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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)

Sharath
May 12, 2024, 6:35 a.m.
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Feb. 7, 2024, 12:34 p.m.
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Vinay Patel
Nov. 29, 2023, 7:05 a.m.
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Nov. 29, 2023, 6:35 a.m.
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Oct. 18, 2023, 6:35 a.m.
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Vishal Rajbhar
Sept. 29, 2023, 10 p.m.
Vishal Rajbhar
Komal
Sept. 20, 2023, 6:35 a.m.
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Komal
Sept. 20, 2023, 6:35 a.m.
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Mukul Joshi
April 19, 2023, 6:35 a.m.
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Mariyam
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.
Nice questions🙋
Priya
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
Ayush
April 15, 2021, 5:56 p.m.
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April 12, 2021, 7:20 p.m.
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ULL THE BULL
March 8, 2021, 5:22 p.m.
Some questions dont have correct solutions
Rohini
Feb. 21, 2021, 11:19 a.m.
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pritam
Feb. 20, 2021, 11:02 a.m.
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Irene
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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Sean McLaughlin
Aug. 16, 2020, 9:07 p.m.
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whateverdoyoureallycare2323
Aug. 16, 2020, 9:06 p.m.
Vivek
Aug. 16, 2020, 9:06 p.m.
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Priyanka
Aug. 15, 2020, 12:33 p.m.
Thanks
Aug. 1, 2020, 12:22 p.m.
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Saptarshi Mukherjee
July 20, 2020, 12:18 p.m.
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B. Shivanireddy1
July 9, 2020, 2 p.m.
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Ayush
July 2, 2020, 7:26 a.m.
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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.