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Limit - JEE Main Previous Year Question with Solutions

JEE Main Previous Year Question of Math with Solutions are available at eSaral. Practicing JEE Main Previous Year Papers Questions of mathematics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas. eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects. Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more. Download eSaral app for free study material and video tutorials.
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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May 12, 2024, 6:35 a.m.
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Nov. 29, 2023, 7:05 a.m.
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Sept. 29, 2023, 10 p.m.
Vishal Rajbhar
Sept. 20, 2023, 6:35 a.m.
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Sept. 20, 2023, 6:35 a.m.
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April 19, 2023, 6:35 a.m.
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March 9, 2023, 6:49 p.m.
Aug. 14, 2021, 2:23 p.m.
to easy
Aug. 9, 2021, 11:49 p.m.
Nice questions🙋
May 2, 2021, 12:20 p.m.
Level of questions are easy
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April 29, 2021, 6:01 p.m.
too easy
April 15, 2021, 5:56 p.m.
I can't understand some answers
D rugved
April 12, 2021, 7:20 p.m.
level of questions is somewhat easy
March 8, 2021, 5:22 p.m.
Some questions dont have correct solutions
Feb. 21, 2021, 11:19 a.m.
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Feb. 20, 2021, 11:02 a.m.
Feb. 14, 2021, 10:22 a.m.
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Dec. 8, 2020, 11:51 a.m.
Hey! Thanks for the questions.But there are no solutions for some questions please look into those question kindly update answers it would be helpful Thank you.
Aug. 31, 2020, 7:57 p.m.
Questions are overlapped we can't understand
Aug. 27, 2020, 2:20 p.m.
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Aug. 25, 2020, 2:54 p.m.
V good revision
Sean McLaughlin
Aug. 16, 2020, 9:08 p.m.
Thanks man!
Sean McLaughlin
Aug. 16, 2020, 9:07 p.m.
Aug. 16, 2020, 9:06 p.m.
Aug. 16, 2020, 9:06 p.m.
For anyone having problems seeing the questions, right-click on the equation part, click "Show Math as", then click "MathML code". Copy all the text in the new window that opens with Ctrl+A. Now open a new tab and type in this URL: . And paste whatever you copied! Woila! Thank me later!
Aug. 15, 2020, 12:33 p.m.
Aug. 1, 2020, 12:22 p.m.
good collection of questions but poor quality
Saptarshi Mukherjee
July 20, 2020, 12:18 p.m.
Good collection of problems but very bad printing of should look after it!!!!
B. Shivanireddy1
July 9, 2020, 2 p.m.
thank you , but if it is pdf it will be more clearity
July 2, 2020, 7:26 a.m.
Thanks for help
May 22, 2020, 9:49 a.m.
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April 29, 2020, 4:03 p.m.
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March 25, 2020, 11:08 p.m.
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