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Q. Let $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ be a positive increasing function with $\lim _{x \rightarrow \infty} \frac{\mathrm{f}(3 \mathrm{x})}{\mathrm{f}(\mathrm{x})}=1 .$ Then $\lim _{\mathrm{x} \rightarrow \infty} \frac{\mathrm{f}(2 \mathrm{x})}{\mathrm{f}(\mathrm{x})}=$
(1) 1
(2) $\frac{2}{3}$
(3) $\frac{3}{2}$
(4) 3
[AIEEE-2010]
Ans. (1)
$\mathrm{f}(\mathrm{x})$ is a positive increasing function
$\Rightarrow 0<\mathrm{f}(\mathrm{x})<\mathrm{f}(2 \mathrm{x})<\mathrm{f}(3 \mathrm{x})$
$\Rightarrow 0<1<\frac{\mathrm{f}(2 \mathrm{x})}{\mathrm{f}(\mathrm{x})}<\frac{\mathrm{f}(3 \mathrm{x})}{\mathrm{f}(\mathrm{x})}$
$\Rightarrow \lim _{x \rightarrow \infty} 1 \leq \lim _{x \rightarrow \infty} \frac{f(2 x)}{f(x)} \leq \lim _{x \rightarrow \infty} \frac{f(3 x)}{f(x)}$
By sandwich theorem.
$\Rightarrow \lim _{x \rightarrow \infty} \frac{f(2 x)}{f(x)}=1$
Q. $\lim _{x \rightarrow 2}\left(\frac{\sqrt{1-\cos \{2(x-2)\}}}{x-2}\right)$
(1) equals $-\sqrt{2}$
(2) equals $\frac{1}{\sqrt{2}}$
(3) does not exist
(4) equals $\sqrt{2}$
[AIEEE-2011]
Ans. (3)
$\lim _{x \rightarrow 2} \frac{\sqrt{1-\cos 2(x-2)}}{(x-2)}=\lim _{x \rightarrow 2} \frac{\sqrt{2 \sin ^{2}(x-2)}}{(x-2)}$
$=\lim _{x \rightarrow 2} \frac{\sqrt{2}|\sin (x-2)|}{(x-2)}$
$\mathrm{RHL}$ at $\mathrm{x}=2, \lim _{\mathrm{h} \rightarrow 0} \frac{\sqrt{2}|\sin (2+\mathrm{h}-2)|}{(2+\mathrm{h})-2}=\lim _{\mathrm{h} \rightarrow 0} \frac{\sqrt{2}|\sinh |}{\mathrm{h}}$
$=\lim _{h \rightarrow 0} \frac{\sqrt{2} \sinh }{-h}=-\sqrt{2}$
LHL at $\mathrm{x}=2, \lim _{\mathrm{h} \rightarrow 0} \frac{\sqrt{2}|\sin (2-\mathrm{h}-2)|}{(2-\mathrm{h})-2}$
$=\lim _{h \rightarrow 0} \frac{\sqrt{2}|\sin (-h)|}{-h}=\lim _{h \rightarrow 0} \frac{\sqrt{2} \sinh }{-h}=-\sqrt{2}$
$\because \mathrm{LHL} \neq \mathrm{RHL}$
Hence, limit does not exist.
Q. Let $f: \mathrm{R} \rightarrow[0, \infty)$ be such that $\lim _{x \rightarrow 5} f(x)$ exists and $\lim _{x \rightarrow 5} \frac{(f(x))^{2}-9}{\sqrt{|x-5|}}=0 .$ Then $\lim _{x \rightarrow 5} \operatorname{Lim}_{x \rightarrow 5}(x)$ equal –
(1) 3 (2) 0 (3) 1 (4) 2
[AIEEE-2011]
Ans. (1)
$\lim _{x \rightarrow 5} \frac{(f(x))^{2}-9}{\sqrt{|x-5|}}=0$
$\therefore \quad$ Question must be in $\frac{0}{0}$ form
$\therefore \quad(f(5))^{2}-9=0$
$\Rightarrow \quad f(5)=3$
Q. $\lim _{x \rightarrow 0} \frac{\sin \left(\pi \cos ^{2} x\right)}{x^{2}}$ is equal to:
(1) $\frac{\pi}{2}$
(2) 1
$(3)-\pi$
(4)$\pi$
[JEE Mains Offline-2014]
Ans. (4)
$\lim _{x \rightarrow 0} \frac{\sin \left(\pi \cos ^{2} x\right)}{x^{2}}$
$\Rightarrow \lim _{x \rightarrow 0} \frac{\sin \left(\pi-\pi \sin ^{2} x\right)}{x^{2}}$
$\Rightarrow \lim _{x \rightarrow 0} \frac{\sin \left(\pi \sin ^{2} x\right)}{\pi \sin ^{2} x} \times \pi \frac{\sin ^{2} x}{x^{2}}$
$\Rightarrow \lim _{x \rightarrow 0} \frac{\sin \left(\pi \sin ^{2} x\right)}{\pi \sin ^{2} x} \times \pi \times \lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)^{2}$
$\Rightarrow 1 \times \pi \times 1$
$=\pi$
Q. If $\lim _{x \rightarrow 2} \frac{\tan (x-2)\left\{x^{2}+(k-2) x-2 k\right\}}{x^{2}-4 x+4}=5$ then $k$ is equal to
(1) 3 (2) 1 (3) 0 (4) 2
[JEE Mains Online-2014]
Ans. (1)
If $\lim _{x \rightarrow 2} \frac{\tan (x-2)\left[x^{2}+k x-2 k-2 x\right]}{(x-2)^{2}}=5$
$\lim _{x \rightarrow 2}\left(\frac{\tan (x-2)}{(x-2)}\right) \frac{(x+k)(x-2)}{(x-2)}=5$
1. (2 + k) = 5
K = 3
Q. Let $\mathrm{p}=\lim _{x \rightarrow 0+}\left(1+\tan ^{2} \sqrt{\mathrm{x}}\right)^{\frac{1}{2 \mathrm{x}}}$ then log $\mathrm{p}$ is equal to :-
(1) $\frac{1}{4}$
(2) 2
(3) 1
(4) $\frac{1}{2}$
[JEE Mains -2016]
Ans. (4)
$\mathrm{p}=\mathrm{e}^{\mathrm{x} \rightarrow 0^{\frac{1}{2}}\left(\frac{\tan \sqrt{x}}{\sqrt{x}}\right)^{2}}=\sqrt{\mathrm{e}}$
$\log \mathrm{p}=\frac{1}{2}$
Q. $\lim _{n \rightarrow \infty}\left(\frac{(n+1)(n+2) \ldots .3 n}{n^{2 n}}\right)^{1 / n}$ is equal to :-
(1) $3 \log 3-2$
(2) $\frac{18}{\mathrm{e}^{4}}$b
(3) $\frac{27}{\mathrm{e}^{2}}$
(4) $\frac{9}{\mathrm{e}^{2}}$
[JEE Mains 2016]
Ans. (3)
Q. $\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cot x-\cos x}{(\pi-2 x)^{3}}$ equals :-
(1) $\frac{1}{4}$
(2) $\frac{1}{24}$
(3) $\frac{1}{16}$
(4) $\frac{1}{8}$
[JEE Mains -2017]
Ans. (3)
$\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cot x(1-\sin x)}{-8\left(x-\frac{\pi}{2}\right)^{3}}$
$=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\tan \left(\frac{\pi}{2}-x\right)}{8\left(\frac{\pi}{2}-x\right)} \frac{\left(1-\cos \left(\frac{\pi}{2}-x\right)\right)}{\left(\frac{\pi}{2}-x\right)^{2}}$
$=\frac{1}{8} \cdot 1 \cdot \frac{1}{2}=\frac{1}{16}$
Q. For each $\mathrm{t} \in \mathrm{R},$ let $[\mathrm{t}]$ be the greatest integer less than or equal to t. Then
$\lim _{\mathrm{x} \rightarrow 0+} \mathrm{x}\left(\left[\frac{1}{\mathrm{x}}\right]+\left[\frac{2}{\mathrm{x}}\right]+\ldots \ldots+\left[\frac{15}{\mathrm{x}}\right]\right)$\
(1) is equal to 15.
(2) is equal to 120.
(3) does not exist (in R).
(4) is equal to 0.
[JEE Mains -2018]
Ans. (2)
Q. $\lim _{x \rightarrow 0} \frac{(27+x)^{1 / 3}-3}{9-(27+x)^{2 / 3}}$ equals :
$(1)-\frac{1}{3}$
(2) $\frac{1}{6}$
$(3)-\frac{1}{6}$
(4) $\frac{1}{3}$
[JEE Mains -2018]
Ans. (3)
Q. $\lim _{x \rightarrow 0} \frac{x \tan 2 x-2 x \tan x}{(1-\cos 2 x)^{2}}$ equals :-
$(1)-\frac{1}{2}$
(2) $\frac{1}{4}$
(3) $\frac{1}{2}$
(4) 1
[JEE Mains -2018]
Ans. (3)
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Aug. 16, 2020, 9:06 p.m.
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