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Limit - JEE Main Previous Year Question with Solutions

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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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Comments

Sharath
May 12, 2024, 6:35 a.m.
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K. Lalith Aditya
Oct. 10, 2024, 6:35 a.m.
Yes
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Nov. 29, 2023, 7:05 a.m.
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March 9, 2023, 6:49 p.m.
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bhisma
Aug. 14, 2021, 2:23 p.m.
to easy
Aarzoo
Aug. 9, 2021, 11:49 p.m.
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Priya
May 2, 2021, 12:20 p.m.
Level of questions are easy
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April 29, 2021, 6:01 p.m.
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Ayush
April 15, 2021, 5:56 p.m.
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D rugved
April 12, 2021, 7:20 p.m.
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ULL THE BULL
March 8, 2021, 5:22 p.m.
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Rohini
Feb. 21, 2021, 11:19 a.m.
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Feb. 20, 2021, 11:02 a.m.
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Feb. 14, 2021, 10:22 a.m.
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Rohith
Dec. 8, 2020, 11:51 a.m.
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Manoj
Aug. 31, 2020, 7:57 p.m.
Questions are overlapped we can't understand
Sam
Aug. 27, 2020, 2:20 p.m.
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Ali
Aug. 25, 2020, 2:54 p.m.
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Sean McLaughlin
Aug. 16, 2020, 9:08 p.m.
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Aug. 16, 2020, 9:07 p.m.
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Aug. 16, 2020, 9:06 p.m.
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Aug. 16, 2020, 9:06 p.m.
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Priyanka
Aug. 15, 2020, 12:33 p.m.
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alphadk
Aug. 1, 2020, 12:22 p.m.
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Saptarshi Mukherjee
July 20, 2020, 12:18 p.m.
Good collection of problems but very bad printing of questions.........you should look after it!!!!
B. Shivanireddy1
July 9, 2020, 2 p.m.
thank you , but if it is pdf it will be more clearity
Ayush
July 2, 2020, 7:26 a.m.
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Akhil
May 22, 2020, 9:49 a.m.
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A
April 29, 2020, 4:03 p.m.
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Ran
March 25, 2020, 11:08 p.m.
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