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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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Vinay Patel
Nov. 29, 2023, 7:05 a.m.
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Vinay Patel
Nov. 29, 2023, 6:35 a.m.
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Sept. 29, 2023, 10 p.m.
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Komal
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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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
these questions are very easy these were in very basic level
April 29, 2021, 6:01 p.m.
too easy
Ayush
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
ULL THE BULL
March 8, 2021, 5:22 p.m.
Some questions dont have correct solutions
Rohini
Feb. 21, 2021, 11:19 a.m.
Very useful 🤠👍
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.
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.
Manoj
Aug. 31, 2020, 7:57 p.m.
Questions are overlapped we can't understand
Sam
Aug. 27, 2020, 2:20 p.m.
Nice matter 🥰
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.
Thanks
alphadk
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 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.
Thanks for help
Akhil
May 22, 2020, 9:49 a.m.
Chill bro
A
April 29, 2020, 4:03 p.m.
Why problem in question upon question boss
Ran
March 25, 2020, 11:08 p.m.
No answer description