frac{\cos 3 x - \cos 5 x}{x^{2}}$

Answer:

$\lim_{x \to 0} [\frac{\cos 3 x - \cos 5 x}{x^{2}}]$

$= \lim_{x \to 0} [\frac{- 2 \sin (\frac{3 x + 5 x}{2}) \sin (\frac{3 x - 5 x}{2})}{x^{2}}] [∵ cosC - cosD = - 2 \sin (\frac{C + D}{2}) \sin (\frac{C - D}{2})] = 2 \lim_{x \to 0} [\frac{2 \sin 4 x \sin x}{x^{2}}] [∵ \sin (- θ) = - \sin θ] = 2 \lim_{x \to 0} [\frac{\sin 4 x}{4 x} \times 4 \times \frac{\sin x}{x}] = 2 \times 4 = 8 .$

Page No 29.50:

Question 25:

$\lim_{x \to 0} \frac{\tan 3 x - 2 x}{3 x - \sin^{2} x}$

Answer:

$\lim_{x \to 0} [\frac{\tan 3 x - 2 x}{3 x - \sin^{2} x}]$

Dividing the numerator and the denominator by x:

$\lim_{x \to 0} [\frac{\frac{\tan 3 x}{x} - 2}{3 - \frac{\sin^{2} x}{x}}] = \lim_{x \to 0} [\frac{\frac{\tan 3 x}{3 x} \times 3 - 2}{3 - (\frac{\sin x}{x}) \times \sin x}] = \frac{3 - 2}{3 - 1 \times 0} [∵ \lim_{x \to 0} \frac{\tan x}{x} = 1, \lim_{x \to 0} \frac{\sin x}{x} = 1] = \frac{1}{3}$

Page No 29.50:

Question 26:

$\lim_{x \to 0} \frac{\sin (2 + x) - \sin (2 - x)}{x}$

Answer:

$\lim_{x \to 0} [\frac{\sin (2 + x) - \sin (2 - x)}{x}]$

$\Rightarrow \lim_{x \to 0} [\frac{2 \cos (\frac{2 + x + 2 - x}{2}) \sin (\frac{2 + x - 2 + x}{2})}{x}] \Rightarrow \lim_{x \to 0} [\frac{2 \cos 2 \sin x}{x}] \Rightarrow 2 \cos 2 [∵ \lim_{x \to 0} \frac{\sin x}{x} = 1]$

Page No 29.50:

Question 27:

$\lim_{h \to 0} \frac{{(a + h)}^{2} \sin (a + h) - a^{2} \sin a}{h}$

Answer:

$\lim_{h \to 0} [\frac{{(a + h)}^{2} \sin (a + h) - a^{2} \sin a}{h}]$

$= \lim_{h \to 0} [\frac{a^{2} \sin (a + h) + h^{2} \sin (a + h) + 2 a h \sin (a + h) - a^{2} \sin a}{h}] = \lim_{h \to 0} [a^{2} \{\frac{\sin (a + h) - \sin (a)}{h}\} + \frac{h^{2} \sin (a + h)}{h} + \frac{2 a h \sin (a + h)}{h}] Div iding and multiplying the denominator by 2: \lim_{h \to 0} [a^{2} \{\frac{2 \cos (\frac{a + h + a}{2}) \sin (\frac{a + h - a}{2})}{2 \times \frac{h}{2}}\} + h \sin (a + h) + 2 a \sin (a + h)] = \lim_{h \to 0} [a^{2} \cos (\frac{a + h + a}{2}) + h \sin (a + h) + 2 a \sin (a + h)] = a^{2} \cos a + 0 + 2 a \sin a = a^{2} \cos a + 2 a \sin a$

Page No 29.50:

Question 28:

$\lim_{x \to 0} \frac{\tan x - \sin x}{\sin 3 x - 3 \sin x}$

Answer:

$\lim_{x \to 0} [\frac{\tan x - \sin x}{\sin 3 x - 3 \sin x}]$

$= \lim_{x \to 0} [\frac{\frac{si n x}{\cos x} - \sin x}{3 \sin x - 4 \sin^{3} x - 3 \sin x}] = \lim_{x \to 0} [\frac{\sin x (1 - \cos x)}{\cos x \times - 4 \sin^{3} x}] = \lim_{x \to 0} [\frac{2 \sin^{2} (\frac{x}{2})}{\cos x \times - 4 \sin^{2} x}] = \lim_{x \to 0} [\frac{2 \sin (\frac{x}{2}) \times \sin (\frac{x}{2})}{\cos x \times (- 4 \sin x) \times \sin x}] = \lim_{x \to 0} [\frac{2 \frac{\sin (\frac{x}{2})}{\frac{x}{2}} \times \frac{x}{2} \times \frac{\sin (\frac{x}{2})}{\frac{x}{2}} \times \frac{x}{2}}{\cos x \times - 4 \frac{\sin x}{x} \times x \times \frac{\sin x}{x} \times x}] = \lim_{x \to 0} [\frac{2 \times \frac{x}{2} \times \frac{x}{2}}{\cos x \times - 4 \times x \times x}] = \frac{- 1}{8 \cos 0} = - \frac{1}{8}$

Page No 29.50:

Question 29:

$\lim_{x \to 0} \frac{\sec 5 x - \sec 3 x}{\sec 3 x - \sec x}$

Answer:

$\lim_{x \to 0} [\frac{\sec 5 x - \sec 3 x}{\sec 3 x - \sec x}]$

$= \lim_{x \to 0} [\frac{\frac{1}{\cos 5 x} \frac{1}{\cos 3 x}}{\frac{1}{\cos 3 x} - \frac{1}{\cos x}}] = \lim_{x \to 0} [\frac{\cos 3 x - \cos 5 x}{\cos 5 x \cos 3 x \{\frac{\cos x - \cos 3 x}{\cos x \cos 3 x}\}}] = \lim_{x \to 0} [\frac{(\cos 3 x - \cos 5 x) \cos x}{\cos 5 x \{\cos x - \cos 3 x\}}] = \lim_{x \to 0} [\frac{- 2 \sin (\frac{3 x + 5 x}{2}) \sin (\frac{3 x - 5 x}{2}) \times \cos x}{\cos (5 x) [- 2 \sin (\frac{x + 3 x}{2}) \sin (\frac{x - 3 x}{2})]}] [∵ cosC - cosD = - 2 \sin (\frac{C + D}{2}) \sin (\frac{C - D}{2})] = \lim_{x \to 0} [\frac{\sin (4 x) \times \sin (- x) \times \cos x}{\cos (5 x) \times \sin (2 x) \times \sin (- x)}] = \lim_{x \to 0} [\frac{\sin (4 x)}{4 x} \times \frac{4 x}{\frac{\sin (2 x)}{2 x} \times 2 x} \times \frac{\cos x}{\cos 5 x}] = \frac{4}{2} \frac{\cos 0}{\cos 0} = 2$

Page No 29.50:

Question 30:

$\lim_{x \to 0} \frac{1 - \cos 2 x}{\cos 2 x - \cos 8 x}$

Answer:

$\lim_{x \to 0} [\frac{1 - \cos 2 x}{\cos 2 x - \cos 8 x}]$

$= \lim_{x \to 0} [\frac{2 \sin^{2} x}{2 \sin (\frac{2 x + 8 x}{2}) \sin (\frac{8 x - 2 x}{2})}] [∵ cosC - cosD = 2 \sin (\frac{C + D}{2}) \sin (\frac{D - C}{2})] = \lim_{x \to 0} [\frac{2 \sin x \times \sin x}{2 \sin 5 x \times \sin 3 x}] = \lim_{x \to 0} [\frac{\sin x \times \sin x}{\sin 5 x \times \sin 3 x}] [∵ \sin (- θ) = - \sin θ] = \lim_{x \to 0} [\frac{\frac{\sin x}{x} \times x \times \frac{\sin x}{x} \times x}{\frac{\sin 5 x}{5 x} \times 5 x \times \frac{\sin 3 x \times 3 x}{3 x}}] = \frac{1}{15}$

Page No 29.50:

Question 31:

$\lim_{x \to 0} \frac{1 - \cos 2 x + \tan^{2} x}{x \sin x}$

Answer:

$\lim_{x \to 0} [\frac{1 - \cos 2 x + \tan^{2} x}{x \sin x}] = \lim_{x \to 0} [\frac{2 \sin^{2} x + \tan^{2} x}{x \sin x}] [∵ 1 - \cos 2 A = 2 \sin^{2} A] Dividing numerator & denominator by x^{2} : \lim_{x \to 0} [\frac{\frac{2 \sin^{2} x}{x^{2}} + \frac{\tan^{2} x}{x^{2}}}{(\frac{\sin x}{x})}] = \frac{2 {(1)}^{2} + {(1)}^{2}}{1} [∵ \lim_{x \to 0} \frac{\sin^{2} x}{x^{2}} = 1, \lim_{x \to 0} \frac{\tan^{2} x}{x^{2}} = 1] = 3$

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Question 32:

$\lim_{x \to 0} \frac{\sin (a + x) + \sin (a - x) - 2 \sin a}{x \sin x}$

Answer:

$\lim_{x \to 0} [\frac{\sin (a + x) + \sin (a - x) - 2 \sin a}{x \sin x}] = \lim_{x \to 0} [\frac{2 \sin (\frac{a + x + a - x}{2}) \cos (\frac{a + x - a + x}{2}) - 2 \sin a}{x \sin x}] \{∵ \sin C + \sin D = 2 \sin (\frac{C + D}{2}) \cos (\frac{C - D}{2})\} = \lim_{x \to 0} [\frac{2 \sin a \cos x - 2 \sin a}{x \sin x}] = \lim_{x \to 0} [\frac{2 \sin a (\cos x - 1)}{x \sin x}] = \lim_{x \to 0} [\frac{2 \sin a (1 - 2 \sin^{2} \frac{x}{2} - 1)}{x \sin x}] = \lim_{x \to 0} [\frac{2 \sin a (- 2 \sin^{2} \frac{x}{2})}{x \sin x}] = - 4 \sin a \lim_{x \to 0} [\frac{1}{\frac{x \sin x}{x^{2}}} \times \frac{\sin (\frac{x}{2})}{\frac{x}{2}} \times \frac{\sin (\frac{x}{2})}{\frac{x}{2}} \times \frac{1}{4}] = - 4 \sin a \times \frac{1}{1} \times 1 \times 1 \times \frac{1}{4} = - \sin a$

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Question 33:

$\lim_{x \to 0} \frac{x^{2} - \tan 2 x}{\tan x}$

Answer:

$\lim_{x \to 0} [\frac{x^{2} - \tan 2 x}{\tan x}] Dividing the numerator and the denominator by x: \lim_{x \to 0} [\frac{x - \frac{\tan 2 x}{x}}{(\frac{\tan x}{x})}] = \lim_{x \to 0} [\frac{x - \frac{\tan 2 x}{2 x} \times 2}{(\frac{\tan x}{x})}] = \frac{0 - 2 (1)}{1} = - 2$

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Question 34:

$\lim_{x \to 0} \frac{\sqrt{2} - \sqrt{1 + \cos x}}{x^{2}}$

Answer:

$\lim_{x \to 0} [\frac{\sqrt{2} - \sqrt{1 + \cos x}}{x^{2}}] Rationalising the numerator : \lim_{x \to 0} [\frac{(\sqrt{2} - \sqrt{1 + \cos x}) (\sqrt{2} + \sqrt{1 + \cos x})}{x^{2} \times (\sqrt{2} + \sqrt{1 + \cos x})}] = \lim_{x \to 0} [\frac{2 - (1 + \cos x)}{x^{2} [\sqrt{2} + \sqrt{1 + \cos x}]}] = \lim_{x \to 0} [\frac{1 - \cos x}{x^{2} [\sqrt{2} + \sqrt{1 + \cos x}]}] = \lim_{x \to 0} [\frac{2 \sin^{2} (\frac{x}{2})}{x^{2} \{\sqrt{2} + \sqrt{1 + \cos x}\}}] = 2 \lim_{x \to 0} [\frac{\sin^{2} (\frac{x}{2})}{4 \times {(\frac{x}{2})}^{2}} \times \frac{1}{\sqrt{2} + \sqrt{1 + \cos x}}] = 2 \times \frac{1}{4} \times \frac{1}{\sqrt{2} + \sqrt{1 + \cos 0}} = \frac{2}{4 (\sqrt{2} + \sqrt{2})} = \frac{1}{4 \sqrt{2}}$

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Question 35:

$\lim_{x \to 0} \frac{x \tan x}{1 - \cos x}$

Answer:

$\lim_{x \to 0} [\frac{x \tan x}{1 - \cos x}] {Dividing the numerator and the denominator by x}^{2} : \lim_{x \to 0} [\frac{\frac{x \tan x}{x^{2}}}{\frac{1 - \cos x}{x^{2}}}] = \lim_{x \to 0} [\frac{\frac{\tan x}{x}}{\frac{2 \sin^{2} \frac{x}{2}}{\frac{x}{2} \times \frac{x}{2} \times 4}}] = \frac{4}{2} = 2$

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Question 36:

$\lim_{x \to 0} \frac{x^{2} + 1 - \cos x}{x \sin x}$

Answer:

$\lim_{x \to 0} [\frac{x^{2} + 1 - \cos x}{x \sin x}] = \lim_{x \to 0} [\frac{x^{2} + 2 \sin^{2} \frac{x}{2}}{x \sin x}] Dividing the numerator and the denominator by x^{2} : \lim_{x \to 0} [\frac{1 + \frac{2 \sin^{2} (\frac{x}{2})}{x^{2}}}{\frac{\sin x}{x}}] = \lim_{x \to 0} [\frac{1 + \frac{2 \sin^{2} (\frac{x}{2})}{{(\frac{x}{2})}^{2} \times 4}}{\frac{\sin x}{x}}] = \frac{1 + \frac{2}{4}}{1} = \frac{\frac{3}{2}}{1} = \frac{3}{2}$

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Question 37:

$\lim_{x \to 0} \frac{\sin 2 x (\cos 3 x - \cos x)}{x^{3}}$

Answer:

$\lim_{x \to 0} [\frac{\sin 2 x (\cos 3 x - \cos x)}{x^{3}}] = \lim_{x \to 0} [\frac{\sin 2 x \times - 2 \sin (\frac{3 x + x}{2}) \sin (\frac{3 x - x}{2})}{x^{3}}] = - 2 \lim_{x \to 0} [\frac{\sin 2 x}{2 x} \times \frac{\sin 2 x}{2 x} \times \frac{\sin x}{x}] \times 2 \times 2 = - 8$

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Question 38:

$\lim_{x \to 0} \frac{2 \sin x^{\circ} - \sin 2 x^{\circ}}{x^{3}}$

Answer:

$\lim_{x \to 0} [\frac{2 \sin x ° - \sin (2 x °)}{x^{3}}] = \lim_{x \to 0} [\frac{2 \sin (\frac{π x}{180}) - \sin (\frac{2 π x}{180})}{x^{3}}] = \lim_{x \to 0} [\frac{2 \sin (\frac{π x}{180}) - 2 \sin (\frac{π x}{180}) \times \cos (\frac{π x}{180})}{x^{3}}] = \lim_{x \to 0} [\frac{2 \sin (\frac{π x}{180}) [1 - \cos (\frac{π x}{180})]}{x^{3}}] = \lim_{x \to 0} [\frac{2 \sin (\frac{π x}{180}) \times 2 \sin^{2} (\frac{π x}{360})}{x \times x^{2}}] = \lim_{x \to 0} [\frac{4 \sin [\frac{π x}{180}] \times \sin^{2} [\frac{π x}{360}]}{\frac{π x}{180} \times \frac{π x}{360} \times \frac{π x}{360}} \times \frac{π}{180} \times {(\frac{π}{360})}^{2}] = 4 \lim_{x \to 0} [\frac{\sin (\frac{π x}{180})}{\frac{π x}{180}} \times \frac{\sin (\frac{π x}{360}) \times \sin (\frac{π x}{360})}{\frac{π x}{360} \times \frac{π x}{360}} \times \frac{π^{3}}{180 \times 360^{2}}] = 4 \times 1 \times 1 \times 1 \times \frac{π^{3}}{180 \times 360 \times 360} = {(\frac{π}{180})}^{3}$

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Question 39:

$\lim_{x \to 0} \frac{x^{3} \cot x}{1 - \cos x}$

Answer:

$\lim_{x \to 0} [\frac{x^{3} \cot x}{1 - \cos x}] = \lim_{x \to 0} [\frac{x^{3}}{\tan x (1 - \cos x)}] = \lim_{x \to 0} [\frac{x^{3}}{\tan x \times 2 \sin^{2} (\frac{x}{2})}] = \lim_{x \to 0} [\frac{x}{\tan x} \times \frac{x^{2}}{2 \sin^{2} (\frac{x}{2})}] = \lim_{x \to 0} [\frac{x}{\tan x} \times \frac{\frac{x^{2}}{4} \times 4}{2 \times \sin^{2} \frac{x}{2}}] = \lim_{x \to 0} [\frac{x}{\tan x} \times {(\frac{\frac{x}{2}}{\sin \frac{x}{2}})}^{2} \times \frac{4}{2}] = 1 \times 1 \times \frac{4}{2} = 2$

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Question 40:

$\lim_{x \to 0} \frac{x \tan x}{1 - \cos 2 x}$

Answer:

$\lim_{x \to 0} [\frac{x \tan x}{1 - \cos (2 x)}] = \lim_{x \to 0} [\frac{x \tan x}{2 \sin^{2} x}] Dividing the numerator and the denominator by x^{2} : \lim_{x \to 0} [\frac{\frac{x \tan x}{x^{2}}}{2 \frac{\sin^{2} x}{x^{2}}}] = \lim_{x \to 0} [\frac{\frac{\tan x}{x}}{2 {(\frac{\sin x}{x})}^{2}}] = \frac{1}{2}$

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Question 41:

$\lim_{x \to 0} \frac{\sin (3 + x) - \sin (3 - x)}{x}$

Answer:

$\lim_{x \to 0} [\frac{\sin (3 + x) - \sin (3 - x)}{x}] = \lim_{x \to 0} [\frac{2 \cos (\frac{3 + x + 3 - x}{2}) \sin (\frac{3 + x - 3 + x}{2})}{x}] \{∵ \sin C - \sin D = 2 \cos (\frac{C + D}{2}) \sin (\frac{C - D}{2})\} = \lim_{x \to 0} [\frac{2 \cos 3 \sin x}{x}] = 2 \cos 3$

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Question 42:

$\lim_{x \to 0} \frac{\cos 2 x - 1}{\cos x - 1}$

Answer:

$\lim_{x \to 0} [\frac{\cos (2 x) - 1}{\cos x - 1}] = \lim_{x \to 0} [\frac{1 - 2 \sin^{2} x - 1}{1 - 2 \sin^{2} (\frac{x}{2}) - 1}] = \lim_{x \to 0} [\frac{\sin^{2} x}{\sin^{2} (\frac{x}{2})}] = \lim_{x \to 0} [\frac{\sin^{2} x}{x^{2}} \times \frac{x^{2}}{\frac{\sin^{2} (\frac{x}{2})}{\frac{x^{2}}{4}} \times \frac{x^{2}}{4}}] = \lim_{x \to 0} [{(\frac{\sin x}{x})}^{2} \times \frac{1}{{(\frac{\sin (\frac{x}{2})}{\frac{x}{2}})}^{2}} \times 4] = 4$

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Question 43:

$\lim_{x \to 0} \frac{3 \sin^{2} x - 2 \sin x^{2}}{3 x^{2}}$

Answer:

$\lim_{x \to 0} [\frac{3 \sin^{2} x - 2 \sin x^{2}}{3 x^{2}}] = \lim_{x \to 0} [\frac{3 \sin^{2} x}{3 x^{2}} - \frac{2 \sin x^{2}}{3 x^{2}}] = 1 - \frac{2}{3} = \frac{1}{3}$

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Question 44:

$\lim_{x \to 0} \frac{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}{x}$

Answer:

$\lim_{x \to 0} [\frac{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}{x}] = \lim_{x \to 0} [\frac{(\sqrt{1 + \sin x} - \sqrt{1 - \sin x}) (\sqrt{1 + \sin x} + \sqrt{1 - \sin x})}{x (\sqrt{1 + \sin x} + \sqrt{1 - \sin x})}] = \lim_{x \to 0} [\frac{(1 + \sin x) - (1 - \sin x)}{x (\sqrt{1 + \sin x} + \sqrt{1 - \sin x})}] = \lim_{x \to 0} [\frac{2 \sin x}{x (\sqrt{1 + \sin x} + \sqrt{1 - \sin x})}] = \frac{2}{\sqrt{1 + 0} + \sqrt{1 - 0}} = \frac{2}{2} = 1$

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Question 45:

$\lim_{x \to 0} \frac{1 - \cos 4 x}{x^{2}}$

Answer:

$\lim_{x \to 0} [\frac{1 - \cos (4 x)}{x^{2}}] = \lim_{x \to 0} [\frac{2 \sin^{2} 2 x}{x^{2}}] = \lim_{x \to 0} 2 [\frac{\sin 2 x}{x} \times \frac{\sin 2 x}{x}] = \lim_{x \to 0} [2 \times \frac{\sin 2 x}{2 x} \times \frac{\sin 2 x}{2 x} \times 4] = 2 \times 1 \times 1 \times 4 = 8$

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Question 46:

$\lim_{x \to 0} \frac{x \cos x + \sin x}{x^{2} + \tan x}$

Answer:

$\lim_{x \to 0} [\frac{x \cos x + \sin x}{x^{2} + \tan x}] Dividing the numerator and the denominator by x: \lim_{x \to 0} [\frac{\cos x + \frac{\sin x}{x}}{x + \frac{\tan x}{x}}] = \frac{\cos 0 + 1}{0 + 1} = 2$

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Question 47:

$\lim_{x \to 0} \frac{1 - \cos 2 x}{3 \tan^{2} x}$

Answer:

$\lim_{x \to 0} [\frac{1 - \cos (2 x)}{3 \tan^{2} x}] = \lim_{x \to 0} [\frac{2 \sin^{2} x}{3 \tan^{2} x}] = \lim_{x \to 0} [\frac{2}{3} \times \frac{\sin^{2} x}{\sin^{2} x} \times \cos^{2} x] = \frac{2}{3} \cos^{2} 0 = \frac{2}{3}$

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Question 48:

$\lim_{θ \to 0} \frac{1 - \cos 4 θ}{1 - \cos 6 θ}$

Answer:

$\lim_{θ \to 0} [\frac{1 - \cos (4 θ)}{1 - \cos 6 θ}] = \lim_{θ \to 0} [\frac{2 \sin^{2} 2 θ}{2 \sin^{2} 3 θ}] \{∵ 1 - \cos A = 2 \sin^{2} (\frac{A}{2})\} = \lim_{θ \to 0} [\frac{\sin^{2} 2 θ}{{(2 θ)}^{2}} \times \frac{{(2 θ)}^{2}}{\frac{\sin^{2} 3 θ}{{(3 θ)}^{2}} \times {(3 θ)}^{2}}] = \lim_{θ \to 0} [{(\frac{\sin 2 θ}{2 θ})}^{2} \times {(\frac{3 θ}{\sin 3 θ})}^{2} \times \frac{4}{9}] = \frac{4}{9} [∵ \lim_{X \to 0} \frac{\sin x}{x} = 1]$

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Question 49:

$\lim_{x \to 0} \frac{a x + x \cos x}{b \sin x}$

Answer:

$\lim_{x \to 0} [\frac{a x + x \cos x}{b \sin x}] Dividing the numerator and the denominator by x: \lim_{x \to 0} [\frac{a + \cos x}{b (\frac{\sin x}{x})}] [∵ \lim_{x \to 0} \frac{\sin x}{x} = 1] = \frac{a + \cos 0}{b} = \frac{a + 1}{b}$

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Question 50:

$\lim_{θ \to 0} \frac{\sin 4 θ}{\tan 3 θ}$

Answer:

$\lim_{θ \to 0} [\frac{\sin 4 θ}{\tan 3 θ}] = \lim_{θ \to 0} [\frac{\sin 4 θ}{4 θ} \times \frac{4 θ}{\frac{\tan 3 θ}{3 θ} \times 3 θ}] = \frac{4}{3}$

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Question 51:

Evaluate the following limits:

$\lim_{x \to 0} \frac{2 \sin x - \sin 2 x}{x^{3}}$

Answer:

$\lim_{x \to 0} \frac{2 \sin x - \sin 2 x}{x^{3}} = \lim_{x \to 0} \frac{2 \sin x - 2 \sin x \cos x}{x^{3}} = \lim_{x \to 0} \frac{2 \sin x (1 - \cos x)}{x^{3}} = \lim_{x \to 0} \frac{2 \sin x \times 2 \sin^{2} \frac{x}{2}}{x^{3}}$
$= \lim_{x \to 0} \frac{\sin x \times \sin^{2} \frac{x}{2}}{x \times \frac{x^{2}}{4}} = \lim_{x \to 0} \frac{\sin x}{x} \times {(\lim_{x \to 0} \frac{\sin \frac{x}{2}}{\frac{x}{2}})}^{2} = 1 \times 1 (\lim_{θ \to 0} \frac{\sin θ}{θ} = 1) = 1$

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Question 52:

$\lim_{x \to 0} \frac{1 - \cos 5 x}{1 - \cos 6 x}$

Answer:

$\lim_{x \to 0} [\frac{1 - \cos (5 x)}{1 - \cos (6 x)}] = \lim_{x \to 0} [\frac{2 \sin^{2} (\frac{5 x}{2})}{2 \sin^{2} 3 x}] = \lim_{x \to 0} [\frac{\sin^{2} (\frac{5 x}{2})}{{(\frac{5 x}{2})}^{2}} \times \frac{{(\frac{5 x}{2})}^{2}}{\frac{\sin^{2} 3 x}{{(3 x)}^{2}} \times {(3 x)}^{2}}] = \lim_{x \to 0} [{\{\frac{\sin (\frac{5 x}{2})}{\frac{5 x}{2}}\}}^{2} \frac{\frac{25}{4} x^{2}}{{\{\frac{\sin (3 x)}{3 x}\}}^{2} \times 9 x^{2}}] = \frac{25}{9 \times 4} = \frac{25}{36}$

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Question 53:

$\lim_{x \to 0} \frac{cosec x - \cot x}{x}$

Answer:

$\lim_{x \to 0} [\frac{cosec x - \cot x}{x}] = \lim_{x \to 0} [\frac{\frac{1}{\sin x} - \frac{\cos x}{\sin x}}{x}] = \lim_{x \to 0} [\frac{1 - \cos x}{x \sin x}] = \lim_{x \to 0} [\frac{2 \sin^{2} (\frac{x}{2})}{x \sin x}] = \lim_{x \to 0} [\frac{2 \sin^{2} (\frac{x}{2})}{{(\frac{x}{2})}^{2}} \times \frac{\frac{x^{2}}{4}}{\frac{x \sin x}{x \times x} \times x^{2}}] = 2 \times \frac{1}{4} = \frac{1}{2}$

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Question 54:

$\lim_{x \to 0} \frac{\sin 3 x + 7 x}{4 x + \sin 2 x}$

Answer:

$\lim_{x \to 0} [\frac{\sin 3 x + 7 x}{4 x + \sin 2 x}] = \lim_{x \to 0} [\frac{\frac{\sin 3 x}{3 x} \times 3 x + 7 x}{4 x + \frac{\sin 2 x}{2 x} \times 2 x}] = \lim_{x \to 0} [\frac{(\frac{\sin 3 x}{3 x} \times 3 + 7) x}{(4 + \frac{\sin 2 x}{2 x} \times 2) x}] = \frac{3 + 7}{4 + 2} = \frac{10}{6} = \frac{5}{3}$

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Question 55:

$\lim_{x \to 0} \frac{5 x + 4 \sin 3 x}{4 \sin 2 x + 7 x}$

Answer:

$\lim_{x \to 0} [\frac{5 x + 4 \sin 3 x}{4 \sin 2 x + 7 x}] = \lim_{x \to 0} [\frac{5 x + 4 \times \frac{\sin 3 x}{3 x} \times 3 x}{\frac{4 \sin 2 x}{2 x} \times 2 x + 7 x}] = \lim_{x \to 0} [\frac{(5 + 4 \frac{\sin 3 x \times 3}{3 x}) x}{(4 \frac{\sin 2 x}{2 x} \times 2 + 7) x}] = \frac{5 + 4 \times 3}{4 \times 2 + 7} [∵ \lim_{x \to 0} \frac{\sin (3 x)}{3 x} = 1] = \frac{17}{15}$

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Question 56:

$\lim_{x \to 0} \frac{3 \sin x - \sin 3 x}{x^{3}}$

Answer:

$\lim_{x \to 0} [\frac{3 \sin x - \sin (3 x)}{x^{3}}] = \lim_{x \to 0} [\frac{3 \sin x - (3 \sin x - 4 \sin^{3} x)}{x^{3}}] = \lim_{x \to 0} [\frac{4 \sin^{3} x}{x^{3}}] = \lim_{x \to 0} [4 {(\frac{\sin x}{x})}^{3}] = 4 \times 1 = 4$

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Question 57:

$\lim_{x \to 0} \frac{\tan 2 x - \sin 2 x}{x^{3}}$

Answer:

$\lim_{x \to 0} [\frac{\tan (2 x) - \sin (2 x)}{x^{3}}] = \lim_{x \to 0} [\frac{\frac{\sin 2 x}{\cos 2 x} - \sin 2 x}{x^{3}}] = \lim_{x \to 0} [\frac{\sin 2 x \{1 - \cos 2 x\}}{co S 2 x \times x^{3}}] = \lim_{x \to 0} [\frac{\sin 2 x \times 2 \sin^{2} x}{\cos 2 x \times x^{3}}] = \lim_{x \to 0} [\frac{\sin 2 x}{2 x} \times \frac{2}{\cos 2 x} \times 2 {(\frac{\sin x}{x})}^{2}] = \frac{2 \times 2}{\cos 0} = 4$

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Question 58:

$\lim_{x \to 0} \frac{\sin a x + b x}{a x + \sin b x}$

Answer:

$\lim_{x \to 0} [\frac{\sin (a x) + b x}{a x + \sin (b x)}] = \lim_{x \to 0} [\frac{\frac{\sin a x}{a x} \times a x + b x}{a x + \frac{\sin b x}{b x} \times b x}] = \lim_{x \to 0} [\frac{(\frac{\sin a x}{a x} \times a + b) x}{(a + \frac{\sin b x}{b x} \times b) x}] = \frac{1 \times a + b}{a + b} = 1$

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Question 59:

$\lim_{x \to 0} (cosec x - \cot x)$

Answer:

$\lim_{x \to 0} [cosec x - \cot x] = \lim_{x \to 0} [\frac{1}{\sin x} - \frac{\cos x}{\sin x}] = \lim_{x \to 0} [\frac{1 - \cos x}{\sin x}] [∵ 1 - \cos A = 2 \sin^{2} (\frac{A}{2})] = \lim_{x \to 0} [\frac{2 \sin^{2} (\frac{x}{2})}{2 \sin \frac{x}{2} \cos \frac{x}{2}}] = \lim_{x \to 0} (\tan (\frac{x}{2})) = 0$

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Question 60:

Evaluate the following limits:

$\lim_{x \to 0} \frac{\sin (α + β) x + \sin (α - β) x + \sin 2 α x}{\cos^{2} β x - \cos^{2} α x}$

Answer:

$\lim_{x \to 0} \frac{\sin (α + β) x + \sin (α - β) x + \sin 2 α x}{\cos^{2} β x - \cos^{2} α x} = \lim_{x \to 0} \frac{2 \sin [\frac{(α + β) x + (α - β) x}{2}] \cos [\frac{(α + β) x - (α - β) x}{2}] + 2 \sin α x \cos α x}{(1 - \sin^{2} β x) - (1 - \sin^{2} α x)} = \lim_{x \to 0} \frac{2 \sin α x \cos β x + 2 \sin α x \cos α x}{\sin^{2} α x - \sin^{2} β x} = \lim_{x \to 0} \frac{2 \sin α x (\cos β x + \cos α x)}{\sin^{2} α x - \sin^{2} β x}$
$= \lim_{x \to 0} \frac{2 α x \times \frac{\sin α x}{α x} \times (\cos β x + \cos α x)}{α^{2} x^{2} \times \frac{\sin^{2} α x}{α^{2} x^{2}} - β^{2} x^{2} \times \frac{\sin^{2} β x}{β^{2} x^{2}}} = \frac{2 α \times \lim_{x \to 0} (\frac{\sin α x}{α x}) \times \lim_{x \to 0} (\cos β x + \cos α x)}{α^{2} \times {(\lim_{x \to 0} \frac{\sin α x}{α x})}^{2} - β^{2} \times {(\lim_{x \to 0} \frac{\sin β x}{β x})}^{2}} \times \lim_{x \to 0} \frac{x}{x^{2}} = \frac{2 α \times 1 \times (1 + 1)}{α^{2} \times 1 - β^{2} \times 1} \times \lim_{x \to 0} \frac{1}{x} = \frac{4 α}{α^{2} - β^{2}} \times \infty = \infty$

Disclaimer: There is some mistake or some misprinting in the question.

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Question 61:

Evaluate the following limits:

$\lim_{x \to 0} \frac{\cos a x - \cos b x}{\cos c x - 1}$

Answer:

$\lim_{x \to 0} \frac{\cos a x - \cos b x}{\cos c x - 1} = \lim_{x \to 0} \frac{- 2 \sin (\frac{a x + b x}{2}) \sin (\frac{a x - b x}{2})}{- 2 \sin^{2} \frac{c x}{2}} (1 - \cos 2 θ = 2 \sin^{2} θ) = \lim_{x \to 0} \frac{(\frac{a + b}{2}) x \times \frac{\sin (\frac{a + b}{2}) x}{(\frac{a + b}{2}) x} \times (\frac{a - b}{2}) x \times \frac{\sin (\frac{a - b}{2}) x}{(\frac{a - b}{2}) x}}{\frac{c^{2} x^{2}}{4} \times \frac{\sin^{2} \frac{c x}{2}}{\frac{c^{2} x^{2}}{4}}} = \frac{(a + b) (a - b)}{c^{2}} \times \frac{\lim_{x \to 0} \frac{\sin (\frac{a + b}{2}) x}{(\frac{a + b}{2}) x} \times \lim_{x \to 0} \frac{\sin (\frac{a - b}{2}) x}{(\frac{a - b}{2}) x}}{{(\lim_{x \to 0} \frac{\sin \frac{c x}{2}}{\frac{c x}{2}})}^{2}}$
$= \frac{a^{2} - b^{2}}{c^{2}} \times \frac{1 \times 1}{1} (\lim_{θ \to 0} \frac{\sin θ}{θ} = 1) = \frac{a^{2} - b^{2}}{c^{2}}$

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Question 62:

Evaluate the following limits:

$\lim_{h \to 0} \frac{{(a + h)}^{2} \sin (a + h) - a^{2} \sin a}{h}$

Answer:

$\lim_{h \to 0} \frac{{(a + h)}^{2} \sin (a + h) - a^{2} \sin a}{h} = \lim_{h \to 0} \frac{(a^{2} + 2 a h + h^{2}) \sin (a + h) - a^{2} \sin a}{h} = \lim_{h \to 0} \frac{(2 a h + h^{2}) \sin (a + h) + a^{2} \sin (a + h) - a^{2} \sin a}{h} = \lim_{h \to 0} \frac{(2 a h + h^{2}) \sin (a + h)}{h} + \lim_{h \to 0} \frac{a^{2} \sin (a + h) - a^{2} \sin a}{h} = \lim_{h \to 0} (2 a + h) \sin (a + h) + a^{2} \lim_{h \to 0} \frac{\sin (a + h) - \sin a}{h}$
$= (2 a + 0) \sin (a + 0) + a^{2} \lim_{h \to 0} \frac{2 \sin (\frac{h}{2}) \cos (\frac{2 a + h}{2})}{h} = 2 a \sin a + a^{2} \times \lim_{h \to 0} \frac{\sin (\frac{h}{2})}{\frac{h}{2}} \times \lim_{h \to 0} \cos (\frac{2 a + h}{2}) = 2 a \sin a + a^{2} \times 1 \times \cos (\frac{2 a + 0}{2}) (\lim_{θ \to 0} \frac{\sin θ}{θ} = 1) = 2 a \sin a + a^{2} \cos a$

Page No 29.51:

Question 63:

If $\lim_{x \to 0} k x cosec x = \lim_{x \to 0} x cosec k x,$ find k.

Answer:

$\lim_{x \to 0} k x . cosec x = \lim_{x \to 0} x cosec k x \Rightarrow \lim_{x \to 0} [\frac{k x}{\sin x}] = \lim_{x \to 0} [\frac{x}{\sin (k x)}] \Rightarrow k \lim_{x \to 0} [\frac{x}{\sin x}] = \lim_{x \to 0} [\frac{k x}{\sin (k x)}] \times \frac{1}{k} \Rightarrow k = \frac{1}{k} \Rightarrow k^{2} = 1 \Rightarrow k = \pm 1$

Page No 29.62:

Question 1:

$\lim_{x \to π} \frac{\sin x}{π - x}$

Answer:

$\lim_{x \to π} \frac{\sin x}{π - x} = \lim_{h \to 0} \frac{\sin (π - h)}{π - (π - h)} [∵ \lim_{x \to a} f (x) = \lim_{h \to 0} f (a - h)] = \lim_{h \to 0} \frac{\sin h}{h} [∵ \sin (π - 0) = \sin 0] \Rightarrow 1$

Page No 29.62:

Question 2:

$\lim_{x \to \frac{π}{2}} \frac{\sin 2 x}{\cos x}$

Answer:

$\lim_{x \to \frac{π}{2}} \frac{\sin 2 x}{\cos x} [\sin 2 x = 2 \sin x \cos x] = \lim_{x \to \frac{π}{2}} \frac{2 \sin x \cos x}{\cos x} = \lim_{x \to \frac{π}{2}} 2 \sin x \Rightarrow 2$

Page No 29.62:

Question 3:

$\lim_{x \to \frac{π}{2}} \frac{\cos^{2} x}{1 - \sin x}$

Answer:

$\lim_{x \to \frac{π}{2}} \frac{\cos^{2} x}{1 - \sin x} = \lim_{x \to \frac{π}{2}} \frac{1 - \sin^{2} x}{1 - \sin x} [\begin{matrix} ∵ \cos^{2} x = 1 - \sin^{2} x \\ a^{2} - b^{2} = (a - b) (a + b) \end{matrix}] = \lim_{x \to \frac{π}{2}} \frac{(1 - \sin x) (1 + \sin x)}{(1 - \sin x)} = \lim_{x \to \frac{π}{2}} (1 + \sin x) \Rightarrow 1 + 1 = 2$

Page No 29.62:

Question 4:

Evaluate the following limits:

$\lim_{x \to \frac{π}{3}} \frac{\sqrt{1 - \cos 6 x}}{\sqrt{2} (\frac{π}{3} - x)}$

Answer:

$\lim_{x \to \frac{π}{3}} \frac{\sqrt{1 - \cos 6 x}}{\sqrt{2} (\frac{π}{3} - x)} = \lim_{x \to \frac{π}{3}} \frac{\sqrt{2 \sin^{2} 3 x}}{\sqrt{2} (\frac{π}{3} - x)} (1 - \cos 2 θ = 2 \sin^{2} θ) = \lim_{x \to \frac{π}{3}} \frac{\sqrt{2} \sin 3 x}{\sqrt{2} (\frac{π}{3} - x)} = \lim_{x \to \frac{π}{3}} \frac{\sin 3 x}{(\frac{π}{3} - x)}$
$= \lim_{h \to 0} \frac{\sin 3 (\frac{π}{3} + h)}{\frac{π}{3} - (\frac{π}{3} + h)} (Put x = \frac{π}{3} + h) = \lim_{h \to 0} \frac{\sin (π + 3 h)}{- h} = \lim_{h \to 0} \frac{- \sin 3 h}{- h} [\sin (π + θ) = - \sin θ] = 3 \times \lim_{h \to 0} \frac{\sin 3 h}{3 h} = 3 \times 1 (\lim_{θ \to 0} \frac{\sin θ}{θ} = 1) = 3$

Page No 29.62:

Question 5:

$\lim_{x \to a} \frac{\cos x - \cos a}{x - a}$

Answer:

$\lim_{x \to a} \frac{\cos x - \cos a}{x - a} = \lim_{x \to a} \frac{- 2 \sin (\frac{x + a}{2}) \sin (\frac{x - a}{2})}{2 (\frac{x - a}{2})} [∵ \cos A - \cos B - 2 \sin (\frac{A - B}{2}) \sin (\frac{A + B}{2})] = \lim_{x \to a} - \sin (\frac{x + a}{2}) [∵ \lim_{θ \to a} \sin \frac{(θ - a)}{(θ - a)} = 1] = - \sin (\frac{2 a}{2}) \Rightarrow - \sin a$

Page No 29.62:

Question 6:

$\lim_{x \to \frac{π}{4}} \frac{1 - \tan x}{x - \frac{π}{4}}$

Answer:

$\lim_{x \to \frac{π}{4}} \frac{1 - \tan x}{x - \frac{π}{4}} = \lim_{h \to 0} \frac{1 - \tan (\frac{π}{4} + h)}{(\frac{π}{4} + h - \frac{π}{4})} = \lim_{h \to 0} \frac{1 - (\frac{1 + \tan h}{1 - \tan h})}{h} [∵ \tan (\frac{π}{4} + 0) = \frac{1 + \tan θ}{1 - \tan θ}] = \lim_{h \to 0} \frac{1 - \tan h - 1 - \tan h}{(1 - \tan h) h} = \lim_{h \to 0} \frac{- 2 \tan h}{h (1 - \tan h)} [∵ \lim_{h \to 0} \frac{\tan h}{h} = 1] \Rightarrow \frac{- 2}{1 - 0} = - 2$

Page No 29.62:

Question 7:

$\lim_{x \to \frac{π}{2}} \frac{1 - \sin x}{{(\frac{π}{2} - x)}^{2}}$

Answer:

$\lim_{x \to \frac{π}{2}} \frac{1 - \sin x}{{(\frac{π}{2} - x)}^{2}} = \lim_{h \to 0} \frac{1 - \sin (\frac{π}{2} - h)}{{(\frac{π}{2} - (\frac{π}{2} - h))}^{2}} = \lim_{h \to 0} \frac{1 - \cos h}{h^{2}} = \lim_{h \to 0} \frac{2 \sin^{2} \frac{h}{2}}{\frac{4 h^{2}}{4}} [∵ \lim_{h \to 0} \frac{\sin h}{h} = 1] = \frac{1}{2} \lim_{h \to 0} {(\frac{\sin \frac{h}{2}}{\frac{h}{2}})}^{2} \Rightarrow \frac{1}{2}$

Page No 29.62:

Question 8:

$\lim_{x \to \frac{π}{3}} \frac{\sqrt{3} - \tan x}{π - 3 x}$

Answer:

$\lim_{x \to \frac{π}{3}} \frac{\sqrt{3} - \tan x}{π - 3 x} = \lim_{h \to 0} \frac{\sqrt{3} - \tan (\frac{π}{3} - h)}{π - 3 (\frac{π}{3} - h)} = \lim_{h \to 0} \frac{\sqrt{3} - (\frac{\tan \frac{π}{3} - \tan h}{1 + \tan \frac{π}{3} \tan h})}{π - 3 (\frac{π}{3} - h)} = \lim_{h \to 0} \frac{\sqrt{3} - (\frac{\sqrt{3} - \tan h}{1 + \sqrt{3} \tan h})}{3 h} = \lim_{h \to 0} \frac{\sqrt{3} + 3 \tan h - \sqrt{3} + \tan h}{(1 + \sqrt{3} \tan h) 3 h} = \lim_{h \to 0} \frac{4 \tan h}{3 h (1 + \sqrt{3} \tan h)} = \frac{4}{3}$

Page No 29.62:

Question 9:

$\lim_{x \to \frac{π}{2}} \frac{\cot x}{\frac{π}{2} - x}$

Answer:

$\lim_{x \to \frac{π}{2}} \frac{\cot x}{\frac{π}{2} - x} = \lim_{h \to 0} \frac{\cot (\frac{π}{2} - h)}{\frac{π}{2} - (\frac{π}{2} - h)} \Rightarrow \lim_{x \to 0} \frac{\tan h}{h} = 1$

Page No 29.62:

Question 10:

$\lim_{x \to \frac{π}{4}} \frac{\sqrt{\cos x} - \sqrt{\sin x}}{x - \frac{π}{4}}$

Answer:

$\lim_{x \to \frac{π}{4}} \frac{\sqrt{\cos x} - \sqrt{\sin x}}{x - \frac{π}{4}} Rationalising the numerator : = \lim_{x \to \frac{π}{4}} (\frac{\sqrt{\cos x} - \sqrt{\sin x}}{x - \frac{π}{4}}) (\frac{\sqrt{\cos x} + \sqrt{\sin x}}{\sqrt{\cos x} + \sqrt{\sin x}}) = \lim_{x \to \frac{π}{4}} \frac{\cos x - \sin x}{(x - \frac{π}{4}) (\sqrt{\cos x} + \sqrt{\sin x})} = \lim_{h \to 0} \frac{\cos (\frac{π}{4} + h) - \sin (\frac{π}{4} + h)}{(\frac{π}{4} + h - \frac{π}{4}) (\sqrt{\cos (\frac{π}{4} + h)} + \sqrt{\sin (\frac{π}{4} + h)})} = \lim_{h \to 0} \frac{\cos \frac{π}{4} \cos h - \sin \frac{π}{4} \sin h - \sin \frac{π}{4} \cos h - \cos \frac{π}{4} \sin h}{h (\sqrt{\cos (\frac{π}{4} + h)} + \sqrt{\sin (\frac{π}{4} + h)})} = \lim_{h \to 0} \frac{- \sqrt{2} \sin h}{h (\sqrt{\cos (\frac{π}{4} + h)} + \sqrt{\sin (\frac{π}{4} + h)})} \Rightarrow \frac{- \sqrt{2} \times 1}{2 {(\frac{1}{2})}^{\frac{1}{4}}} = \frac{- 1}{2^{\frac{1}{4}}} = - 2^{- \frac{1}{4}}$

Page No 29.62:

Question 11:

$\lim_{x \to \frac{π}{2}} \frac{\sqrt{2 - \sin x} - 1}{{(\frac{π}{2} - x)}^{2}}$

Answer:

$\lim_{x \to \frac{π}{2}} \frac{\sqrt{2 - \sin x} - 1}{{(\frac{π}{2} - x)}^{2}} = \lim_{h \to 0} \frac{\sqrt{2 - \sin (\frac{π}{2} - h)} - 1}{{(\frac{π}{2} - (\frac{π}{2} - h))}^{2}} = \lim_{h \to 0} \frac{\sqrt{2 - \cos h} - 1}{h^{2}} Dividing the numerator and the denominator by \sqrt{2 - \cos h} + 1 : \lim_{h \to 0} \frac{(\sqrt{2 - \cos h} - 1) (\sqrt{2 - \cos h} + 1)}{(\sqrt{2 - \cos h} + 1) h^{2}} = \lim_{h \to 0} \frac{2 - \cos h - 1}{h^{2} (\sqrt{2 - \cos h} + 1)} = \lim_{h \to 0} \frac{1 - \cos h}{h^{2} [\sqrt{2 - \cos h} + 1]} = \lim_{h \to 0} \frac{2 \sin^{2} \frac{h}{2}}{\frac{4 h^{2}}{4} (\sqrt{2 - \cos h} + 1)} = \frac{1}{2} \lim_{h \to 0} {(\frac{\sin \frac{h}{2}}{\frac{h}{2}})}^{2} \times \lim_{h \to 0} (\frac{1}{\sqrt{2 - \cos h} + 1}) = \frac{1}{2 (\sqrt{2 - 1} + 1)} = \frac{1}{4}$

Page No 29.62:

Question 12:

$\lim_{x \to \frac{π}{4}} \frac{\sqrt{2} - \cos x - \sin x}{{(\frac{π}{4} - x)}^{2}}$

Answer:

$\lim_{x \to \frac{π}{4}} \frac{\sqrt{2} - \cos x - \sin x}{{(\frac{π}{4} - x)}^{2}} Dividing the numerator and the denominator by \sqrt{2} : \lim_{x \to \frac{π}{4}} \frac{1 - (\frac{1}{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x)}{\frac{1}{\sqrt{2}} {(\frac{π}{4} - x)}^{2}} = \lim_{x \to \frac{π}{4}} \frac{1 - (\cos \frac{π}{4} \cos x + \sin \frac{π}{4} \sin x)}{\frac{1}{\sqrt{2}} {(\frac{π}{4} - x)}^{2}} = \lim_{x \to \frac{π}{4}} \frac{\sqrt{2} (1 - \cos (\frac{π}{4} - x))}{{(\frac{π}{4} - x)}^{2}} = \lim_{x \to \frac{π}{4}} \frac{\sqrt{2} (2 \sin^{2} (\frac{\frac{π}{4} - x}{2}))}{{(\frac{π}{4} - x)}^{2}} [∵ 1 - \cos θ = 2 \sin \frac{2 θ}{2}] = 2 \sqrt{2} \lim_{x \to \frac{π}{4}} \frac{[\sin^{2} (\frac{\frac{π}{4} - x}{2})]}{4 {(\frac{\frac{π}{4} - x}{2})}^{2}} = \frac{2 \sqrt{2}}{4} = \frac{1}{\sqrt{2}}$

Page No 29.62:

Question 13:

$\lim_{x \to \frac{π}{8}} \frac{\cot 4 x - \cos 4 x}{{(π - 8 x)}^{3}}$

Answer:

$\lim_{x \to \frac{π}{8}} \frac{\cot 4 x - \cos 4 x}{{(π - 8 x)}^{3}} = \lim_{h \to 0} [\frac{\cot 4 (\frac{π}{8} - h) - \cos 4 (\frac{π}{8} - h)}{{(π - 8 (\frac{π}{8} - h))}^{3}}] = \lim_{h \to 0} [\frac{\tan 4 h - \sin 4 h}{{(8 h)}^{3}}] = \lim_{h \to 0} [\frac{\sin 4 h - \cos 4 h \sin 4 h}{512 (\cos 4 h) h^{3}}] = \lim_{h \to 0} [\frac{\sin 4 h (1 - \cos 4 h)}{(\cos 4 h) 512 h^{3}}] = \lim_{h \to 0} [\frac{\tan 4 h}{4 h} \times \frac{2 \sin^{2} 2 h}{32 \times 4 h^{2}}] = \frac{1}{16} \lim_{h \to 0} (\frac{\tan 4 h}{4 h}) \times \lim_{h \to 0} {(\frac{\sin 2 h}{2 h})}^{2} = \frac{1}{16} \times 1 \times 1 = \frac{1}{16}$

Page No 29.62:

Question 14:

$\lim_{x \to a} \frac{\cos x - \cos a}{\sqrt{x} - \sqrt{a}}$

Answer:

$\lim_{x \to a} \frac{\cos x - \cos a}{\sqrt{x} - \sqrt{a}} = \lim_{x \to a} \frac{- 2 \sin (\frac{x + a}{2}) \sin (\frac{x - a}{2})}{\sqrt{x} - \sqrt{a}} Dividing the numerator and the denominator by \sqrt{x} + \sqrt{a} : - 2 \lim_{x \to a} \frac{(\sin (\frac{x + a}{2}) \sin (\frac{x - a}{2}))}{2 (\frac{x - a}{2})} (\sqrt{x} + \sqrt{a}) = - 2 \lim_{x \to a} \sin (\frac{x + a}{2}) \times \frac{\sin (\frac{x - a}{2})}{2 (\frac{x - a}{2})} (\sqrt{x} + \sqrt{a}) \Rightarrow - 2 \sin (\frac{2 a}{2}) (\frac{\sqrt{a} + \sqrt{a}}{2}) = - 2 \sqrt{a} \sin a$

Page No 29.62:

Question 15:

$\lim_{x \to π} \frac{\sqrt{5 + \cos x} - 2}{{(π - x)}^{2}}$

Answer:

$\lim_{x \to π} \frac{\sqrt{5 + \cos x} - 2}{{(π - x)}^{2}} = \lim_{h \to 0} \frac{\sqrt{5 + \cos (π - h)} - 2}{{(π - (π - h))}^{2}} = \lim_{h \to 0} \frac{\sqrt{5 - \cos h} - 2}{h^{2}} Rationalising the numerator, we get : \lim_{h \to 0} \frac{(\sqrt{5 - \cos h} - 2) (\sqrt{5 - \cos h} + 2)}{h^{2} (\sqrt{5 - \cos h} + 2)} = \lim_{h \to 0} \frac{5 - \cos h - 4}{h^{2} (\sqrt{5 - \cos h} + 2)} = \lim_{h \to 0} \frac{1 - \cos h}{h^{2} [\sqrt{5 - \cos h} + 2]} = \lim_{h \to 0} \frac{2 \sin^{2} \frac{h}{2}}{4 (\frac{h^{2}}{4}) (\sqrt{5 - \cos h} + 2)} = \frac{1}{2 (\sqrt{5 - 1} + 2)} = \frac{1}{2 (4)} = \frac{1}{8}$

Page No 29.62:

Question 16:

$\lim_{x \to a} \frac{\cos \sqrt{x} - \cos \sqrt{a}}{x - a}$

Answer:

$\lim_{x \to a} \frac{\cos \sqrt{x} - \cos \sqrt{a}}{x - a} = \lim_{x \to a} \frac{- 2 \sin (\frac{\sqrt{x} + \sqrt{a}}{2}) \sin (\frac{\sqrt{x} - \sqrt{a}}{2})}{2 (\frac{\sqrt{x} - \sqrt{a}}{2}) (\sqrt{x} + \sqrt{a})} = \lim_{x \to a} \frac{- \sin (\frac{\sqrt{x} + \sqrt{a}}{2})}{\sqrt{x} + \sqrt{a}} \times \frac{\sin (\frac{\sqrt{x} - \sqrt{a}}{2})}{(\frac{\sqrt{x} - \sqrt{a}}{2})} = \frac{- 1 \sin (\sqrt{a})}{2 \sqrt{a}} \times 1 = \frac{- \sin \sqrt{a}}{2 \sqrt{a}}$

Page No 29.62:

Question 17:

$\lim_{x \to a} \frac{\sin \sqrt{x} - \sin \sqrt{a}}{x - a}$

Answer:

$\lim_{x \to a} \frac{\sin \sqrt{x} - \sin \sqrt{a}}{x - a} = \lim_{x \to a} \frac{2 \cos (\frac{\sqrt{x} + \sqrt{a}}{2}) \sin (\frac{\sqrt{x} - \sqrt{a}}{2})}{(\sqrt{x} + \sqrt{a}) (\sqrt{x} - \sqrt{a})} = \lim_{x \to a} \frac{2 \cos (\frac{\sqrt{x} + \sqrt{a}}{2})}{\sqrt{x} + \sqrt{a}} \times \frac{\sin (\frac{\sqrt{x} - \sqrt{a}}{2})}{2 (\frac{\sqrt{x} - \sqrt{a}}{2})} = \frac{1}{\sqrt{a} + \sqrt{a}} \cos (\frac{2 \sqrt{a}}{2}) \Rightarrow \frac{1}{2 \sqrt{a}} \cos \sqrt{a}$

Page No 29.62:

Question 18:

$\lim_{x \to 1} \frac{1 - x^{2}}{\sin 2 π x}$

Answer:

$\lim_{x \to 1} \frac{1 - x^{2}}{\sin 2 π x} = \lim_{h \to 0} \frac{1 - {(1 - h)}^{2}}{\sin 2 π (1 - h)} = \lim_{h \to 0} \frac{2 h - h^{2}}{- \sin 2 π h} = \lim_{h \to 0} \frac{- (2 - h)}{\frac{\sin 2 π h}{h}} = \lim_{h \to 0} \frac{(h - 2)}{\frac{2 π \sin 2 π h}{(2 π h)}} = \frac{0 - 2}{2 π \times 1} = \frac{- 1}{π}$

Page No 29.62:

Question 19:

$\lim_{x \to \frac{π}{4}} \frac{f (x) - f (\frac{π}{4})}{x - \frac{π}{4}},$ where f(x) = sin 2x

Answer:

$\lim_{x \to \frac{π}{4}} \frac{f (x) - f (\frac{π}{4})}{x - \frac{π}{4}} = \lim_{h \to 0} \frac{f (\frac{π}{4} + h) - f (\frac{π}{4})}{\frac{π}{4} + h - \frac{π}{4}} It is given that f (x) = \sin 2 x . \Rightarrow \lim_{h \to 0} \frac{\sin (\frac{π}{2} + 2 h) - \sin (\frac{π}{2})}{h} = \lim_{h \to 0} \frac{\cos 2 h - 1}{h} = \lim_{h \to 0} - 2 (\frac{\sin^{2} h}{h \times h}) (h) \Rightarrow \lim_{h \to 0} (- 2 h) = 0$

Page No 29.62:

Question 20:

$\lim_{x \to 1} \frac{1 + \cos π x}{{(1 - x)}^{2}}$

Answer:

$\lim_{x \to 1} \frac{1 + \cos π x}{{(1 - x)}^{2}} = \lim_{h \to 0} \frac{1 + \cos π (1 - h)}{{(1 - (1 - h))}^{2}} = \lim_{h \to 0} \frac{1 - \cos π h}{h^{2}} = \lim_{h \to 0} \frac{2 \sin^{2} \frac{π h}{2}}{\frac{4}{π^{2}} (\frac{π^{2} h^{2}}{4})} = \lim_{h \to 0} \frac{\sin^{2} \frac{π h}{2}}{\frac{2}{π^{2}} {(\frac{π h}{2})}^{2}} \Rightarrow \frac{π^{2}}{2}$

Page No 29.62:

Question 21:

$\lim_{x \to 1} \frac{1 - x^{2}}{\sin π x}$

Answer:

$\lim_{x \to 1} \frac{1 - x^{2}}{\sin π x} \lim_{h \to 0} \frac{1 - {(1 - h)}^{2}}{\sin π (1 - h)} \lim_{h \to 0} \frac{2 h - h^{2}}{\sin π h} = \lim_{h \to 0} \frac{h (2 - h)}{\sin π h} = \lim_{h \to 0} \frac{(2 - h)}{π \times \frac{\sin π h}{π h}} \Rightarrow \frac{2 - 0}{π} = \frac{2}{π}$

Page No 29.62:

Question 22:

$\lim_{x \to \frac{π}{4}} \frac{1 - \sin 2 x}{1 + \cos 4 x}$

Answer:

$\lim_{x \to \frac{π}{4}} \frac{1 - \sin 2 x}{1 + \cos 4 x} = \lim_{h \to 0} \frac{1 - \sin 2 (\frac{π}{4} - h)}{1 + \cos 4 (\frac{π}{4} - h)} = \lim_{h \to 0} \frac{1 - \cos 2 h}{1 - \cos 4 h} = \lim_{h \to 0} \frac{2 \sin^{2} h}{2 \sin^{2} 2 h} \Rightarrow \lim_{h \to 0} \frac{\frac{\sin^{2} h}{h^{2}}}{\frac{4 \sin^{2} 2 h}{4 h^{2}}} \Rightarrow \frac{1}{4}$

Page No 29.62:

Question 23:

$\lim_{x \to 1} \frac{1 - \frac{1}{x}}{\sin π (x - 1)}$

Answer:

$\lim_{x \to 1} [\frac{1 - \frac{1}{x}}{sinπ (x - 1)}] = \lim_{x \to 1} [\frac{x - 1}{x sinπ (x - 1)}]$

Let y = x – 1
If x → 1, then y → 0.

$= \lim_{y \to 0} [\frac{y}{(y + 1) sinπ y}] = \lim_{y \to 0} [\frac{1}{π (y + 1) \frac{\sin πy}{πy}}] = \frac{1}{π (0 + 1) \times 1} = \frac{1}{π}$

Page No 29.62:

Question 24:

$\lim_{n \to \infty} n \sin (\frac{π}{4 n}) \cos (\frac{π}{4 n})$

Answer:

$\lim_{n \to \infty} n \sin \frac{π}{4 n} \cos \frac{π}{4 n} = \lim_{n \to \infty} n \sin \frac{π}{4 n} \times \lim_{n \to \infty} \cos \frac{π}{4 n} = \lim_{n \to \infty} \frac{n \sin \frac{π}{4 n}}{\frac{π}{4 n}} \times \frac{π}{4 n} \times 1 = \frac{π}{4} \lim_{n \to \infty} \frac{\sin \frac{π}{4 n}}{\frac{π}{4 n}} Let y = \frac{π}{4 n} If n \to \infty, then y \to 0 . = \frac{π}{4} \lim_{y \to 0} \frac{\sin y}{y} = \frac{π}{4}$

Page No 29.62:

Question 25:

$\lim_{n \to \infty} 2^{n - 1} \sin (\frac{a}{2^{n}})$

Answer:

$\lim_{n \to \infty} 2^{n - 1} \sin (\frac{a}{2^{n}}) = \lim_{n \to \infty} \frac{2^{n}}{2} \times \frac{\sin (\frac{a}{2^{n}})}{(\frac{a}{2^{n}})} \times (\frac{a}{2^{n}}) Let y = \frac{a}{2^{n}} If n \to \infty, then y \to 0 . = \lim_{y \to 0} \frac{a}{2} \times (\frac{\sin y}{y}) \Rightarrow \frac{a}{2}$

Page No 29.62:

Question 26:

$\lim_{n \to \infty} \frac{\sin (\frac{a}{2^{n}})}{\sin (\frac{b}{2^{n}})}$

Answer:

$\lim_{n \to \infty} \frac{\sin (\frac{a}{2^{n}})}{\sin (\frac{b}{2^{n}})} = \lim_{n \to \infty} \frac{(\frac{a}{2^{n}}) \sin (\frac{a}{2^{n}})}{(\frac{a}{2^{n}}) \times (\frac{b}{2^{n}}) \times (\frac{\sin (\frac{b}{2^{n}})}{\frac{b}{2^{n}}})} Let : y = \frac{a}{2^{n}} z = \frac{b}{2^{n}} If n \to \infty, then y \to 0 and z \to 0 . = \frac{y}{z} \lim_{y \to 0} \frac{\sin y}{y} \times \frac{1}{\lim_{z \to 0} \frac{\sin z}{z}} = \frac{y}{z} \times 1 \times \frac{1}{1} = \frac{\frac{a}{2^{n}} \times 1}{\frac{b}{2^{n}} \times 1} = \frac{a}{b}$

Page No 29.62:

Question 27:

$\lim_{x \to - 1} \frac{x^{2} - x - 2}{(x^{2} + x) + \sin (x + 1)}$

Answer:

$\lim_{x \to - 1} \frac{x^{2} - x - 2}{x^{2} + x + \sin (x + 1)} = \lim_{x \to - 1} \frac{(x - 2) (x + 1)}{x (x + 1) + \sin (x + 1)} Let y = x + 1 If x \to - 1, then y \to 0 . = \lim_{y \to 0} \frac{(y - 3) y}{(y - 1) y + \sin y} Dividing the numerator and the denominator by y : = \lim_{y \to 0} \frac{(y - 3)}{(y - 1) + \frac{\sin y}{y}} = \frac{0 - 3}{(0 - 1) + 1} \Rightarrow \frac{- 3}{0} = \infty$

Page No 29.62:

Question 28:

$\lim_{x \to 2} \frac{x^{2} - x - 2}{x^{2} - 2 x + \sin (x - 2)}$

Answer:

$\lim_{x \to 2} \frac{x^{2} - x - 2}{x^{2} - 2 x + \sin (x - 2)} = \lim_{x \to 2} \frac{(x - 2) (x + 1)}{x (x - 2) + \sin (x - 2)} Let y = x - 2 x \to 2 ∴ y \to 0 = \lim_{y \to 0} \frac{y (y + 3)}{(y + 2) y + \sin y} Dividing the numerator and the denominator by y: = \lim_{y \to 0} \frac{(y + 3)}{(y + 2) + \frac{\sin y}{y}} = \frac{3}{2 + 1} = \frac{3}{3} = 1$

Page No 29.63:

Question 29:

$\lim_{x \to 1} (1 - x) \tan (\frac{π x}{2})$

Answer:

$\lim_{x \to 1} (1 - x) \tan \frac{π x}{2} = \lim_{h \to 0} \{1 - (1 - h)\} \tan \frac{π}{2} (1 - h) = \lim_{h \to 0} h \tan (\frac{π}{2} - \frac{π h}{2}) = \lim_{h \to 0} h \cot \frac{π h}{2} = \lim_{h \to 0} \frac{h}{\tan \frac{π h}{2}} = \lim_{h \to 0} \frac{1}{\frac{\tan \frac{π h}{2} \times \frac{π}{2}}{\frac{πh}{2}}} = \frac{1}{\frac{π}{2}} [∵ \lim_{h \to 0} \tan \frac{h}{h} = 1] = \frac{2}{π}$

Page No 29.63:

Question 30:

$\lim_{x \to \frac{π}{4}} \frac{1 - \tan x}{1 - \sqrt{2} \sin x}$

Answer:

$\lim_{x \to \frac{π}{4}} [\frac{1 - \tan x}{1 - \sqrt{2} \sin x}] It is of \frac{0}{0} form .$

Rationalising the denominator, we get:

$\lim_{x \to \frac{π}{4}} [\frac{(1 - \tan x) (1 + \sqrt{2} \sin x)}{(1 - \sqrt{2} \sin x) (1 + \sqrt{2} \sin x)}] = \lim_{x \to \frac{π}{4}} [\frac{(1 - \tan x) (1 + \sqrt{2} \sin x)}{1 - 2 \sin^{2} x}] = \lim_{x \to \frac{π}{4}} [\frac{(1 - \frac{\sin x}{\cos x}) (1 + \sqrt{2} \sin x)}{\cos 2 x}] = \lim_{x \to \frac{π}{4}} [\frac{(\cos x - \sin x) (1 + \sqrt{2} \sin x)}{\cos x \cos 2 x}] = \lim_{x \to \frac{π}{4}} [\frac{(\cos x - \sin x) (1 + \sqrt{2} \sin x)}{\cos x \cdot (\cos^{2} x - \sin^{2} x)}] = \lim_{x \to \frac{π}{4}} [\frac{(\cos x - \sin x) (1 + \sqrt{2} \sin x)}{\cos x [\cos x - \sin x] [\cos x + \sin x]}] = \frac{(1 + \sqrt{2} \times \frac{1}{\sqrt{2}})}{(\frac{1}{\sqrt{2}}) (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}})} = \frac{2}{\frac{1}{\sqrt{2}} \times \sqrt{2}} = 2$

Page No 29.63:

Question 31:

$\lim_{x \to π} \frac{\sqrt{2 + \cos x} - 1}{{(π - x)}^{2}}$

Answer:

$\lim_{x \to π} \frac{\sqrt{2 + \cos x} - 1}{{(π - x)}^{2}} = \lim_{h \to 0} \frac{\sqrt{2 + \cos (π - h)} - 1}{{(π - (π - h))}^{2}} = \lim_{h \to 0} (\frac{\sqrt{2 - \cos h} - 1}{h^{2}}) \times \frac{\sqrt{2 - \cos h} + 1}{\sqrt{2 - \cos h} + 1} = \lim_{h \to 0} \frac{2 - \cos h - 1}{h^{2} (\sqrt{2 - \cos h} + 1)} = \lim_{h \to 0} \frac{1 - \cos h}{h^{2} (\sqrt{2 - \cos h} + 1)} = \lim_{h \to 0} \frac{2 \sin^{2} \frac{h}{2}}{4 \times \frac{h^{2}}{4} \times [\sqrt{2 - \cos h} + 1]} = \frac{1 \times 1^{2}}{2 (\sqrt{2 - 1} + 1)} = \frac{1}{4}$

Page No 29.63:

Question 32:

$\lim_{x \to π} \frac{1 + \cos x}{\tan^{2} x}$

Answer:

$\lim_{x \to π} \frac{1 + \cos x}{\tan^{2} x} = \lim_{h \to 0} \frac{1 + \cos (π + h)}{\tan^{2} (π + h)} = \lim_{h \to 0} \frac{1 - \cos h}{\tan^{2} h} = \lim_{h \to 0} \frac{2 \sin^{2} \frac{h}{2}}{\tan^{2} h} = \lim_{h \to 0} \frac{2 \sin^{2} \frac{h}{2}}{4 (\frac{h^{2}}{4}) \times \frac{\tan^{2} h}{h^{2}}} = \frac{1}{2} \frac{\lim_{h \to 0} {(\frac{\sin \frac{h}{2}}{\frac{h}{2}})}^{2}}{{\lim_{h \to 0} (\frac{\tan h}{h})}^{2}} = \frac{1}{2} \times \frac{{(1)}^{2}}{{(1)}^{2}} = \frac{1}{2}$

Page No 29.63:

Question 33:

$\lim_{x \to \frac{π}{2}} (\frac{π}{2} - x) \tan x$

Answer:

$\lim_{x \to 1} \frac{1 - \frac{1}{x}}{\sin π (x - 1)} = \lim_{x \to 1} \frac{x - 1}{x \sin π (x - 1)} Let y = x - 1 x \to 1 ∴ y \to 0 = \lim_{y \to 0} \frac{y}{(y + 1) \sin π y} = \lim_{y \to 0} \frac{1}{π (y + 1) \times \frac{\sin π y}{π y}} = \frac{1}{π (0 + 1) \times 1} = \frac{1}{π}$

Page No 29.63:

Question 34:

$\lim_{x \to \frac{π}{6}} \frac{\cot^{2} x - 3}{cosec x - 2}$

Answer:

$\lim_{x \to \frac{π}{6}} [\frac{\cot^{2} x - 3}{cosec x - 2}] = \lim_{x \to \frac{π}{6}} [\frac{{cosec}^{2} x - 1 - 3}{cosec x - 2}] = \lim_{x \to \frac{π}{6}} [\frac{{cosec}^{2} x - 4}{cosec x - 2}] = \lim_{x \to \frac{π}{6}} [\frac{(cosec x - 2) (cosec x + 2)}{(cosec x - 2)}] = cosec \frac{π}{6} + 2 = 4$

Page No 29.63:

Question 35:

$\lim_{x \to \frac{π}{4}} \frac{\sqrt{2} - \cos x - \sin x}{{(4 x - π)}^{2}}$

Answer:

$\lim_{x \to \frac{π}{4}} \frac{\sqrt{2} - \cos x - \sin x}{{(4 x - π)}^{2}} = \lim_{h \to 0} \frac{\sqrt{2} - \{\cos (\frac{π}{4} + h) + \sin (\frac{π}{4} + h)\}}{{(4 (\frac{π}{4} + h) - π)}^{2}} = \lim_{h \to 0} \frac{\sqrt{2} - \{\cos \frac{π}{4} \cos h - \sin \frac{π}{4} \sin h + \sin \frac{π}{4} \cos h + \cos \frac{π}{4} \sin h\}}{{(4 h)}^{2}} = \lim_{h \to 0} \frac{\sqrt{2} - \sqrt{2} \cos h}{{(4 h)}^{2}} = \lim_{h \to 0} \frac{\sqrt{2} (1 - \cos h)}{16 h^{2}} = \lim_{h \to 0} \frac{2 \sqrt{2} \sin^{2} \frac{h}{2}}{64 \times \frac{h^{2}}{4}} = \frac{\sqrt{2}}{32} \times {(1)}^{2} = \frac{1}{16 \sqrt{2}}$

Page No 29.63:

Question 36:

$\lim_{x \to \frac{π}{2}} \frac{(\frac{π}{2} - x) \sin x - 2 \cos x}{(\frac{π}{2} - x) + \cot x}$

Answer:

$\lim_{x \to \frac{π}{2}} \frac{(\frac{π}{2} - x) \sin x - 2 \cos x}{(\frac{π}{2} - x) + \cot x} = \lim_{h \to 0} \frac{[\frac{π}{2} - (\frac{π}{2} - h)] \sin (\frac{π}{2} - h) - 2 \cos (\frac{π}{2} - h)}{[\frac{π}{2} - (\frac{π}{2} - h)] + \cot (\frac{π}{2} - h)} (Plugging x = \frac{π}{2} - h) = \lim_{h \to 0} \frac{h \cos h - 2 \sin h}{h + \tan h} Dividing the numerator and the denominator by h: \lim_{h \to 0} \frac{\cos h - 2 \frac{\sin h}{h}}{1 + \frac{\tan h}{h}} \Rightarrow \frac{1 - 2 (1)}{1 + 1} = \frac{- 1}{2}$

Page No 29.63:

Question 37:

$\lim_{x \to \frac{π}{4}} \frac{\cos x - \sin x}{(\frac{π}{4} - x) (\cos x + \sin x)}$

Answer:

$\lim_{x \to \frac{π}{4}} \frac{\cos x - \sin x}{(\frac{π}{4} - x) (\cos x + \sin x)} Dividing the numerator and the denominator by \sqrt{2} : \lim_{x \to \frac{π}{4}} \frac{\frac{1}{\sqrt{2}} \cos x - \frac{1}{\sqrt{2}} \sin x}{(\frac{π}{4} - x) \frac{(\cos x + \sin x)}{\sqrt{2}}} = \lim_{x \to \frac{π}{4}} \frac{\sqrt{2} (\sin \frac{π}{4} \cos x - \cos \frac{π}{4} \sin x)}{(\frac{π}{4} - x) (\cos x + \sin x)} = \lim_{x \to \frac{π}{4}} \frac{\sqrt{2} (\sin (\frac{π}{4} - x))}{(\frac{π}{4} - x) (\cos x + \sin x)} = \lim_{x \to \frac{π}{4}} \frac{\sqrt{2}}{\sin x + \cos x} \times \lim_{x \to \frac{π}{4}} \frac{\sin (\frac{π}{4} - x)}{(\frac{π}{4} - x)} \Rightarrow \frac{\sqrt{2}}{\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}} \times 1 = 1$

Page No 29.63:

Question 38:

Evaluate the following limits:

$\lim_{x \to π} \frac{1 - \sin \frac{x}{2}}{\cos \frac{x}{2} (\cos \frac{x}{4} - \sin \frac{x}{4})}$ [NCERT EXEMPLAR]

Answer:

$\lim_{x \to π} \frac{1 - \sin \frac{x}{2}}{\cos \frac{x}{2} (\cos \frac{x}{4} - \sin \frac{x}{4})}$

Put $x = π + h$

When $x \to π, h \to 0$

$∴ \lim_{x \to π} \frac{1 - \sin \frac{x}{2}}{\cos \frac{x}{2} (\cos \frac{x}{4} - \sin \frac{x}{4})} = \lim_{h \to 0} \frac{1 - \sin (\frac{π + h}{2})}{\cos (\frac{π + h}{2}) [\cos (\frac{π + h}{4}) - \sin (\frac{π + h}{4})]} = \lim_{h \to 0} \frac{1 - \sin (\frac{π}{2} + \frac{h}{2})}{\cos (\frac{π}{2} + \frac{h}{2}) [\cos (\frac{π}{4} + \frac{h}{4}) - \sin (\frac{π}{4} + \frac{h}{4})]} = \lim_{h \to 0} \frac{1 - \cos (\frac{h}{2})}{- \sin (\frac{h}{2}) [(\cos \frac{π}{4} \cos \frac{h}{4} - \sin \frac{π}{4} \sin \frac{h}{4}) - (\sin \frac{π}{4} \cos \frac{h}{4} + \cos \frac{π}{4} \sin \frac{h}{4})]}$
$= \lim_{h \to 0} \frac{2 \sin^{2} \frac{h}{4}}{- 2 \sin \frac{h}{4} \cos \frac{h}{4} (\frac{1}{\sqrt{2}} \cos \frac{h}{4} - \frac{1}{\sqrt{2}} \sin \frac{h}{4} - \frac{1}{\sqrt{2}} \cos \frac{h}{4} - \frac{1}{\sqrt{2}} \sin \frac{h}{4})} = \lim_{h \to 0} \frac{\sin \frac{h}{4}}{- \cos \frac{h}{4} \times (- \sqrt{2} \sin \frac{h}{4})} = \frac{1}{\sqrt{2}} \times \frac{1}{\lim_{h \to 0} \cos \frac{h}{4}} = \frac{1}{\sqrt{2}} \times 1 (\lim_{θ \to 0} \cos θ = 1) = \frac{1}{\sqrt{2}}$

Page No 29.65:

Question 1:

$\lim_{x \to π} \frac{1 + \cos x}{\tan^{2} x}$

Answer:

$\lim_{x \to π} (\frac{1 + \cos x}{\tan^{2} x}) [\frac{0}{0} form] = \lim_{x \to π} [\frac{(1 + \cos x)}{\sin^{2} x} \times \cos^{2} x] = \lim_{x \to π} [\frac{(1 + \cos x)}{1 - \cos^{2} x} \times \cos^{2} x] = \lim_{x \to π} [\frac{(1 + \cos x)}{(1 - \cos x) (1 + \cos x)} \times \cos^{2} x] = \lim_{x \to π} [\frac{\cos^{2} x}{(1 - \cos x)}] = \frac{\cos^{2} π}{1 - \cos π} = \frac{{(- 1)}^{2}}{1 - (- 1)} = \frac{1}{2}$

Page No 29.65:

Question 2:

$\lim_{x \to \frac{π}{4}} \frac{{cosec}^{2} x - 2}{\cot x - 1}$

Answer:

$\lim_{x \to \frac{π}{4}} [\frac{{cosec}^{2} x - 2}{\cot x - 1}] = \lim_{x \to \frac{π}{4}} [\frac{1 + \cot^{2} x - 2}{\cot x - 1}] = \lim_{x \to \frac{π}{4}} [\frac{\cot^{2} x - 1}{\cot x - 1}] = \lim_{x \to \frac{π}{4}} [\frac{(\cot x - 1) (\cot x + 1)}{(\cot x - 1)}] = \cot \frac{π}{4} + 1 = 2$

Page No 29.65:

Question 3:

$\lim_{x \to \frac{π}{6}} \frac{\cot^{2} x - 3}{cosec x - 2}$

Answer:

$\lim_{x \to \frac{π}{6}} [\frac{\cot^{2} x - 3}{cosec x - 2}] = \lim_{x \to \frac{π}{6}} [\frac{{cosec}^{2} x - 1 - 3}{cosec x - 2}] = \lim_{x \to \frac{π}{6}} [\frac{{cosec}^{2} x - 4}{cosec x - 2}] = \lim_{x \to \frac{π}{6}} [\frac{(cosec x - 2) (cosec x + 2)}{(cosec x - 2)}] = cosec \frac{π}{6} + 2 = 2 + 2 = 4$

Page No 29.65:

Question 4:

$\lim_{x \to \frac{π}{4}} \frac{2 - {cosec}^{2} x}{1 - \cot x}$

Answer:

$\lim_{x \to \frac{π}{4}} [\frac{2 - {cosec}^{2} x}{1 - \cot x}] = \lim_{x \to \frac{π}{4}} [\frac{2 - (1 + \cot^{2} x)}{1 - \cot x}] = \lim_{x \to \frac{π}{4}} [\frac{1 - \cot^{2} x}{1 - \cot x}] = \lim_{x \to \frac{π}{4}} [\frac{(1 - \cot x) (1 + \cot x)}{(1 - \cot x)}] = 1 + \cot (\frac{π}{4}) = 1 + 1 = 2$

Page No 29.65:

Question 5:

$\lim_{x \to π} \frac{\sqrt{2 + \cos x} - 1}{{(π - x)}^{2}}$

Answer:

$\lim_{x \to π} [\frac{\sqrt{2 + \cos x} - 1}{{(π - x)}^{2}}] Rationalising the numerator, we get : \lim_{x \to π} [\frac{(\sqrt{2 + \cos x} - 1) \times (\sqrt{2 + \cos x} + 1)}{{(π - x)}^{2} (\sqrt{2 + \cos x} + 1)}] = \lim_{x \to π} [\frac{2 + \cos x - 1}{{(π - x)}^{2} (\sqrt{2 + \cos x} + 1)}] = \lim_{x \to π} [\frac{1 + \cos x}{{(π - x)}^{2} [\sqrt{2 + \cos x} + 1]}]$

Let x = π $-$ h
when x → π, then h → 0

$= \lim_{h \to 0} [\frac{1 + \cos (π - h)}{{[π - (π - h)]}^{2} [\sqrt{2 + \cos (π - h)} + 1]}] = \lim_{h \to 0} [\frac{1 - \cos h}{h^{2} [\sqrt{2 - \cos h} + 1]}] \{∵ \cos (π - θ) = - \cos θ\} = \lim_{h \to 0} [\frac{2 \sin^{2} (\frac{h}{2})}{4 \times \frac{h^{2}}{4} [\sqrt{2 - \cos h} + 1]}] = \frac{1}{2} \lim_{h \to 0} [{(\frac{\sin \frac{h}{2}}{\frac{h}{2}})}^{2} \times \frac{1}{[\sqrt{2 - \cos h} + 1]}] = \frac{1}{2} \times 1 \times \frac{1}{(\sqrt{2 - \cos 0} + 1)} = \frac{1}{2} \times \frac{1}{(\sqrt{1} + 1)} = \frac{1}{2 \times 2} = \frac{1}{4}$

Page No 29.65:

Question 6:

$\lim_{x \to \frac{3 π}{2}} \frac{1 + {cosec}^{3} x}{\cot^{2} x}$

Answer:

$\lim_{x \to \frac{3 π}{3}} [\frac{1 + {cosec}^{3} x}{\cot^{2} x}] = \lim_{x \to \frac{3 π}{2}} [\frac{(1 + cosec x) (1^{2} + {cosec}^{2} x - cosec x)}{({cosec}^{2} x - 1)}] = \lim_{x \to \frac{3 π}{2}} [\frac{(1 + cosec x) (1 + {cosec}^{2} x - cosec x)}{(cosec x - 1) (cosec x + 1)}] = \frac{1 + {cosec}^{2} (\frac{3 π}{2}) - cosec (\frac{3 π}{2})}{cosec (\frac{3 π}{2}) - 1} = \frac{1 + 1 + 1}{- 1 - 1} = - \frac{3}{2}$

Page No 29.71:

Question 1:

$\lim_{x \to 0} \frac{5^{x} - 1}{\sqrt{4 + x} - 2}$

Answer:

$\lim_{x \to 0} [\frac{5^{x} - 1}{\sqrt{4 + x} - 2}]$

Rationalising the denominator, we get:

$= \lim_{x \to 0} [\frac{(5^{x} - 1) (\sqrt{4 + x} + 2)}{(\sqrt{4 + x} - 2) (\sqrt{4 + x} + 2)}] = \lim_{x \to 0} [\frac{(5^{x} - 1) (\sqrt{4 + x} + 2)}{4 + x - 4}] = \lim_{x \to 0} [(\frac{5^{x} - 1}{x}) (\sqrt{4 + x} + 2)] \{∵ \lim_{x \to 0} (\frac{a^{x} - 1}{x}) = \log a\} = \log 5 \times (\sqrt{4 + 0} + 2) = 4 \log 5$

Page No 29.71:

Question 2:

$\lim_{x \to 0} \frac{\log (1 + x)}{3^{x} - 1}$

Answer:

$\lim_{x \to 0} [\frac{\log (1 + x)}{3^{x} - 1}]$

Dividing the numerator and the denominator by x:

$\lim_{x \to 0} [\frac{\log (1 + x)}{x \cdot (\frac{3^{x} - 1}{x})}] [∵ \lim_{x \to 0} \frac{\log (1 + x)}{x} = 1 \lim_{x \to 0} (\frac{a^{x} - 1}{x}) = \log a] = \frac{1}{\log 3}$

Page No 29.71:

Question 3:

$\lim_{x \to 0} \frac{a^{x} + a^{- x} - 2}{x^{2}}$

Answer:

$\lim_{x \to 0} [\frac{a^{x} + a^{- x} - 2}{x^{2}}] = \lim_{x \to 0} [\frac{{(a^{\frac{x}{2}})}^{2} + {(a^{- \frac{x}{2}})}^{2} - 2 a^{\frac{x}{2}} \cdot a^{- \frac{x}{2}}}{x^{2}}] = \lim_{x \to 0} [\frac{{(a^{\frac{x}{2}} - a^{- \frac{x}{2}})}^{2}}{x^{2}}] = \lim_{x \to 0} [\frac{{(a^{\frac{x}{2}} - \frac{1}{a^{\frac{x}{2}}})}^{2}}{x^{2}}] = \lim_{x \to 0} [\frac{{(a^{x} - 1)}^{2}}{x^{2}} \times \frac{1}{{(a^{\frac{x}{2}})}^{2}}] = \lim_{x \to 0} [{(\frac{a^{x} - 1}{x})}^{2} \times \frac{1}{a^{x}}] = \frac{{(\log a)}^{2}}{a^{0}} = {(\log a)}^{2}$

Page No 29.71:

Question 4:

$\lim_{x \to 0} \frac{a^{m x} - 1}{b^{n x} - 1}, n \neq 0$

Answer:

$\lim_{x \to 0} (\frac{a^{m x} - 1}{b^{n x} - 1}) = \lim_{x \to 0} ((\frac{a^{m x} - 1}{m x}) \times \frac{m x}{(\frac{b^{n x} - 1}{n x}) \times n x}) = \frac{\log_{e} (a)}{\log_{e} (b)} \times \frac{m}{n} = \frac{m}{n} \frac{\log a}{\log b}$

Page No 29.71:

Question 5:

$\lim_{x \to 0} \frac{a^{x} + b^{x} - 2}{x}$

Answer:

$\lim_{x \to 0} [\frac{a^{x} + b^{x} - 2}{x}] = \lim_{x \to 0} [\frac{a^{x} - 1 + b^{x} - 1}{x}] = \lim_{x \to 0} [\frac{a^{x} - 1}{x}] + \lim_{x \to 0} [\frac{b^{x} - 1}{x}] = \log a + \log b = \log (a b)$

Page No 29.71:

Question 6:

$\lim_{x \to 0} \frac{9^{x} - 2 . 6^{x} + 4^{x}}{x^{2}}$

Answer:

$\lim_{x \to 0} [\frac{9^{x} - 2 . 6^{x} + 4^{x}}{x^{2}}] = \lim_{x \to 0} [\frac{{(3^{x})}^{2} - 2 \cdot 3^{x} \cdot 2^{x} + {(2^{x})}^{2}}{x^{2}}] = \lim_{x \to 0} [\frac{{(3^{x} - 2^{x})}^{2}}{x^{2}}] = \lim_{x \to 0} [{(\frac{3^{x} - 2^{x}}{2^{x}})}^{2} \times \frac{{(2^{x})}^{2}}{x^{2}}] = \lim_{x \to 0} {[\frac{{(\frac{3}{2})}^{x} - 1}{x}]}^{2} \times 2^{2 x} = {[\log (\frac{3}{2})]}^{2} \times 2^{0} = {[\log (\frac{3}{2})]}^{2}$

Page No 29.71:

Question 7:

$\lim_{x \to 0} \frac{8^{x} - 4^{x} - 2^{x} + 1}{x^{2}}$

Answer:

$\lim_{x \to 0} [\frac{8^{x} - 4^{x} - 2^{x} + 1}{x^{2}}] = \lim_{x \to 0} [\frac{{(2^{x})}^{3} - {(2^{x})}^{2} - 2^{x} + 1}{x^{2}}] = \lim_{x \to 0} [\frac{{(2^{x})}^{2} (2^{x} - 1) - 1 (2^{x} - 1)}{x^{2}}] = \lim_{x \to 0} [\frac{(2^{2 x} - 1) (2^{x} - 1)}{x^{2}}] = \lim_{x \to 0} [\frac{2 (2^{2 x} - 1)}{2 x} \times (\frac{2^{x} - 1}{x})] = 2 \log 2 \times \log 2 = \log {(2)}^{2} \times \log 2 = (\log 4) \times (\log 2)$

Page No 29.71:

Question 8:

$\lim_{x \to 0} \frac{a^{m x} - b^{n x}}{x}$

Answer:

$\lim_{x \to 0} [\frac{a^{m x} - b^{n x}}{x}] = \lim_{x \to 0} [\frac{a^{m x} - 1 - b^{n x} + 1}{x}] = \lim_{x \to 0} [\frac{(a^{m x} - 1) - (b^{n x} - 1)}{x}] = \lim_{x \to 0} [(\frac{a^{m x} - 1}{m x}) \times m - (\frac{b^{n x} - 1}{n x}) \times n] = m \log a - n \log b = \log {(a)}^{m} - \log {(b)}^{n} = \log (\frac{a^{m}}{b^{n}})$

Page No 29.71:

Question 9:

$\lim_{x \to 0} \frac{a^{x} + b^{x} + c^{x} - 3}{x}$

Answer:

$\lim_{x \to 0} [\frac{a^{n} + b^{n} + c^{n} - 3}{x}] = \lim_{x \to 0} [\frac{(a^{n} - 1)}{x} + \frac{(b^{n} - 1)}{x} + \frac{(c^{n} - 1)}{x}] = \log a + \log b + \log c = \log (a b c)$

Page No 29.71:

Question 10:

$\lim_{x \to 2} \frac{x - 2}{\log_{a} (x - 1)}$

Answer:

$\lim_{x \to 2} [\frac{x - 2}{\log_{a} (x - 1)}]$

Let x = 2 + h

x → 2
∴ h → 0

$= \lim_{h \to 0} [\frac{(2 + h) - 2}{\frac{\log \{(2 + h) - 1\}}{\log a}}] = \log a \lim_{h \to 0} [\frac{h}{\log (1 + h)}] = \log a \times 1$

Page No 29.71:

Question 11:

$\lim_{x \to 0} \frac{5^{x} + 3^{x} + 2^{x} - 3}{x}$

Answer:

$\lim_{x \to 0} [\frac{5^{x} + 3^{x} + 2^{x} - 3}{x}] = \lim_{x \to 0} [\frac{5^{x} - 1}{x} + \frac{3^{x} - 1}{x} + \frac{2^{x} - 1}{x}] = \log (5) + \log (3) + \log (2) = \log (5 \times 3 \times 2) = \log 30$

Page No 29.71:

Question 12:

$\lim_{x \to \infty} (a^{1 / x} - 1) x$

Answer:

$\lim_{x \to \infty} (a^{\frac{1}{x}} - 1) x Let y = \frac{1}{x} x \to \infty ∴ y \to 0 \Rightarrow \lim_{y \to 0} (\frac{a^{y} - 1}{y}) = \log a$

Page No 29.71:

Question 13:

$\lim_{x \to 0} \frac{a^{m x} - b^{n x}}{\sin k x}$

Answer:

$\lim_{x \to 0} [\frac{a^{m x} - b^{n x}}{\sin (k x)}] = \lim_{x \to 0} [\frac{(a^{m x} - 1) - (b^{n x} - 1)}{\sin k x}] Dividing the numerator and the denomiantor by x: = \lim_{x \to 0} [\frac{m (\frac{a^{m x} - 1}{m x}) - n (\frac{b^{n x} - 1}{n x})}{k \times \frac{\sin k x}{k x}}] = \frac{(m \log a - n \log b)}{k \times 1} = \frac{1}{k} [\log {(a)}^{m} - \log {(b)}^{n}] = \frac{1}{k} \log (\frac{a^{m}}{b^{n}})$

Page No 29.71:

Question 14:

$\lim_{x \to 0} \frac{a^{x} + b^{x} - c^{x} - d^{x}}{x}$

Answer:

$\lim_{x \to 0} [\frac{a^{n} + b^{n} - c^{n} - d^{n}}{x}] \lim_{x \to 0} [\frac{a^{n} + b^{n} - 2 - c^{n} - d^{n} + 2}{x}] = \lim_{x \to 0} [\frac{(a^{n} - 1) + (b^{n} - 1) - (c^{n} - 1) - (d^{n} - 1)}{x}] = \lim_{x \to 0} [(\frac{a^{n} - 1}{x}) + (\frac{b^{n} - 1}{x}) - (\frac{c^{n} - 1}{x}) - (\frac{d^{n} - 1}{x})] = \log a + \log b - \log c - \log d = (\log a + \log b) - (\log c + \log d) = \log (a b) - \log (c d) = \log (\frac{a b}{c d})$

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Question 15:

$\lim_{x \to 0} \frac{e^{x} - 1 + \sin x}{x}$

Answer:

$\lim_{x \to 0} [\frac{e^{x} - 1 + \sin x}{x}] = \lim_{x \to 0} [\frac{e^{x} - 1}{x} + \frac{\sin x}{x}] = 1 + 1 = 2$

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Question 16:

$\lim_{x \to 0} \frac{\sin 2 x}{e^{x} - 1}$

Answer:

$\lim_{x \to 0} [\frac{\sin (2 x)}{e^{x} - 1}]$

Dividing the numerator and the denominator by x:

$= \lim_{x \to 0} [\frac{\sin 2 x}{x \times \frac{e^{x} - 1}{x}}] = \lim_{x \to 0} [\frac{\sin 2 x}{2 x} \times \frac{2}{(\frac{e^{x} - 1}{x})}] = 1 \times \frac{2}{1} = 2$

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Question 17:

$\lim_{x \to 0} \frac{e^{\sin x} - 1}{x}$

Answer:

$\lim_{x \to 0} [\frac{e^{\sin x} - 1}{x}] = \lim_{x \to 0} [\frac{e^{\sin x} - 1}{\sin x} \times \frac{\sin x}{x}]$

x → 0
∴ sin x → 0

Let y=sin x

x → 0
∴ y → 0

$\Rightarrow \lim_{y \to 0} (\frac{e^{y} - 1}{y}) \times \lim_{x \to 0} (\frac{\sin x}{x}) = 1 \times 1$

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Question 18:

$\lim_{x \to 0} \frac{e^{2 x} - e^{x}}{\sin 2 x}$

Answer:

$\lim_{x \to 0} [\frac{e^{2 x} - e^{x}}{\sin (2 x)}] = \lim_{x \to 0} [\frac{e^{x} (e^{x} - 1)}{\sin 2 x}] = \lim_{x \to 0} [\frac{e^{x} (e^{x} - 1)}{x} \times \frac{2 x}{\sin 2 x} \times \frac{1}{2}] = e^{0} \times 1 \times \frac{1}{1} \times \frac{1}{2} = \frac{1}{2}$

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Question 19:

$\lim_{x \to a} \frac{\log x - \log a}{x - a}$

Answer:

$\lim_{x \to a} [\frac{\log x - \log a}{x - a}] = \lim_{x \to a} [\frac{\log (\frac{x}{a})}{a (\frac{x}{a} - 1)}] = \lim_{x \to a} [\frac{\log [1 + (\frac{x}{a} - 1)]}{a (\frac{x}{a} - 1)}] x \to a ∴ \frac{x}{a} \to 1 \Rightarrow \frac{x}{a} - 1 \to 0 Let y = \frac{x}{a} - 1 x \to a ∴ y \to 0 = \lim_{y \to 0} [\frac{\log (1 + y)}{a \times y}] = \frac{1}{a} \times 1 = \frac{1}{a}$

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Question 20:

$\lim_{x \to 0} \frac{\log (a + x) - \log (a - x)}{x}$

Answer:

$\lim_{x \to 0} [\frac{\log (a + x) - \log (a - x)}{x}] = \lim_{x \to 0} [\frac{\log (\frac{a + x}{a - x})}{x}] = \lim_{x \to 0} [\frac{\log (1 + \frac{a + x}{a - x} - 1)}{x}] = \lim_{x \to 0} [\frac{\log (1 + \frac{a + x - a + x}{a - x})}{x}] = \lim_{x \to 0} [\frac{\log (1 + \frac{2 x}{a - x})}{\frac{2 x}{a - x} \times (\frac{a - x}{2})}] x \to 0 ∴ \frac{2 x}{a - x} \to 0 Let y = \frac{2 x}{a - x} = \lim_{y \to 0} [\frac{\log (1 + y)}{y}] \times \lim_{x \to 0} (\frac{1}{\frac{a - x}{2}}) = 1 \times \frac{2}{a}$

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Question 21:

$\lim_{x \to 0} \frac{\log (2 + x) + \log 0.5}{x}$

Answer:

$\lim_{x \to 0} [\frac{\log (2 + x) + \log (0.5)}{x}] = \lim_{x \to 0} [\frac{\log ((2 + x) \times 0.5)}{x}] = \lim_{x \to 0} [\frac{\log (1 + \frac{x}{2})}{\frac{x}{2} \times 2}] = \frac{1}{2} \times 1 = \frac{1}{2}$

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Question 22:

$\lim_{x \to 0} \frac{\log (a + x) - \log a}{x}$

Answer:

$\lim_{x \to 0} [\frac{\log (a + x) - \log a}{x}] = \lim_{x \to 0} [\frac{\log (\frac{a + x}{a})}{x}] = \lim_{x \to 0} [\log (1 +)]$