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limEvaluate:_{x--> -2}(x^{3}-7x - 6) / (x^{4}+5x - 6)f(x) = [x] sin(πx) atx = k, kis an integer is(a) (–1)

^{k}(k– 1)π (b) (–1)^{k – 1}(k– 1)π (c) (–1)^{k}kπ (d) (–1)^{k – 1}kπ^{- }e^{(1/xlogx)}1. g is not defined at x=0

2. g is not continuous at x=0

3. The limit of g(x) as x approaches 0 equals 1

4. g'(0)=1

$\left(1\right)(2x+1)(x-1{)}^{2}\phantom{\rule{0ex}{0ex}}(2)\frac{2x+1}{(1-x{)}^{2}}\phantom{\rule{0ex}{0ex}}(3)2x+1\phantom{\rule{0ex}{0ex}}(4)\frac{3-x}{(1-x{)}^{3}}$

(cosx +cos 3x + cos5x…cos(2n-1)x)/nwhere n is turning to infinity(n[a.0b.infinity c. 1 d.-1]^{p}sin^{2}(n!))/(n+1) n-> infinity^{-}x-[x] isA. lim h tends to 0 k-h-[k-h]

= what to do next sir

in book they are doing like this

= lim htends to 0 k-h-(k-1) = 1

how are they puting 1 over ther ? plz explain.

$Find\underset{x\to {2}^{-}}{\mathrm{lim}}\frac{\left[x\right]-x}{2-x},where\left[x\right]isthegreatestintegervalueofx.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}1.\u25cb-1/2\phantom{\rule{0ex}{0ex}}2.\u25cb1/2\phantom{\rule{0ex}{0ex}}3.\u25cb\infty \phantom{\rule{0ex}{0ex}}4.\u25cb-\infty $

(a) 0

(b) $\frac{64}{27}$

(c) $\frac{8}{3}$

(d) none of these

^{+ }(sinx)^{(tanx)}I don't know why it is a function as the graph is not continuous. For the value of x to be π/2 i.e., tan π/2, the value of the y goes to positive infinity from left-hand side and the negative infinity from the right-hand side. Moreover, there is asymptote at x = π/2, and thus even there does not exist any limit.

Q). $\underset{x\to 2}{\mathrm{lim}}\frac{{x}^{2}-3x+2}{x-2}$

1. 0

2. -2/3

3 3

4. 1

^{2}+x^{3}+...+x^{n}is equal to 1-x^{n}/1-x_{d}denotes the right hand derivative of f(x) at x = I, then isquestion no.8

$\underset{x\to \frac{\mathrm{\pi}}{4}}{\mathrm{lim}}\frac{1-\mathrm{tan}x}{1-\sqrt{2}\mathrm{sin}x}\phantom{\rule{0ex}{0ex}}\left(A\right)0\left(B\right)\frac{1}{2}\left(C\right)1\left(D\right)2$

$Q.1Thevalueof\underset{x\to 0}{L\mathrm{im}}\left({\mathrm{sin}}^{-1}\left[\mathrm{sin}x\right]+{\mathrm{cos}}^{-1}\left[\mathrm{cos}x\right]-2{\mathrm{tan}}^{-1}\left[\mathrm{tan}x\right]\right)isequalto\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}[Note:\left[k\right]denoteslargestintegerfunctionlessthanorequaltok.]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(A\right)\mathrm{\pi}\left(\mathrm{B}\right)\frac{\mathrm{\pi}}{2}\left(\mathrm{C}\right)\frac{3\mathrm{\pi}}{2}\left(\mathrm{D}\right)\mathrm{non}-\mathrm{existent}$

Q. Determine the intervals on which the given function is continuous:

$f\left(x\right)=\left\{\begin{array}{l}\frac{{x}^{2}+3x-10}{x-2},x\ne 2\\ 10,x\ne 2\end{array}\right.$

1. $\left(-\infty ,2\right)$

2. $\left(2,\infty \right)$

3. $\left(-\infty ,\infty \right)$

4. $\left(-\infty ,2\right)and\left(2,\infty \right)$

(a) 0

(b) $\frac{64}{27}$

(c) $\frac{8}{3}$

(d) none of these

$Find\frac{dy}{dx},ify=12\left(1-cott\right),x=10\left(t-\mathrm{sin}t\right),-\frac{\mathrm{\pi}}{2}t\frac{\mathrm{\pi}}{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}1.\u25cb\mathrm{tan}\frac{t}{4}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}2.\u25cb3\mathrm{cos}\frac{t}{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}3.\u25cb\frac{6}{5}cot\frac{t}{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}4.\u25cb2\mathrm{sin}t$

$f\left(x\right)=\left\{\begin{array}{l}\frac{{x}^{2}+3x-10}{x-2},x\ne 2\\ 10,x=2\end{array}\right.$

1. $\left(-\infty ,2\right)$

2. $\left(2,\infty \right)$

3. $\left(-\infty ,\infty \right)$

4. $\left(-\infty ,2\right)and\left(2,\infty \right)$

1. -1

2. 1

3. 2

4. -2

$4.\underset{x\to 0}{L\mathrm{im}}\frac{{27}^{x}-{9}^{x}-{3}^{x}+1}{\sqrt{2}-\sqrt{1+\mathrm{cos}x}}\phantom{\rule{0ex}{0ex}}\left(A\right)4\sqrt{2}{\left(\mathrm{ln}3\right)}^{2}\left(B\right)8\sqrt{2}(\mathrm{ln}3{)}^{2}(C)4\sqrt{2}\left(\mathrm{ln}3\right)(D)8\sqrt{2}\left(\mathrm{ln}3\right)$

$\left(1\right)-1/2\phantom{\rule{0ex}{0ex}}\left(2\right)1/2\phantom{\rule{0ex}{0ex}}\left(3\right)\infty \phantom{\rule{0ex}{0ex}}\left(4\right)-\infty $