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Question 1:

Show that $\underset{x\to 0}{\mathrm{lim}}\frac{x}{\left|x\right|}$ does not exist.

Left hand limit:

Right hand limit:

Left hand limit ≠ Right hand limit

Question 2:

Find k so that $\underset{x\to 2}{\mathrm{lim}}f\left(x\right)$ may exist, where $f\left(x\right)=\left\{\begin{array}{ll}2x+3,& x\le 2\\ x+k,& x>2\end{array}\right\.$

Now,  exists if the left hand limit is equal to the right hand limit.

⇒7 = 2 + k
k = 5

Question 3:

Show that does not exist.

Question 4:

Let f(x) be a function defined by
Show that does not exist.

Question 5:

Let Prove that does not exist.

Question 6:

Let . Prove that does not exist.

We have,

LHL of f(x) at x = 0

=

RHL of f(x) at x = 0

=

Clearly, $\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)\ne \underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)$

Hence, does not exist.

Question 7:

Find , where

We have,

LHL of f(x) at x = 3

=

RHL of f(x) at x = 3

=

Clearly, $\underset{x\to {3}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {3}^{+}}{\mathrm{lim}}f\left(x\right)=4$

If Find and .

(ii)

Question 9:

Find , if $f\left(x\right)=\left\{\begin{array}{ll}{x}^{2}-1,& x\le 1\\ -{x}^{2}-1,& x>1\end{array}\right\$

Question 10:

Evaluate , where $f\left(x\right)=\left\{\begin{array}{ll}\frac{\left|x\right|}{x},& x\ne 0\\ 0,& x=0\end{array}\right\$

Question 11:

Let a1, a2, ..., an be fixed real numbers such that
f(x) = (xa1) (xa2) ... (xan)
What is For a ≠ a1, a2, ..., an. Compute

Find

Question 13:

Evaluate the following one sided limits:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

(xi)

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

(xi)

Question 14:

Show that does not exist.

Find:

(i)

(ii)

(iii)

Question 16:

Prove that for all aR. Also, prove that

Show that

Question 18:

Find Is it equal to

Question 19:

Find

We know:
$\underset{x\to \frac{5}{2}}{\mathrm{lim}}\left[x\right]=\underset{x\to {\frac{5}{2}}^{-}}{\mathrm{lim}}\left[x\right]=\underset{x\to {\frac{5}{2}}^{+}}{\mathrm{lim}}\left[x\right]$

$\therefore \underset{x\to \frac{5}{2}}{\mathrm{lim}}\left[x\right]=2$

Question 20:

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

Question 21:

Show that does not exist.

= – (An oscillating number that oscillates between –1 and 1)

Question 22:

Let and if , find the value of k.

We have,

It is given that,

Hence, the value of k is 6.

Question 1:

$\underset{x\to 1}{\mathrm{lim}}\left(\frac{{x}^{2}+1}{x+1}\right)\phantom{\rule{0ex}{0ex}}=\frac{{1}^{2}+1}{1+1}\phantom{\rule{0ex}{0ex}}=1$

Question 2:

$\underset{x\to 0}{\mathrm{lim}}\left(\frac{2{x}^{2}+3x+4}{{x}^{2}+3x+2}\right)\phantom{\rule{0ex}{0ex}}=\frac{2×0+3×0+4}{0+3×0+2}\phantom{\rule{0ex}{0ex}}=\frac{4}{2}\phantom{\rule{0ex}{0ex}}=2$

Question 3:

$\underset{x\to 3}{\mathrm{lim}}\left[\frac{\sqrt{2x+3}}{x+3}\right]\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{2×3+3}}{3+3}\phantom{\rule{0ex}{0ex}}=\frac{3}{6}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}$

Question 4:

$\underset{x\to 1}{\mathrm{lim}}\left(\frac{\sqrt{x+8}}{\sqrt{x}}\right)\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{1+8}}{1}\phantom{\rule{0ex}{0ex}}=3$

Question 5:

$\underset{x\to a}{\mathrm{lim}}\left(\frac{\sqrt{x}+\sqrt{a}}{x+a}\right)\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{a}+\sqrt{a}}{a+a}\phantom{\rule{0ex}{0ex}}=\frac{2\sqrt{a}}{2a}\phantom{\rule{0ex}{0ex}}=\frac{1}{\sqrt{a}}$

Question 6:

$\underset{x\to 1}{\mathrm{lim}}\left[\frac{1+{\left(x-1\right)}^{2}}{1+{x}^{2}}\right]\phantom{\rule{0ex}{0ex}}=\frac{1+{\left(1-1\right)}^{2}}{1+{1}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}$

Question 7:

$\underset{x\to 0}{\mathrm{lim}}\left[\frac{{x}^{2/3}-9}{x-27}\right]\phantom{\rule{0ex}{0ex}}=\frac{0-9}{0-27}\phantom{\rule{0ex}{0ex}}=\frac{1}{3}$

Question 8:

$\underset{x\to 0}{\mathrm{lim}}\left(9\right)\phantom{\rule{0ex}{0ex}}=9$

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

Question 9:

$\underset{x\to 2}{\mathrm{lim}}\left(3-x\right)\phantom{\rule{0ex}{0ex}}=3-2\phantom{\rule{0ex}{0ex}}=1$

Question 10:

$\underset{x\to -1}{\mathrm{lim}}\left(4{x}^{2}+2\right)\phantom{\rule{0ex}{0ex}}=4{\left(-1\right)}^{2}+2\phantom{\rule{0ex}{0ex}}=4+2\phantom{\rule{0ex}{0ex}}=6$

Question 11:

$\underset{x\to -1}{\mathrm{lim}}\left(\frac{{x}^{3}-3x+1}{x-1}\right)\phantom{\rule{0ex}{0ex}}=\frac{{\left(-1\right)}^{3}-3\left(-1\right)+1}{-1-1}\phantom{\rule{0ex}{0ex}}=\frac{-1+3+1}{-2}\phantom{\rule{0ex}{0ex}}=\frac{-3}{2}$

Question 12:

$\underset{x\to 0}{\mathrm{lim}}\left(\frac{3x+1}{x+3}\right)\phantom{\rule{0ex}{0ex}}=\frac{3×0+1}{0+3}\phantom{\rule{0ex}{0ex}}=\frac{1}{3}$

Question 13:

$\underset{x\to 3}{\mathrm{lim}}\left(\frac{{x}^{2}-9}{x+2}\right)\phantom{\rule{0ex}{0ex}}=\frac{{3}^{2}-9}{3+2}\phantom{\rule{0ex}{0ex}}=\frac{9-9}{5}\phantom{\rule{0ex}{0ex}}=0$

Question 14:

$\underset{x\to 0}{\mathrm{lim}}\left(\frac{ax+b}{cx+d}\right)\phantom{\rule{0ex}{0ex}}=\frac{a×0+b}{c×0+d}\phantom{\rule{0ex}{0ex}}=\frac{b}{d}$

Question 14:

$\underset{x\to 2}{\mathrm{lim}}\left[\frac{x}{x-2}-\frac{4}{{x}^{2}-2x}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 2}{\mathrm{lim}}\left[\frac{x}{x-2}-\frac{4}{x\left(x-2\right)}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 2}{\mathrm{lim}}\left[\frac{{x}^{2}-4}{x\left(x-2\right)}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 2}{\mathrm{lim}}\left[\frac{\left(x-2\right)\left(x+2\right)}{x\left(x-2\right)}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 2}{\mathrm{lim}}\left[\frac{\left(x+2\right)}{x}\right]\phantom{\rule{0ex}{0ex}}=\frac{2+2}{2}\phantom{\rule{0ex}{0ex}}=2$

Question 15:

$\underset{x\to 1}{\mathrm{lim}}\left[\frac{\left({x}^{3}-1\right)-x\left({x}^{2}+x-2\right)}{\left({x}^{2}+x-2\right)\left({x}^{3}-1\right)}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\underset{x\to 1}{\mathrm{lim}}\left[\frac{\left({x}^{3}-1\right)-{x}^{3}-{x}^{2}+2x}{\left({x}^{2}+x-2\right)\left({x}^{3}-1\right)}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\underset{x\to 1}{\mathrm{lim}}\left[\frac{-{x}^{2}+2x-1}{\left({x}^{2}+x-2\right)\left(x-1\right)\left({x}^{2}+x+1\right)}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\underset{x\to 1}{\mathrm{lim}}\left[\frac{-\left({x}^{2}-2x+1\right)}{\left({x}^{2}+x-2\right)\left(x-1\right)\left({x}^{2}+x+1\right)}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\underset{x\to 1}{\mathrm{lim}}\left[\frac{-{\left(x-1\right)}^{2}}{\left({x}^{2}+x-2\right)\left(x-1\right)\left({x}^{2}+x+1\right)}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\underset{x\to 1}{\mathrm{lim}}\left[\frac{-\left(x-1\right)}{\left({x}^{2}+2x-x-2\right)\left({x}^{2}+x+1\right)}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\underset{x\to 1}{\mathrm{lim}}\left[\frac{-\left(x-1\right)}{\left\{x\left(x+2\right)-1\left(x+2\right)\right\}\left({x}^{2}+x+1\right)}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\underset{x\to 1}{\mathrm{lim}}\left[\frac{-\left(x-1\right)}{\left(x-1\right)\left(x+2\right)\left({x}^{2}+x+1\right)}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{-1}{\left(1+2\right)\left(1+1+1\right)}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{-1}{9}$

Question 16:

$\underset{x\to 3}{\mathrm{lim}}\left[\frac{1}{x-3}-\frac{2}{{x}^{2}-4x+3}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\underset{x\to 3}{\mathrm{lim}}\left[\frac{1}{x-3}-\frac{2}{{x}^{2}-3x-x+3}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\underset{x\to 3}{\mathrm{lim}}\left[\frac{1}{x-3}-\frac{2}{x\left(x-3\right)-1\left(x-3\right)}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\underset{x\to 3}{\mathrm{lim}}\left[\frac{1}{x-3}-\frac{2}{\left(x-1\right)\left(x-3\right)}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\underset{x\to 3}{\mathrm{lim}}\left[\frac{x-1-2}{\left(x-3\right)\left(x-1\right)}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\underset{x\to 3}{\mathrm{lim}}\left[\frac{1}{x-1}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{1}{3-1}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}$

Question 17:

$\underset{x\to 2}{\mathrm{lim}}\left[\frac{1}{x-2}-\frac{2}{{x}^{2}-2x}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 2}{\mathrm{lim}}\left[\frac{1}{x-2}-\frac{2}{x\left(x-2\right)}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 2}{\mathrm{lim}}\left[\frac{x-2}{x\left(x-2\right)}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 2}{\mathrm{lim}}\left[\frac{1}{x}\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}$

Question 21:

$\underset{x\to 2}{\mathrm{lim}}\left[\frac{1}{x-2}-\frac{4}{{x}^{3}-2{x}^{2}}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 2}{\mathrm{lim}}\left[\frac{1}{x-2}-\frac{4}{{x}^{2}\left(x-2\right)}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 2}{\mathrm{lim}}\left[\frac{{x}^{2}-4}{{x}^{2}\left(x-2\right)}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 2}{\mathrm{lim}}\left[\frac{\left(x-2\right)\left(x+2\right)}{{x}^{2}\left(x-2\right)}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 2}{\mathrm{lim}}\left[\frac{x+2}{{x}^{2}}\right]\phantom{\rule{0ex}{0ex}}=\frac{2+2}{{2}^{2}}\phantom{\rule{0ex}{0ex}}=1$

Question 22:

$\underset{x\to 3}{\mathrm{lim}}\left[\frac{1}{x-3}-\frac{3}{{x}^{2}-3x}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 3}{\mathrm{lim}}\left[\frac{1}{x-3}-\frac{3}{x\left(x-3\right)}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 3}{\mathrm{lim}}\left[\frac{x-3}{x\left(x-3\right)}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 3}{\mathrm{lim}}\left[\frac{1}{x}\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{3}$

Question 23:

$\underset{x\to 1}{\mathrm{lim}}\left[\frac{1}{x-1}-\frac{2}{{x}^{2}-1}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 1}{\mathrm{lim}}\left[\frac{1}{x-1}-\frac{2}{\left(x-1\right)\left(x+1\right)}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 1}{\mathrm{lim}}\left[\frac{x+1-2}{\left(x-1\right)\left(x+1\right)}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 1}{\mathrm{lim}}\left[\frac{\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 1}{\mathrm{lim}}\left[\frac{1}{x+1}\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{1+1}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}$

Question 24:

$\underset{x\to 3}{\mathrm{lim}}\left[\left({x}^{2}-9\right)\left\{\frac{1}{x+3}+\frac{1}{x-3}\right\}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 3}{\mathrm{lim}}\left[\left({x}^{2}-9\right)\left\{\frac{x-3+x+3}{\left(x+3\right)\left(x-3\right)}\right\}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 3}{\mathrm{lim}}\left[\left({x}^{2}-9\right)\left(\frac{2x}{{x}^{2}-9}\right)\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 3}{\mathrm{lim}}\left(2x\right)\phantom{\rule{0ex}{0ex}}=2×3\phantom{\rule{0ex}{0ex}}=6$

Question 25:

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

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

$⇒\underset{x\to 1}{\mathrm{lim}}\left[\frac{{x}^{4}-3{x}^{3}+2}{{x}^{3}-5{x}^{2}+3x+1}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 1}{\mathrm{lim}}\left[\frac{\left(x-1\right)\left({x}^{3}-2{x}^{2}-2x-2\right)}{\left(x-1\right)\left({x}^{2}-4x-1\right)}\right]\phantom{\rule{0ex}{0ex}}=\frac{\left(1{\right)}^{3}-2{\left(1\right)}^{2}-2\left(1\right)-2}{{\left(1\right)}^{2}-4×1-1}\phantom{\rule{0ex}{0ex}}=\frac{1-2-2-2}{1-4-1}\phantom{\rule{0ex}{0ex}}=\frac{-5}{-4}\phantom{\rule{0ex}{0ex}}=\frac{5}{4}$

Question 26:

$\underset{x\to 2}{\mathrm{lim}}\left[\frac{{x}^{3}+3{x}^{2}-9x-2}{{x}^{3}-x-6}\right]$

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

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

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

$⇒\underset{x\to 2}{\mathrm{lim}}\left[\frac{{x}^{3}+3{x}^{2}-9x-2}{{x}^{3}-x-6}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 2}{\mathrm{lim}}\left[\frac{\left(x-2\right)\left({x}^{2}+5x+1\right)}{\left(x-2\right)\left({x}^{2}+2x+3\right)}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 2}{\mathrm{lim}}\left[\frac{{x}^{2}+5x+1}{{x}^{2}+2x+3}\right]\phantom{\rule{0ex}{0ex}}=\frac{\left(2{\right)}^{2}+5×2+1}{{\left(2\right)}^{2}+2×2+3}\phantom{\rule{0ex}{0ex}}=\frac{4+10+1}{4+4+3}\phantom{\rule{0ex}{0ex}}=\frac{15}{11}$

Question 29:

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

p(x) = x3 + x2 + 4x + 12
= (x + 2)(x2x + 6)

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

q(x) = (x + 2)(x2 – 2x + 1)

$⇒\underset{x\to -2}{\mathrm{lim}}\left[\frac{{x}^{3}+{x}^{2}+4x+12}{{x}^{3}-3x+2}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to -2}{\mathrm{lim}}\left[\frac{\left(x+2\right)\left({x}^{2}-x+6\right)}{\left(x+2\right)\left({x}^{2}-2x+1\right)}\right]\phantom{\rule{0ex}{0ex}}=\frac{\left(–2{\right)}^{2}-\left(-2\right)+6}{{\left(-2\right)}^{2}-2\left(-2\right)+1}\phantom{\rule{0ex}{0ex}}=\frac{4+2+6}{4+4+1}\phantom{\rule{0ex}{0ex}}=\frac{12}{9}\phantom{\rule{0ex}{0ex}}=\frac{4}{3}$

Question 30:

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

p(x) = (x $-$ 1)(x2 + 4x $-$ 2)

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

$⇒\underset{x\to 1}{\mathrm{lim}}\left[\frac{{x}^{3}+3{x}^{2}-6x+2}{{x}^{3}+3{x}^{2}-3x-1}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 1}{\mathrm{lim}}\left[\frac{\left(x-1\right)\left({x}^{2}+4x-2\right)}{\left(x-1\right)\left({x}^{2}+4x+1\right)}\right]\phantom{\rule{0ex}{0ex}}=\frac{\left(1{\right)}^{2}+4×1-2}{{\left(1\right)}^{2}+4×1+1}\phantom{\rule{0ex}{0ex}}=\frac{1+4-2}{1+4+1}\phantom{\rule{0ex}{0ex}}=\frac{3}{6}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}$

Question 31:

$\underset{x\to 2}{\mathrm{lim}}\left[\frac{1}{x-2}-\frac{2\left(2x-3\right)}{{x}^{3}-3{x}^{2}+2x}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 2}{\mathrm{lim}}\left[\frac{1}{x-2}-\frac{2\left(2x-3\right)}{x\left({x}^{2}-3x+2\right)}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 2}{\mathrm{lim}}\left[\frac{1}{x-2}-\frac{-2\left(2x-3\right)}{x\left({x}^{2}-2x-x+2\right)}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 2}{\mathrm{lim}}\left[\frac{1}{x-2}-\frac{2\left(2x-3\right)}{x\left\{x\left(x-2\right)-1\left(x-2\right)\right\}}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 2}{\mathrm{lim}}\left[\frac{1}{x-2}-\frac{2\left(2x-3\right)}{x\left(x-1\right)\left(x-2\right)}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 2}{\mathrm{lim}}\left[\frac{x\left(x-1\right)-2\left(2x-3\right)}{x\left(x-1\right)\left(x+2\right)}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 2}{\mathrm{lim}}\left[\frac{{x}^{2}-x-4x+6}{x\left(x-1\right)\left(x+2\right)}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 2}{\mathrm{lim}}\left[\frac{{x}^{2}-5x+6}{x\left(x-1\right)\left(x+2\right)}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 2}{\mathrm{lim}}\left[\frac{{x}^{2}-2x-3x+6}{x\left(x-1\right)\left(x-2\right)}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 2}{\mathrm{lim}}\left[\frac{x\left(x-2\right)-3\left(x-2\right)}{x\left(x-1\right)\left(x-2\right)}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 2}{\mathrm{lim}}\left[\frac{\left(x-3\right)\left(x-2\right)}{x\left(x-1\right)\left(x-2\right)}\right]\phantom{\rule{0ex}{0ex}}=\frac{2-3}{2\left(2-1\right)}\phantom{\rule{0ex}{0ex}}=-\frac{1}{2}$

Question 33:

$\underset{x\to 1}{\mathrm{lim}}\left[\frac{x-2}{{x}^{2}-x}-\frac{1}{{x}^{3}-3{x}^{2}+2x}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 1}{\mathrm{lim}}\left[\frac{x-2}{x\left(x-1\right)}-\frac{1}{x\left({x}^{2}-3x+2\right)}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 1}{\mathrm{lim}}\left[\frac{x-2}{x\left(x-1\right)}-\frac{1}{x\left({x}^{2}-2x-x+2\right)}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 1}{\mathrm{lim}}\left[\frac{x-2}{x\left(x-1\right)}-\frac{1}{x\left\{x\left(x-2\right)-1\left(x-2\right)\right\}}\right]$
$=\underset{x\to 1}{\mathrm{lim}}\left[\frac{x-2}{x\left(x-1\right)}-\frac{1}{x\left(x-1\right)\left(x-2\right)}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 1}{\mathrm{lim}}\left[\frac{{\left(x-2\right)}^{2}-1}{x\left(x-1\right)\left(x-2\right)}\right]\phantom{\rule{0ex}{0ex}}$

$=\underset{x\to 1}{\mathrm{lim}}\left[\frac{\left(x-1\right)\left(x-3\right)}{x\left(x-1\right)\left(x-2\right)}\right]\phantom{\rule{0ex}{0ex}}=\frac{\left(1-3\right)}{1\left(1-2\right)}\phantom{\rule{0ex}{0ex}}=\frac{-2}{-1}\phantom{\rule{0ex}{0ex}}=2$

Question 34:

Evaluate the following limits:

$\underset{x\to 1}{\mathrm{lim}}\frac{{x}^{7}-2{x}^{5}+1}{{x}^{3}-3{x}^{2}+2}$

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

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

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

$\therefore \underset{x\to 1}{\mathrm{lim}}\frac{{x}^{7}-2{x}^{5}+1}{{x}^{3}-3{x}^{2}+2}\phantom{\rule{0ex}{0ex}}=\underset{x\to 1}{\mathrm{lim}}\frac{\left(x-1\right)\left({x}^{6}+{x}^{5}-{x}^{4}-{x}^{3}-{x}^{2}-x-1\right)}{\left(x-1\right)\left({x}^{2}-2x-2\right)}\phantom{\rule{0ex}{0ex}}=\underset{x\to 1}{\mathrm{lim}}\frac{{x}^{6}+{x}^{5}-{x}^{4}-{x}^{3}-{x}^{2}-x-1}{{x}^{2}-2x-2}\phantom{\rule{0ex}{0ex}}=\frac{1+1-1-1-1-1-1}{1-2-2}\phantom{\rule{0ex}{0ex}}=\frac{-3}{-3}\phantom{\rule{0ex}{0ex}}=1\phantom{\rule{0ex}{0ex}}$

Question 1:

$\underset{x\to 0}{\mathrm{lim}}\left[\frac{\sqrt{1+x+{x}^{2}}-1}{x}\right]$

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

$\underset{x\to 0}{\mathrm{lim}}\left[\frac{\left(\sqrt{1+x+{x}^{2}}-1\right)\left(\sqrt{1+x+{x}^{2}}+1\right)}{x\left(\sqrt{1+x+{x}^{2}}+1\right)}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 0}{\mathrm{lim}}\left[\frac{1+x+{x}^{2}-1}{x\left(\sqrt{1+x+{x}^{2}}+1\right)}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 0}{\mathrm{lim}}\left[\frac{x\left(1+x\right)}{x\left(\sqrt{1+x+{x}^{2}}+1\right)}\right]\phantom{\rule{0ex}{0ex}}=\frac{1+0}{\sqrt{1+0+0}+1}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}$

Question 2:

$\underset{x\to 0}{\mathrm{lim}}\left[\frac{2x}{\sqrt{a+x}-\sqrt{a-x}}\right]$

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

Rationalising the denominator:

$\underset{x\to 0}{\mathrm{lim}}\left[\frac{2x}{\left(\sqrt{a+x}-\sqrt{a-x}\right)}×\frac{\left(\sqrt{a+x}+\sqrt{a-x}\right)}{\left(\sqrt{a+x}+\sqrt{a-x}\right)}\right]$

$\underset{x\to 0}{\mathrm{lim}}\left[\frac{\left(2x\right)\left(\sqrt{a+x}+\sqrt{a-x}\right)}{\left(a+x\right)-\left(a-x\right)}\right]$

$\underset{x\to 0}{\mathrm{lim}}\left[\frac{2x\left(\sqrt{a+x}+\sqrt{a-x}\right)}{2x}\right]$

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

Question 3:

$\underset{x\to 0}{\mathrm{lim}}\left[\frac{\sqrt{{a}^{2}+{x}^{2}}-a}{{x}^{2}}\right]$

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

$\underset{x\to 0}{\mathrm{lim}}\left[\frac{\left(\sqrt{{a}^{2}+{x}^{2}}-a\right)\left(\sqrt{{a}^{2}+{x}^{2}}+a\right)}{{x}^{2}\left(\sqrt{{a}^{2}+{x}^{2}}+a\right)}\right]$

$\underset{x\to 0}{\mathrm{lim}}\left[\frac{{a}^{2}+{x}^{2}-{a}^{2}}{{x}^{2}\left(\sqrt{{a}^{2}+{x}^{2}}+a\right)}\right]$

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

$\frac{1}{2a}$

Question 4:

$\underset{x\to 0}{\mathrm{lim}}\left[\frac{\sqrt{1+x}-\sqrt{1-x}}{2x}\right]$

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

Rationalising the numerator:

$\underset{x\to 0}{\mathrm{lim}}\left[\frac{\left(\sqrt{1+x}-\sqrt{1-x}\right)\left(\sqrt{1+x}+\sqrt{1-x}\right)}{2x\left(\sqrt{1+x}+\sqrt{1-x}\right)}\right]$

$\underset{x\to 0}{\mathrm{lim}}\left[\frac{2x}{2x\left(\sqrt{1+x}+\sqrt{1-x}\right)}\right]$

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

Question 5:

$\underset{x\to 2}{\mathrm{lim}}\left[\frac{\sqrt{3-x}-1}{2-x}\right]$

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

$=\underset{x\to 2}{\mathrm{lim}}\left[\frac{3-x-1}{\left(2-x\right)\left(\sqrt{3-x}+1\right)}\right]$

$=\underset{x\to 2}{\mathrm{lim}}\left[\frac{\left(2-x\right)}{\left(2-x\right)\left(\sqrt{3-x}+1\right)}\right]$

$=\frac{1}{\sqrt{3-2}+1}\phantom{\rule{0ex}{0ex}}=\frac{1}{1+1}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}$

Question 6:

$\underset{x\to 3}{\mathrm{lim}}\left[\frac{x-3}{\sqrt{x-2}-\sqrt{4-x}}\right]$

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

Rationalising the denominator:

$\underset{x\to 3}{\mathrm{lim}}\left[\frac{\left(x-3\right)\left(\sqrt{x-2}+\sqrt{4-x}\right)}{\left(\sqrt{x-2}-\sqrt{4-x}\right)\left(\sqrt{x-2}+\sqrt{4-x}\right)}\right]$

$\underset{x\to 3}{\mathrm{lim}}\left[\frac{\left(x-3\right)\left(\sqrt{x-2}+\sqrt{4-x}\right)}{\left(x-2\right)-\left(4-x\right)}\right]$

$\underset{x\to 3}{\mathrm{lim}}\left[\frac{\left(x-3\right)\left(\sqrt{x-2}+\sqrt{4-x}\right)}{2x-6}\right]$

$\underset{x\to 3}{\mathrm{lim}}\left[\frac{\left(x-3\right)\left(\sqrt{x-2}+\sqrt{4-x}\right)}{2\left(x-3\right)}\right]$

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

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

Question 7:

$\underset{x\to 0}{\mathrm{lim}}\left[\frac{x}{\sqrt{1+x}-\sqrt{1-x}}\right]$

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

Rationalising the denominator:

$\underset{x\to 0}{\mathrm{lim}}\left[\frac{x}{\left(\sqrt{1+x}-\sqrt{1-x}\right)}×\frac{\left(\sqrt{1+x}+\sqrt{1-x}\right)}{\left(\sqrt{1+x}+\sqrt{1-x}\right)}\right]$

$\underset{x\to 0}{\mathrm{lim}}\left[\frac{x\left(\sqrt{1+x}+\sqrt{1-x}\right)}{\left(1+x\right)-\left(1-x\right)}\right]$

$\underset{x\to 0}{\mathrm{lim}}\left[\frac{x\left(\sqrt{1+x}+\sqrt{1-x}\right)}{2x}\right]$

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

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

Question 8:

$\underset{x\to 1}{\mathrm{lim}}\left[\frac{\sqrt{5x-4}-\sqrt{x}}{x-1}\right]$

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

$\underset{x\to 1}{\mathrm{lim}}\left[\left(\frac{\sqrt{5x-4}-\sqrt{x}}{x-1}\right)\left(\frac{\sqrt{5x-4}+\sqrt{x}}{\sqrt{5x-4}+\sqrt{x}}\right)\right]$

$\underset{x\to 1}{\mathrm{lim}}\left[\frac{5x-4-x}{\left(x-1\right)\left(\sqrt{5x-4}+\sqrt{x}\right)}\right]$

$\underset{x\to 1}{\mathrm{lim}}\left[\frac{4\left(x-1\right)}{\left(x-1\right)\left(\sqrt{5x-4}+\sqrt{x}\right)}\right]$

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

$\frac{4}{2}$

= 2

Question 9:

$\underset{x\to 1}{\mathrm{lim}}\left[\frac{x-1}{\sqrt{{x}^{2}+3}-2}\right]$

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

$\underset{x\to 1}{\mathrm{lim}}\left[\frac{\left(x-1\right)\left(\sqrt{{x}^{2}+3}+2\right)}{\left(\sqrt{{x}^{2}+3}-2\right)\left(\sqrt{{x}^{2}+3}+2\right)}\right]$

$\underset{x\to 1}{\mathrm{lim}}\left[\frac{\left(x-1\right)\left(\sqrt{{x}^{2}+3}+2\right)}{{x}^{2}+3-4}\right]$

$\underset{x\to 1}{\mathrm{lim}}\left[\frac{\left(x-1\right)\left(\sqrt{{x}^{2}+3}+2\right)}{\left({x}^{2}-1\right)}\right]$

$\underset{x\to 1}{\mathrm{lim}}\left[\frac{\left(x-1\right)\left(\sqrt{{x}^{2}+3}+2\right)}{\left(x-1\right)\left(x+1\right)}\right]$

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

$\frac{4}{2}$

= 2

Question 10:

$\underset{x\to 3}{\mathrm{lim}}\left[\frac{\sqrt{x+3}-\sqrt{16}}{{x}^{2}-9}\right]$

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

$\underset{x\to 3}{\mathrm{lim}}\left[\frac{\left(\sqrt{x+3}-\sqrt{6}\right)\left(\sqrt{x+3}+\sqrt{6}\right)}{\left({x}^{2}-9\right)\left(\sqrt{x+3}+\sqrt{6}\right)}\right]$

$\underset{x\to 3}{\mathrm{lim}}\left[\frac{\left(x+3-6\right)}{\left(x-3\right)\left(x+3\right)\left(\sqrt{x+3}+\sqrt{6}\right)}\right]$

$\frac{1}{6×2\sqrt{6}}$

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

Question 11:

$\underset{x\to 1}{\mathrm{lim}}\left[\frac{\sqrt{5x-4}-\sqrt{x}}{{x}^{2}-1}\right]$

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

$\underset{x\to 1}{\mathrm{lim}}\left[\frac{\left(\sqrt{5x-4}-\sqrt{x}\right)\left(\sqrt{5x-4}+\sqrt{x}\right)}{\left(\sqrt{5x-4}+\sqrt{x}\right)\left({x}^{2}-1\right)}\right]$

$\underset{x\to 1}{\mathrm{lim}}\left[\frac{5x-4-x}{\left(\sqrt{5x-4}+\sqrt{x}\right)\left(x-1\right)\left(x+1\right)}\right]$

$\underset{x\to 1}{\mathrm{lim}}\left[\frac{4\left(x-1\right)}{\left(\sqrt{5x-4}+\sqrt{x}\right)\left(x-1\right)\left(x+1\right)}\right]$

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

$\frac{4}{2×2}$

= 1

Question 12:

$\underset{x\to 0}{\mathrm{lim}}\left[\frac{\sqrt{1+x}-1}{x}\right]$

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

$\underset{x\to 0}{\mathrm{lim}}\left[\frac{\left(\sqrt{1+x}-1\right)\left(\sqrt{1+x}+1\right)}{x\left(\sqrt{1+x}+1\right)}\right]$

=$\underset{x\to 0}{\mathrm{lim}}\left[\frac{1+x-1}{x\left(\sqrt{1+x}+1\right)}\right]$

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

$\frac{1}{2}$

Question 13:

$\underset{x\to 2}{\mathrm{lim}}\left[\frac{\sqrt{{x}^{2}+1}-\sqrt{5}}{x-2}\right]$

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

$\underset{x\to 2}{\mathrm{lim}}\left[\frac{\left(\sqrt{{x}^{2}+1}-\sqrt{5}\right)\left(\sqrt{{x}^{2}+1}+\sqrt{5}\right)}{\left(x-2\right)\left(\sqrt{{x}^{2}+1}+\sqrt{5}\right)}\right]$

$\underset{x\to 2}{\mathrm{lim}}\left[\frac{{x}^{2}+1-5}{\left(x-2\right)\left(\sqrt{{x}^{2}+1}+\sqrt{5}\right)}\right]$

$\underset{x\to 2}{\mathrm{lim}}\left[\frac{{x}^{2}-4}{\left(x-2\right)\left(\sqrt{{x}^{2}+1}+\sqrt{5}\right)}\right]$

$\underset{x\to 2}{\mathrm{lim}}\left[\frac{\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(\sqrt{{x}^{2}+1}+\sqrt{5}\right)}\right]$

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

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

Question 14:

$\underset{x\to 2}{\mathrm{lim}}\left[\frac{x-2}{\sqrt{x}-\sqrt{2}}\right]$

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

$\underset{x\to 2}{\mathrm{lim}}\left[\frac{{\left(\sqrt{x}\right)}^{2}-{\left(\sqrt{2}\right)}^{2}}{x-\sqrt{2}}\right]$

$\underset{x\to 2}{\mathrm{lim}}\left[\frac{\left(\sqrt{x}-\sqrt{2}\right)\left(\sqrt{x}+\sqrt{2}\right)}{\left(x-\sqrt{2}\right)}\right]$

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

$2\sqrt{2}$

Question 15:

$\underset{x\to 7}{\mathrm{lim}}\left[\frac{4-\sqrt{9+x}}{1-\sqrt{8-x}}\right]$

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

$\underset{x\to 7}{\mathrm{lim}}\left[\frac{16-\left(9+x\right)}{\left(4+\sqrt{9+x}\right)}×\frac{\left(1+\sqrt{8-x}\right)}{1-\left(8-x\right)}\right]$

$\underset{x\to 7}{\mathrm{lim}}\left[\frac{-1\left(-7+x\right)\left(1+\sqrt{8-x}\right)}{\left(4+\sqrt{9+x}\right)\left(-7+x\right)}\right]$

$\underset{x\to 7}{\mathrm{lim}}\left[\frac{-\left(1+\sqrt{8-x}\right)}{4+\sqrt{9+x}}\right]$

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

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

$\frac{-1}{4}$

Question 16:

$\underset{x\to 0}{\mathrm{lim}}\left[\frac{\sqrt{a+x}-\sqrt{a}}{x\sqrt{{a}^{2}+ax}}\right]$

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

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

Question 17:

$\underset{x\to 5}{\mathrm{lim}}\left[\frac{x-5}{\sqrt{6x-5}-\sqrt{4x+5}}\right]$

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

$\underset{x\to 5}{\mathrm{lim}}\left[\frac{\left(x-5\right)\left(\sqrt{6x-5}+\sqrt{4x+5}\right)}{2\left(x-5\right)}\right]$

$\frac{\sqrt{6×5-5}+\sqrt{4×5+5}}{2}$

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

= 5

Question 18:

$\underset{x\to 1}{\mathrm{lim}}\left[\frac{\sqrt{5x-4}-\sqrt{x}}{{x}^{3}-1}\right]$

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

$\underset{x\to 1}{\mathrm{lim}}\left[\frac{5x-4-x}{\left(x-1\right)\left({x}^{2}+x+1\right)\left(\sqrt{5x-4}+\sqrt{x}\right)}\right]$

$\underset{x\to 1}{\mathrm{lim}}\left[\frac{4\left(x-1\right)}{\left(x-1\right)\left({x}^{2}+x+1\right)\left(\sqrt{5x-4}+\sqrt{x}\right)}\right]$

$\frac{4}{3\left(1+1\right)}$

$\frac{2}{3}$

Question 19:

$\underset{x\to 2}{\mathrm{lim}}\left[\frac{\sqrt{1+4x}-\sqrt{5+2x}}{x-2}\right]$

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

$\underset{x\to 2}{\mathrm{lim}}\left[\frac{\left(1+4x\right)-\left(5+2x\right)}{\left(x-2\right)\left(\sqrt{1+4x}+\sqrt{5+2x}\right)}\right]$

$\underset{x\to 2}{\mathrm{lim}}\left[\frac{2\left(x-2\right)}{\left(x-2\right)\left(\sqrt{1+4x}+\sqrt{5+2x}\right)}\right]$

$\frac{2}{\left(\sqrt{1+4×2}+\sqrt{5+2×2}\right)}$

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

$\frac{1}{3}$

Question 20:

$\underset{x\to 1}{\mathrm{lim}}\left[\frac{\sqrt{3+x}-\sqrt{5-x}}{{x}^{2}-1}\right]$

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

$\underset{x\to 1}{\mathrm{lim}}\left[\frac{\left(\sqrt{3+x}-\sqrt{5-x}\right)\left(\sqrt{3+x}+\sqrt{5-x}\right)}{\left(x-1\right)\left(x+1\right)\left(\sqrt{3+x}+\sqrt{5-x}\right)}\right]$

$\underset{x\to 1}{\mathrm{lim}}\left[\frac{\left(3+x\right)-\left(5-x\right)}{\left(x-1\right)\left(x+1\right)\left\{\sqrt{3+x}+\sqrt{5-x}\right\}}\right]$

$\underset{x\to 1}{\mathrm{lim}}\left[\frac{2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)\left\{\sqrt{3+x}+\sqrt{5-x}\right\}}\right]$

$\frac{1}{4}$

Question 21:

$\underset{x\to 0}{\mathrm{lim}}\left[\frac{\sqrt{1+{x}^{2}}-\sqrt{1-{x}^{2}}}{x}\right]$

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

$\underset{x\to 0}{\mathrm{lim}}\left[\frac{\left(\sqrt{1+{x}^{2}}-\sqrt{1-{x}^{2}}\right)\left(\sqrt{1+{x}^{2}}+\sqrt{1-{x}^{2}}\right)}{\left(\sqrt{1+{x}^{2}}+\sqrt{1-{x}^{2}}\right)x}\right]$

$\underset{x\to 0}{\mathrm{lim}}\left[\frac{\left(1+{x}^{2}\right)-\left(1-{x}^{2}\right)}{x\left\{\sqrt{1+{x}^{2}}+\sqrt{1-{x}^{2}}\right\}}\right]$

$\underset{x\to 0}{\mathrm{lim}}\left[\frac{2{x}^{2}}{x\left\{\sqrt{1+{x}^{2}}+\sqrt{1-{x}^{2}}\right\}}\right]$

$\frac{2×0}{\sqrt{1+0}+\sqrt{1-0}}$

= 0

Question 22:

$\underset{x\to 0}{\mathrm{lim}}\left[\frac{\sqrt{1+x+{x}^{2}}-\sqrt{x+1}}{2{x}^{2}}\right]$

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

$\underset{x\to 0}{\mathrm{lim}}\left[\frac{\left(\sqrt{1+x+{x}^{2}}-\sqrt{x+1}\right)\left(\sqrt{1+x+{x}^{2}}+\sqrt{x+1}\right)}{\left(\sqrt{1+x+{x}^{2}}+\sqrt{x+1}\right)2{x}^{2}}\right]$

$\underset{x\to 0}{\mathrm{lim}}\left[\frac{\left(1+x+{x}^{2}\right)-\left(x+1\right)}{\left(\sqrt{1+x+{x}^{2}}+\sqrt{x+1}\right)2{x}^{2}}\right]$

$\underset{x\to 0}{\mathrm{lim}}\left[\frac{{x}^{2}}{\left(\sqrt{1+x+{x}^{2}}+\sqrt{x+1}\right)\left(2{x}^{2}\right)}\right]$

$\frac{1}{\left(\sqrt{1+0+0}+\sqrt{0+1}\right)×2}$

$\frac{1}{2}×\frac{1}{2}$

$\frac{1}{4}$

Question 23:

$\underset{x\to 4}{\mathrm{lim}}\left[\frac{2-\sqrt{x}}{4-x}\right]$

$\underset{x\to 4}{\mathrm{lim}}\left[\frac{2-\sqrt{x}}{2-{\left(\sqrt{x}\right)}^{2}}\right]$

$\underset{x\to 4}{\mathrm{lim}}\left[\frac{\left(2-\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right]$

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

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

$\frac{1}{4}$

Question 24:

$\underset{x\to a}{\mathrm{lim}}\left[\frac{x-a}{\sqrt{x}-\sqrt{a}}\right]$

$\underset{x\to a}{\mathrm{lim}}\left[\frac{{\left(\sqrt{x}\right)}^{2}-{a}^{2}}{\sqrt{x}-\sqrt{a}}\right]$

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

$2\sqrt{a}$

Question 25:

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

Rationalising the numerator:

Question 26:

$\underset{x\to 0}{\mathrm{lim}}\left[\frac{\sqrt{2-x}-\sqrt{2+x}}{x}\right]$

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

$\underset{x\to 0}{\mathrm{lim}}\left[\frac{\left(\sqrt{2-x}-\sqrt{2+x}\right)\left(\sqrt{2-x}+\sqrt{2+x}\right)}{x\left(\sqrt{2-x}+\sqrt{2+x}\right)}\right]$

$\underset{x\to 0}{\mathrm{lim}}\left[\frac{\left(2-x\right)-\left(2+x\right)}{x\left(\sqrt{2-x}+\sqrt{2+x}\right)}\right]$

$\underset{x\to 0}{\mathrm{lim}}\left[\frac{-2x}{x\left(\sqrt{2-x}+\sqrt{2+x}\right)}\right]$

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

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

Question 27:

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

Rationalising the numerator:

Question 28:

$\underset{x\to 1}{\mathrm{lim}}\left[\frac{\left(2x-3\right)\left(\sqrt{x}-1\right)}{3{x}^{2}+3x-6}\right]$

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

⇒ $\underset{x\to 1}{\mathrm{lim}}\left[\frac{\left(2x-3\right)\left(\sqrt{x}-1\right)}{3\left({x}^{2}+x-2\right)}\right]$

$\underset{x\to 1}{\mathrm{lim}}\left[\frac{\left(2x-3\right)\left(\sqrt{x}-1\right)}{3\left({x}^{2}+2x-x-2\right)}\right]$

$\underset{x\to 1}{\mathrm{lim}}\left[\frac{\left(2x-3\right)\left(\sqrt{x}-1\right)}{3\left(x\left(x+2\right)-1\left(x+2\right)\right)}\right]$

$\underset{x\to 1}{\mathrm{lim}}\left[\frac{\left(2x-3\right)\left(\sqrt{x}-1\right)}{3\left(x-1\right)\left(x+2\right)}\right]$

$\underset{x\to 1}{\mathrm{lim}}\left[\frac{\left(2x-3\right)\left(\sqrt{x}-1\right)}{3\left({\left(\sqrt{x}\right)}^{2}-{1}^{2}\right)\left(x+2\right)}\right]$

$\underset{x\to 1}{\mathrm{lim}}\left[\frac{\left(2x-3\right)\left(\sqrt{x}-1\right)}{3\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x+2\right)}\right]$

$\frac{-1}{3\left(2\right)×3}$

$\frac{-1}{18}$

Question 29:

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

Question 30:

$\underset{x\to 1}{\mathrm{lim}}\left[\frac{{x}^{2}-\sqrt{x}}{\sqrt{x}-1}\right]$

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

$\underset{x\to 1}{\mathrm{lim}}\left[\frac{\sqrt{x}\left(x\sqrt{x}-1\right)}{\sqrt{x}-1}\right]$

$\underset{x\to 1}{\mathrm{lim}}\left[\frac{\sqrt{x}\left({x}^{3/2}-1\right)}{{x}^{\frac{1}{2}}-1}\right]$

$\underset{x\to 1}{\mathrm{lim}}\left[\frac{\left(\sqrt{x}\right)\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)}\right]$

= 1 (1 + 1 + 1)

= 3

Question 31:

$\underset{h\to 0}{\mathrm{lim}}\left[\frac{\sqrt{x+h}-\sqrt{x}}{h}\right]$

It is of the form $\frac{0}{0}$.
Rationalising the numerator:
$\underset{h\to 0}{\mathrm{lim}}\left[\frac{\left(\sqrt{x+h}-\sqrt{x}\right)}{h}×\frac{\left(\sqrt{x+h}+\sqrt{x}\right)}{\left(\sqrt{x+h}+\sqrt{x}\right)}\right]$

$\underset{h\to 0}{\mathrm{lim}}\left[\frac{\left(x+h\right)-x}{h\left(\sqrt{x+h}+\sqrt{x}\right)}\right]$

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

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

Question 32:

$\underset{x\to \sqrt{10}}{\mathrm{lim}}\left[\frac{\sqrt{7+2x}-\left(\sqrt{5}+\sqrt{2}\right)}{{x}^{2}-{\left(\sqrt{10}\right)}^{2}}\right]$

$\underset{x\to \sqrt{10}}{\mathrm{lim}}\left[\frac{\sqrt{7+2x}-\sqrt{{\left(\sqrt{5}+\sqrt{2}\right)}^{2}}}{\left(x-\sqrt{10}\right)\left(x+\sqrt{10}\right)}\right]$

$\underset{x\to \sqrt{10}}{\mathrm{lim}}\left[\frac{\sqrt{7+2x}-\sqrt{5+2+2\sqrt{5}\sqrt{2}}}{\left(x-\sqrt{10}\right)\left(x+\sqrt{10}\right)}\right]$

$\underset{x\to \sqrt{10}}{\mathrm{lim}}\left[\frac{\sqrt{7+2x}-\sqrt{7+2\sqrt{10}}}{\left(x-\sqrt{10}\right)\left(x+\sqrt{10}\right)}\right]$

Rationalising the numerator:

$=\frac{2}{\left(2\sqrt{10}\right)×2\sqrt{7+2\sqrt{10}}}\phantom{\rule{0ex}{0ex}}=\frac{1}{2\sqrt{10}\sqrt{{\left(\sqrt{5}+\sqrt{2}\right)}^{2}}}$

$\frac{1}{2\sqrt{10}\left(\sqrt{5}+\sqrt{2}\right)}×\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}$

$=\frac{1}{2\sqrt{10}}\left[\frac{\sqrt{5}-\sqrt{2}}{{\left(\sqrt{5}\right)}^{2}-{\left(\sqrt{2}\right)}^{2}}\right]\phantom{\rule{0ex}{0ex}}$

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

Question 33:

$\underset{x\to \sqrt{6}}{\mathrm{lim}}\left[\frac{\sqrt{5+2x}-\left(\sqrt{3}+\sqrt{2}\right)}{{x}^{2}-6}\right]$

$\underset{x\to \sqrt{6}}{\mathrm{lim}}\left[\frac{\sqrt{5+2x}-\sqrt{{\left(\sqrt{3}+\sqrt{2}\right)}^{2}}}{{x}^{2}-{\left(\sqrt{6}\right)}^{2}}\right]$

$\underset{x\to \sqrt{6}}{\mathrm{lim}}\left[\frac{\sqrt{5+2x}-\sqrt{3+2+2\sqrt{6}}}{\left(x-\sqrt{6}\right)\left(x+\sqrt{6}\right)}\right]$

$\underset{x\to \sqrt{6}}{\mathrm{lim}}\left[\frac{\sqrt{5+2x}-\sqrt{5+2\sqrt{6}}}{\left(x-\sqrt{6}\right)\left(x+\sqrt{6}\right)}\right]$

Rationalising the numerator:

$\underset{x\to \sqrt{6}}{\mathrm{lim}}\left[\frac{\left(\sqrt{5+2x}-\sqrt{5+2\sqrt{6}}\right)\left(\sqrt{5+2x}+\sqrt{5+2\sqrt{6}}\right)}{\left(x-\sqrt{6}\right)\left(x+\sqrt{6}\right)\left(\sqrt{5+2x}+\sqrt{5+2\sqrt{6}}\right)}\right]$

$\underset{x\to \sqrt{6}}{\mathrm{lim}}\left[\frac{\left(5+2x\right)-\left(5+2\sqrt{6}\right)}{\left(x-\sqrt{6}\right)\left(x+\sqrt{6}\right)\left(\sqrt{5+2x}+\sqrt{5+2\sqrt{6}}\right)}\right]$

$\underset{x\to \sqrt{6}}{\mathrm{lim}}\left[\frac{2\left(x-\sqrt{6}\right)}{\left(x-\sqrt{6}\right)\left(x+\sqrt{6}\right)\left(\sqrt{5+2x}+\sqrt{5+2\sqrt{6}}\right)}\right]$

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

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

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

$\frac{1}{2\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)}×\frac{\left(\sqrt{3}-\sqrt{2}\right)}{\left(\sqrt{3}-\sqrt{2}\right)}$

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

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

Question 34:

$\underset{x\to \sqrt{2}}{\mathrm{lim}}\left[\frac{\sqrt{3+2x}-\left(\sqrt{2}+1\right)}{{x}^{2}-2}\right]$

$\underset{x\to \sqrt{2}}{\mathrm{lim}}\left[\frac{\sqrt{3+2x}-\sqrt{{\left(\sqrt{2}+1\right)}^{2}}}{\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)}\right]$

$\underset{x\to \sqrt{2}}{\mathrm{lim}}\left[\frac{\sqrt{3+2x}-\sqrt{2+1+2\sqrt{2}}}{\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)}\right]$

$\underset{x\to \sqrt{2}}{\mathrm{lim}}\left[\frac{\left(\sqrt{3+2x}-\sqrt{3+2\sqrt{2}}\right)}{\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)}\right]$

Rationalising the numerator:

$\underset{x\to \sqrt{2}}{\mathrm{lim}}\left[\frac{\left(\sqrt{3+2x}-\sqrt{3+2\sqrt{2}}\right)\left(\sqrt{3+2x}+\sqrt{3+2\sqrt{2}}\right)}{\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\left(\sqrt{3+2x}+\sqrt{3+2\sqrt{2}}\right)}\right]$

$\underset{x\to \sqrt{2}}{\mathrm{lim}}\left[\frac{\left(3+2x\right)-\left(3+2\sqrt{2}\right)}{\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\left(\sqrt{3+2x}+\sqrt{3+2\sqrt{2}}\right)}\right]$

$\underset{x\to \sqrt{2}}{\mathrm{lim}}\left[\frac{2\left(x-\sqrt{2}\right)}{\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\left(\sqrt{3+2x}+\sqrt{3+2\sqrt{2}}\right)}\right]$

$\frac{2}{\left(\sqrt{2}+\sqrt{2}\right)\left(\sqrt{3+2\sqrt{2}}+\sqrt{3+2\sqrt{2}}\right)}$

$\frac{2}{\left(2\sqrt{2}\right)\left(2\sqrt{3+2\sqrt{2}}\right)}$

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

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

$\frac{1}{2\sqrt{2}\left(\sqrt{2}+1\right)}×\frac{\sqrt{2}-1}{\sqrt{2}-1}$

$\frac{\sqrt{2}-1}{2\sqrt{2}\left(2-1\right)}$

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

Question 1:

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

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

$\underset{y\to b}{\mathrm{lim}}\left[\frac{{y}^{\frac{5}{2}}-{b}^{\frac{5}{2}}}{y-b}\right]\phantom{\rule{0ex}{0ex}}=\frac{5}{2}{\left(b\right)}^{\frac{5}{2}-1}\phantom{\rule{0ex}{0ex}}=\frac{5}{2}{b}^{\frac{3}{2}}\phantom{\rule{0ex}{0ex}}=\frac{5}{2}{\left(a+2\right)}^{\frac{3}{2}}$

Question 2:

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

When x ​→ a and x + 2 ​→ a + 2.
$⇒$y ​→ b

Question 3:

Let y = 1 + x

When x ​→ 0, then 1 + x → 1.
$⇒$y ​→ 1

$\underset{y\to 1}{\mathrm{lim}}\left[\left(\frac{{y}^{6}-{1}^{6}}{y-1}\right)×\frac{\left(y-1\right)}{{y}^{2}-{1}^{2}}\right]\phantom{\rule{0ex}{0ex}}=\frac{6×{\left(1\right)}^{6-1}}{2×{\left(1\right)}^{2-1}}\phantom{\rule{0ex}{0ex}}=\frac{6}{2}\phantom{\rule{0ex}{0ex}}=3$

Let y = 2x

Question 12:

If find the value of n.

x(3)n – 1 = 108
x(3)n – 1 = 4 × 33

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

Question 13:

If find all possible values of a.

Question 14:

If find all possible values of a.

Question 15:

If find all possible values of a.

Question 16:

If find all possible values of a.

Question 4:

It is of the form ∞ - ∞.

Rationalising the numerator:

Question 5:

$\underset{x\to \infty }{\mathrm{lim}}\sqrt{x+1}-\sqrt{x}$

It is of the form ∞–​​∞.

On rationalising, we get:

Question 6:

$\underset{x\to \infty }{\mathrm{lim}}\sqrt{{x}^{2}+7x-x}$

Question 7:

$\underset{x\to \infty }{\mathrm{lim}}\frac{x}{\sqrt{4{x}^{2}+1}-1}$

Question 8:

$\underset{n\to \infty }{\mathrm{lim}}\frac{{n}^{2}}{1+2+3+...+n}$

Question 9:

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

Question 10:

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

Question 11:

$\underset{n\to \infty }{\mathrm{lim}}\left[\frac{\left(n+2\right)!+\left(n+1\right)!}{\left(n+2\right)!-\left(n+1\right)!}\right]$

Dividing the numerator and the denominator by n:

Question 12:

$\underset{x\to \infty }{\mathrm{lim}}\left[x\left\{\sqrt{{x}^{2}+1}-\sqrt{{x}^{2}-1}\right\}\right]$

Question 14:

$\underset{n\to \infty }{\mathrm{lim}}\left[\frac{{1}^{2}+{2}^{2}+...+{n}^{2}}{{n}^{3}}\right]$

Question 15:

$\underset{n\to \infty }{\mathrm{lim}}\left[\frac{1+2+3......n-1}{{n}^{2}}\right]$

Question 16:

$\underset{n\to \infty }{\mathrm{lim}}\left[\frac{{1}^{3}+{2}^{3}+....{n}^{3}}{{n}^{4}}\right]$

Question 17:

$\underset{n\to \infty }{\mathrm{lim}}\left[\frac{{1}^{3}+{2}^{3}+...{n}^{3}}{{\left(n-1\right)}^{4}}\right]$

Dividing the numerator and the denominator by n4:

Question 18:

$\underset{x\to \infty }{\mathrm{lim}}\left[\sqrt{x}\left\{\sqrt{x+1}-\sqrt{x}\right\}\right]$

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

Question 19:

$\underset{n\to \infty }{\mathrm{lim}}\left[\frac{1}{3}+\frac{1}{{3}^{2}}+\frac{1}{{3}^{3}}+...+\frac{1}{{3}^{n}}\right]$

Question 20:

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

$\underset{x\to \infty }{\mathrm{lim}}\left[\frac{{x}^{4}+7{x}^{3}+46x+a}{{x}^{4}+6}\right]$

Dividing the numerator and the denominator by x4:

Question 21:

and $\underset{x\to \infty }{\mathrm{lim}}f\left(x\right)=1,$ then prove that f(−2) = f(2) = 1

Dividing the numerator and the denominator by x2:

Question 22:

Show that $\underset{x\to \infty }{\mathrm{lim}}\left(\sqrt{{x}^{2}+x+1}-x\right)\ne \underset{x\to \infty }{\mathrm{lim}}\left(\sqrt{{x}^{2}+1}-x\right)$

Rationalising the numerator:

Question 23:

$\underset{x\to -\infty }{\mathrm{lim}}\left(\sqrt{4{x}^{2}-7x}+2x\right)$

$\underset{x\to -\infty }{\mathrm{lim}}\left(\sqrt{4{x}^{2}-7x}+2x\right)$

Let x =$-$m
When n → – ∞, then m → ∞.

Dividing the numerator and the denominator by m:

Question 24:

$\underset{x\to -\infty }{\mathrm{lim}}\left(\sqrt{{x}^{2}-8x}+x\right)$

$\underset{x\to -\infty }{\mathrm{lim}}\left(\sqrt{{x}^{2}-8x}+x\right)$

Let x = –m

When x → –∞, then m → ∞.

Dividing the numerator and the denominator by m:

Question 25:

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

Consider the identity

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

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

${\left(n+1\right)}^{5}-1=5\sum _{k=1}^{n}{k}^{4}+10\sum _{k=1}^{n}{k}^{3}+10\sum _{k=1}^{n}{k}^{2}+5\sum _{k=1}^{n}k+\sum _{k=1}^{n}1\phantom{\rule{0ex}{0ex}}⇒{n}^{5}+5{n}^{4}+10{n}^{3}+10{n}^{2}+5n=5\sum _{k=1}^{n}{k}^{4}+\frac{10{n}^{2}{\left(n+1\right)}^{2}}{4}+\frac{10n\left(n+1\right)\left(2n+1\right)}{6}+\frac{5n\left(n+1\right)}{2}+n\phantom{\rule{0ex}{0ex}}⇒5\sum _{k=1}^{n}{k}^{4}={n}^{5}+5{n}^{4}+10{n}^{3}+10{n}^{2}+4n-\frac{5{n}^{2}{\left(n+1\right)}^{2}}{2}-\frac{5n\left(n+1\right)\left(2n+1\right)}{3}-\frac{5n\left(n+1\right)}{2}\phantom{\rule{0ex}{0ex}}⇒5\sum _{k=1}^{n}{k}^{4}={n}^{5}+\frac{5{n}^{4}}{2}+\frac{5{n}^{3}}{3}-\frac{n}{6}$

This expression on further simplification gives

$\sum _{k=1}^{n}{k}^{4}=\frac{n\left(n+1\right)\left(2n+1\right)\left(3{n}^{2}+3n-1\right)}{30}$

$=\frac{1}{30}×6-0\phantom{\rule{0ex}{0ex}}=\frac{1}{5}$

Question 26:

Evaluate: $\underset{n\to \infty }{\mathrm{lim}}\frac{1.2+2.3+3.4+...+n\left(n+1\right)}{{n}^{3}}$

$\underset{n\to \infty }{\mathrm{lim}}\frac{1.2+2.3+3.4+...+n\left(n+1\right)}{{n}^{3}}\phantom{\rule{0ex}{0ex}}=\underset{n\to \infty }{\mathrm{lim}}\frac{\sum _{k=1}^{n}k\left(k+1\right)}{{n}^{3}}\phantom{\rule{0ex}{0ex}}=\underset{n\to \infty }{\mathrm{lim}}\frac{\sum _{k=1}^{n}{k}^{2}+\sum _{k=1}^{n}k}{{n}^{3}}\phantom{\rule{0ex}{0ex}}=\underset{n\to \infty }{\mathrm{lim}}\frac{\frac{n\left(n+1\right)\left(2n+1\right)}{6}+\frac{n\left(n+1\right)}{2}}{{n}^{3}}$

Question 1:

=$\frac{1}{5}\underset{x\to 0}{\mathrm{lim}}\left[\frac{\mathrm{sin}3x}{3x}×3\right]$         $\left[\because \underset{x\to 0}{\mathrm{lim}}\left(\frac{\mathrm{sin}3x}{3x}\right)=1\right]$

$\frac{1}{5}×1×3$

$\frac{3}{5}$

Question 2:

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

Question 3:

$\underset{x\to 0}{\mathrm{lim}}\left[\frac{{x}^{2}}{\mathrm{sin}{x}^{2}}\right]$

Let $\theta ={x}^{2}$

$\underset{\theta \to 0}{\mathrm{lim}}\left(\frac{\theta }{\mathrm{sin}\theta }\right)$

= 1

Question 4:

$\frac{1}{3}\underset{x\to 0}{\mathrm{lim}}\left(\frac{\mathrm{sin}x}{x}\right)×\mathrm{cos}x$

$\frac{1}{3}×1×\mathrm{cos}0$

$\frac{1}{3}×1$

= $\frac{1}{3}$

Question 5:

$\underset{x\to 0}{\mathrm{lim}}\left[\frac{3\mathrm{sin}x-4{\mathrm{sin}}^{3}x}{x}\right]$

$\left[\because {\mathrm{sin}}^{3}A=3\mathrm{sinA}-4{\mathrm{sin}}^{3}\mathrm{A}\right]$

= 1 × 3

= 3

Question 9:

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

Question 10:

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

Question 11:

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

Question 19:

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

Question 20:

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

Question 21:

Dividing the numerator and the denominator by x:

Question 25:

Dividing the numerator and the denominator by x:

Question 51:

Evaluate the following limits:

$\underset{x\to 0}{\mathrm{lim}}\frac{2\mathrm{sin}x-\mathrm{sin}2x}{{x}^{3}}$

$\underset{x\to 0}{\mathrm{lim}}\frac{2\mathrm{sin}x-\mathrm{sin}2x}{{x}^{3}}\phantom{\rule{0ex}{0ex}}=\underset{x\to 0}{\mathrm{lim}}\frac{2\mathrm{sin}x-2\mathrm{sin}x\mathrm{cos}x}{{x}^{3}}\phantom{\rule{0ex}{0ex}}=\underset{x\to 0}{\mathrm{lim}}\frac{2\mathrm{sin}x\left(1-\mathrm{cos}x\right)}{{x}^{3}}\phantom{\rule{0ex}{0ex}}=\underset{x\to 0}{\mathrm{lim}}\frac{2\mathrm{sin}x×2{\mathrm{sin}}^{2}\frac{x}{2}}{{x}^{3}}$

Question 60:

Evaluate the following limits:

$\underset{x\to 0}{\mathrm{lim}}\frac{\mathrm{sin}\left(\alpha +\beta \right)x+\mathrm{sin}\left(\alpha -\beta \right)x+\mathrm{sin}2\alpha x}{{\mathrm{cos}}^{2}\beta x-{\mathrm{cos}}^{2}\alpha x}$

$=\underset{x\to 0}{\mathrm{lim}}\frac{2\alpha x×\frac{\mathrm{sin}\alpha x}{\alpha x}×\left(\mathrm{cos}\beta x+\mathrm{cos}\alpha x\right)}{{\alpha }^{2}{x}^{2}×\frac{{\mathrm{sin}}^{2}\alpha x}{{\alpha }^{2}{x}^{2}}-{\beta }^{2}{x}^{2}×\frac{{\mathrm{sin}}^{2}\beta x}{{\beta }^{2}{x}^{2}}}\phantom{\rule{0ex}{0ex}}=\frac{2\alpha ×\underset{x\to 0}{\mathrm{lim}}\left(\frac{\mathrm{sin}\alpha x}{\alpha x}\right)×\underset{x\to 0}{\mathrm{lim}}\left(\mathrm{cos}\beta x+\mathrm{cos}\alpha x\right)}{{\alpha }^{2}×{\left(\underset{x\to 0}{\mathrm{lim}}\frac{\mathrm{sin}\alpha x}{\alpha x}\right)}^{2}-{\beta }^{2}×{\left(\underset{x\to 0}{\mathrm{lim}}\frac{\mathrm{sin}\beta x}{\beta x}\right)}^{2}}×\underset{x\to 0}{\mathrm{lim}}\frac{x}{{x}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{2\alpha ×1×\left(1+1\right)}{{\alpha }^{2}×1-{\beta }^{2}×1}×\underset{x\to 0}{\mathrm{lim}}\frac{1}{x}\phantom{\rule{0ex}{0ex}}=\frac{4\alpha }{{\alpha }^{2}-{\beta }^{2}}×\infty \phantom{\rule{0ex}{0ex}}=\infty$

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

Question 61:

Evaluate the following limits:

$\underset{x\to 0}{\mathrm{lim}}\frac{\mathrm{cos}ax-\mathrm{cos}bx}{\mathrm{cos}cx-1}$

Question 62:

Evaluate the following limits:

$\underset{h\to 0}{\mathrm{lim}}\frac{{\left(a+h\right)}^{2}\mathrm{sin}\left(a+h\right)-{a}^{2}\mathrm{sin}a}{h}$

$\underset{h\to 0}{\mathrm{lim}}\frac{{\left(a+h\right)}^{2}\mathrm{sin}\left(a+h\right)-{a}^{2}\mathrm{sin}a}{h}\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}\frac{\left({a}^{2}+2ah+{h}^{2}\right)\mathrm{sin}\left(a+h\right)-{a}^{2}\mathrm{sin}a}{h}\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}\frac{\left(2ah+{h}^{2}\right)\mathrm{sin}\left(a+h\right)+{a}^{2}\mathrm{sin}\left(a+h\right)-{a}^{2}\mathrm{sin}a}{h}\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}\frac{\left(2ah+{h}^{2}\right)\mathrm{sin}\left(a+h\right)}{h}+\underset{h\to 0}{\mathrm{lim}}\frac{{a}^{2}\mathrm{sin}\left(a+h\right)-{a}^{2}\mathrm{sin}a}{h}\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}\left(2a+h\right)\mathrm{sin}\left(a+h\right)+{a}^{2}\underset{h\to 0}{\mathrm{lim}}\frac{\mathrm{sin}\left(a+h\right)-\mathrm{sin}a}{h}\phantom{\rule{0ex}{0ex}}$

If  find k.

Question 4:

Evaluate the following limits:

$\underset{x\to \frac{\mathrm{\pi }}{3}}{\mathrm{lim}}\frac{\sqrt{1-\mathrm{cos}6x}}{\sqrt{2}\left(\frac{\mathrm{\pi }}{3}-x\right)}$

Question 19:

where f(x) = sin 2x

Question 23:

$\underset{x\to 1}{\mathrm{lim}}\left[\frac{1-\frac{1}{x}}{\mathrm{sin\pi }\left(x-1\right)}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 1}{\mathrm{lim}}\left[\frac{x-1}{x\mathrm{sin\pi }\left(x-1\right)}\right]$

Let y = x – 1
If x → 1, then y → 0.

Question 30:

Rationalising the denominator, we get:

Question 38:

Evaluate the following limits:

$\underset{x\to \mathrm{\pi }}{\mathrm{lim}}\frac{1-\mathrm{sin}\frac{x}{2}}{\mathrm{cos}\frac{x}{2}\left(\mathrm{cos}\frac{x}{4}-\mathrm{sin}\frac{x}{4}\right)}$                                         [NCERT EXEMPLAR]

$\underset{x\to \mathrm{\pi }}{\mathrm{lim}}\frac{1-\mathrm{sin}\frac{x}{2}}{\mathrm{cos}\frac{x}{2}\left(\mathrm{cos}\frac{x}{4}-\mathrm{sin}\frac{x}{4}\right)}$

Put $x=\mathrm{\pi }+h$

When

$\therefore \underset{x\to \mathrm{\pi }}{\mathrm{lim}}\frac{1-\mathrm{sin}\frac{x}{2}}{\mathrm{cos}\frac{x}{2}\left(\mathrm{cos}\frac{x}{4}-\mathrm{sin}\frac{x}{4}\right)}\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}\frac{1-\mathrm{sin}\left(\frac{\mathrm{\pi }+h}{2}\right)}{\mathrm{cos}\left(\frac{\mathrm{\pi }+h}{2}\right)\left[\mathrm{cos}\left(\frac{\mathrm{\pi }+h}{4}\right)-\mathrm{sin}\left(\frac{\mathrm{\pi }+h}{4}\right)\right]}\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}\frac{1-\mathrm{sin}\left(\frac{\mathrm{\pi }}{2}+\frac{h}{2}\right)}{\mathrm{cos}\left(\frac{\mathrm{\pi }}{2}+\frac{h}{2}\right)\left[\mathrm{cos}\left(\frac{\mathrm{\pi }}{4}+\frac{h}{4}\right)-\mathrm{sin}\left(\frac{\mathrm{\pi }}{4}+\frac{h}{4}\right)\right]}\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}\frac{1-\mathrm{cos}\left(\frac{h}{2}\right)}{-\mathrm{sin}\left(\frac{h}{2}\right)\left[\left(\mathrm{cos}\frac{\mathrm{\pi }}{4}\mathrm{cos}\frac{h}{4}-\mathrm{sin}\frac{\mathrm{\pi }}{4}\mathrm{sin}\frac{h}{4}\right)-\left(\mathrm{sin}\frac{\mathrm{\pi }}{4}\mathrm{cos}\frac{h}{4}+\mathrm{cos}\frac{\mathrm{\pi }}{4}\mathrm{sin}\frac{h}{4}\right)\right]}$

Question 5:

Let x = π $-$ h
when x → π, then h → 0

Question 1:

Rationalising the denominator, we get:

Question 2:

Dividing the numerator and the denominator by x:

Let x = 2 + h

x → 2
h → 0

Question 16:

Dividing the numerator and the denominator by x:

x → 0
∴ sin x → 0

Let y=sin x

x → 0
y → 0