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]

Sol. (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]

Sol. (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]

Sol. (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]

Sol. (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]

Sol. (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]

Sol. (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]

Sol. (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]

Sol. (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]

Sol. (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]

Sol. (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]

Sol. (3)


Administrator

Leave a comment

Please enter comment.
Please enter your name.


Comments
  • May 22, 2020 at 9:49 am

    Chill bro

  • A
    April 29, 2020 at 4:03 pm

    Why problem in question upon question boss

  • Ran
    March 25, 2020 at 11:08 pm

    No answer description