datavorous/entrance-exam-dataset huggingface dataset:
$\lim _{x \rightarrow 0} \frac{x \tan 2 x-2 x \tan x}{(1-\cos 2 x)^2}$ equals.

Options:
A
1

B
$-\frac{1}{2}$

C
$\frac{1}{4}$

D
$\frac{1}{2}$

The correct answer is:
$\frac{1}{2}$

Solution:
Let,
$$
\begin{aligned}
&L=\lim _{x \rightarrow 0} \frac{(x \tan 2 x-2 x \tan x)}{(1-\cos 2 x)^2}=\lim _{x \rightarrow 0} K \text { (say) } \\
&\Rightarrow K=\frac{x\left[\frac{2 \tan x}{1-(\tan x)^2}\right]-2 x \tan x}{\left(1-\left(1-2 \sin ^2 x\right)\right)^2} \\
&=\frac{2 x \tan x-\left[2 x \tan x-2 x \tan ^3 x\right]}{4 \sin ^4 x \times\left(1-\tan ^2 x\right)} \\
&=\frac{2 x \tan ^3 x}{4 \sin ^4 x \times\left(1-\tan ^2 x\right)} \\
&=\frac{2 x \tan ^3 x}{4 \sin ^4 x \times\left(\frac{\cos ^2 x-\sin ^2 x}{\cos ^2 x}\right)}
\end{aligned}
$$
$$
\begin{aligned}
&=\frac{2 x \frac{\sin ^3 x}{\cos ^3 x}}{4 \sin ^4 x \times\left(\frac{\cos ^2 x-\sin ^2 x}{\cos ^2 x}\right)} \\
& \Rightarrow K=\frac{x}{2 \sin x \times\left(\cos ^2 x-\sin ^2 x\right) \cos x} \\
\therefore L &=\lim _{x \rightarrow 0} \frac{x}{2 \sin x} \times \lim _{x \rightarrow 0} \frac{1}{\cos x\left(\cos ^2 x-\sin ^2 x\right)} \\
&=\lim _{x \rightarrow 0} \frac{x}{2 \sin x} \times \lim _{x \rightarrow 0} \frac{1}{\cos 0\left(\cos ^2 0-\sin ^2 0\right)} \\
&=\frac{1}{2}
\end{aligned}
$$

Topic: Mathematics, Limits, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
$\lim _{x \rightarrow 0} \frac{1-\cos m x}{1-\cos n x}=$

Options:
A $\mathrm{m} / \mathrm{n}$
B $n / m$
C $\frac{m^2}{n^2}$
D $\frac{n^2}{m^2}$

The correct answer is:
$\frac{m^2}{n^2}$

Solution:
$\begin{gathered}
\lim _{x \rightarrow 0} \frac{1-\cos m x}{1-\cos n x}=\lim _{x \rightarrow 0}\left\{\frac{2 \sin ^2 \frac{m x}{2}}{2 \sin ^2 \frac{n x}{2}}\right\} \\
=\lim _{x \rightarrow 0}\left[\left\{\frac{\sin \frac{m x}{2}}{\frac{m x}{2}}\right\}^2 \frac{m^2 x^2}{4} \cdot \frac{1}{\left\{\frac{\sin \frac{n x}{2}}{\frac{n x}{2}}\right\}^2} \cdot \frac{4}{n^2 x^2}\right] \\
=\frac{m^2}{n^2} \times 1=\frac{m^2}{n^2} .
\end{gathered}$
Aliter : Apply L-Hospital's rule,
$\lim _{x \rightarrow 0} \frac{1-\cos m x}{1-\cos n x}=\lim _{x \rightarrow 0} \frac{m \sin m x}{n \sin n x}=\lim _{x \rightarrow 0} \frac{m^2 \cos m x}{n^2 \cos n x}=\frac{m^2}{n^2} .$

Topic: Mathematics, Limits, JEE Main
$\lim _{x \rightarrow 0} \frac{x\left(e^x-1\right)}{1-\cos x}=$

Options:
A $0$
B $\infty$
C $-2$
D $2$

The correct answer is:
$2$

Solution:
$\begin{aligned} \lim _{x \rightarrow 0} \frac{x\left(e^x-1\right)}{1-\cos x} & =\lim _{x \rightarrow 0} \frac{2 x\left(e^x-1\right)}{4 \cdot \sin ^2 \frac{x}{2}} =2 \lim _{x \rightarrow 0}\left[\frac{(x / 2)^2}{\sin ^2 \frac{x}{2}}\right]\left(\frac{e^x-1}{x}\right)=2 .\end{aligned}$

Topic: Mathematics, Limits, JEE Main
$\lim _{x \rightarrow 0} \frac{x\left(e^x-1\right)}{1-\cos x}=$

Options:
A $0$
B $\infty$
C $-2$
D $2$

The correct answer is:
$2$

Solution:
$\begin{aligned} \lim _{x \rightarrow 0} \frac{x\left(e^x-1\right)}{1-\cos x} & =\lim _{x \rightarrow 0} \frac{2 x\left(e^x-1\right)}{4 \cdot \sin ^2 \frac{x}{2}} \\ & =2 \lim _{x \rightarrow 0}\left[\frac{(x / 2)^2}{\sin ^2 \frac{x}{2}}\right]\left(\frac{e^x-1}{x}\right)=2 .\end{aligned}$

Topic: Mathematics, Limits, JEE Main
limx→01-cos2(3x)cos3(4x)sin3(4x)loge2x+15 is equal to

Options:
A 15
B 9
C 18
D 24

The correct answer is:
18

Solution:
Given,

limx→01-cos2(3x)cos3(4x)sin3(4x)loge2x+15

Now we now that,

limx→0sinxx=1, limx→01-cosxx2=12 & limx→0log1+xx=1

Now using the above formula we get,

limx→01-cos23xcos34xsin34xloge2x+15

=limx→01-cos3x1+cos3x9x2cos34x9x2sin4x364x32x564x3loge(2x+1)52x5

=limx→01-cos3x9x21+cos3xcos34xsin4x4x3loge(2x+1)2x5×9×6432

=12×211315×9×6432

=9×21=18

Topic: Mathematics, Limits, JEE Main, JEE Main 2023 (08 Apr Shift 1)
limx→0⁡1-cos⁡2x22xtan⁡x-xtan⁡2x is

Options:
A 2
B -12
C -2
D 12

The correct answer is:
-2

Solution:
limx→0⁡1-cos⁡2x22xtan⁡x-xtan⁡2x
= limx→0⁡2sin2⁡x22xx+x33+2x515+…-x2x+23 x33+225 x515+…
= limx→0⁡4x-x33!+x55!-….4x423-83+x6415-6415

(dividing numerator and denominator by x4 )
= limx→0⁡41-x23!+x45!-….4-2+x2-6015+…
= -2

Topic: Mathematics, Limits, JEE Main, JEE Main 2016 (10 Apr Online)
limx→01-cos2x3+cosxxtan4x=

Options:
A 1 2
B 4
C 3
D 2

The correct answer is:
2

Solution:
limx→0⁡(1-cos⁡2x) (3+cos⁡x)xtan⁡4x
= limx→0⁡2sin2⁡x ⋅(3+cos⁡x)x⋅tan⁡4x
= limx→0⁡2⋅sin⁡xx2⋅3+cosxtan⁡4x4x ⋅4
= 2⋅12 ⋅(3+1)1 ⋅4
=2 .

Topic: Mathematics, Limits, JEE Main, JEE Main 2015 (04 Apr)
$\lim _{x \rightarrow 0}\left(\frac{1+\tan x}{1+\sin x}\right)^{\operatorname{cosec} x}$ is equal to

Options:
A $e$
B $\frac{1}{e}$
C $1$
D None of these

The correct answer is:
$1$

Solution:
$\begin{aligned} & \text { Given limit }=\lim _{x \rightarrow 0}\left[(1+\tan x)^{\operatorname{cosec} x} \times 1 /(1+\sin x)^{\operatorname{cosec} x}\right] \\ & \left.=\lim _{x \rightarrow 0}\left[\{1+\tan x)^{\cot x}\right\}^{\sec x} \times\left\{1 /(1+\sin x)^{\operatorname{cosec} x}\right\}\right] \\ & =e^{\sec (0)} \cdot \frac{1}{e}=e \cdot \frac{1}{e}=1\end{aligned}$

Topic: Mathematics, Limits, JEE Main
$\lim _{x \rightarrow 0} \frac{\pi^{x}-1}{\sqrt{1+x}-1}$

Options:
A does not exist
B equals $\log _{e}\left(\pi^{2}\right)$
C equals 1
D lies between 10 and 11

The correct answer is:
equals $\log _{e}\left(\pi^{2}\right)$

Solution:
$\lim _{x \rightarrow 0} \frac{\pi^{x}-1}{\sqrt{1+x}-1}$
$\frac{0}{0}$ form
$=\lim _{x \rightarrow 0} \frac{\pi^{x} \log _{e} \pi}{\frac{1}{2 \sqrt{1+x}}}$
$=\lim _{x \rightarrow 0} 2 \sqrt{1+x}\left(\pi^{x} \log _{e} \pi\right)$
$=2 \sqrt{1}\left(\pi^{0} \log _{e} \pi\right)=2 \log _{e} \pi=\log _{e} \pi^{2}$

Topic: Mathematics, Limits, JEE Main
$\lim _{x \rightarrow 0} \frac{\sqrt{1+x^2}-\sqrt{1-x+x^2}}{3^x-1}$ is equal to

Options:
A $\frac{1}{\log _e 3}$
B $\log _e 9$
C $\frac{1}{\log _e 9}$
D $\log _e 3$

The correct answer is:
$\frac{1}{\log _e 9}$

Solution:
$\begin{aligned} & \lim _{x \rightarrow 0} \frac{\sqrt{1+x^2}-\sqrt{1-x+x^2}}{3^x-1} \\ & =\lim _{x \rightarrow 0} \frac{\sqrt{1+x^2}-\sqrt{1-x+x^2}}{3^x-1} \\ & \times \frac{\sqrt{1+x^2}+\sqrt{1-x+x^2}}{\sqrt{1+x^2}+\sqrt{1-x+x^2}} \\ & =\lim _{x \rightarrow 0} \frac{\left(1+x^2\right)-\left(1-x+x^2\right)}{3^x-1\left(\sqrt{1-x^2}+\sqrt{1-x+x^2}\right)} \\ & =\lim _{x \rightarrow 0} \frac{x}{\left(3^x-1\right)} \times \frac{1}{\left(\sqrt{1-x^2}+\sqrt{1-x+x^2}\right)} \\ & =\lim _{x \rightarrow 0} \frac{1}{\frac{3^x-1}{x}} \times \frac{1}{\sqrt{1-x^2}+\sqrt{1-x+x^2}} \\ & =\log _e \frac{1}{3} \times \frac{1}{\sqrt{1-0}+\sqrt{1-0+0}} \\ & {\left[\because \log \frac{\left(a^x-1\right)}{x}=1\right]} \\ & =\frac{1}{2 \log _e 3}=\frac{1}{\log _e 9} \\ & \end{aligned}$

Topic: Mathematics, Limits, JEE Main
limx→03x2+27x2+21x2 is equal to

Options:
A 1e
B 1e2
C e2
D e

The correct answer is:
1e2

Solution:
Let L=limx→03x2+27x2+21x2

=elimx→01x23x2+27x2+2-1

=elimx→0-4x27x2+2

=e-42

=e-2

=1e2

Topic: Mathematics, Limits, JEE Main, JEE Main 2020 (08 Jan Shift 1)
limx→0(27+x)13-39-(27+x)23 equals

Options:
A -16
B 16
C 13
D -13

The correct answer is:
-16

Solution:
Given limit is limx→027+x13-3 9-27+x23 00 form
Applying L'Hospital's rule,
⇒limx→0 1327+x-23 -2327+x-13=27-23-227-13
=-122713 =-16.

Topic: Mathematics, Limits, JEE Main, JEE Main 2018 (16 Apr Online)
limx→048x4∫0xt3t6+1dt is equal to

Options:

The correct answer is:
12

Solution:
Let

L=limx→048∫0xt31+t6dtx4→00

Applying L'hospital rule, we get

L=limx→048x31+x64x3

⇒L=limx→0121+x6

⇒L=12

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2023 (30 Jan Shift 1)
$\lim _{x \rightarrow 0} \frac{a^{\sin x}-1}{b^{\sin x}-1}=$

Options:
A $\frac{a}{b}$
B $\frac{b}{a}$
C $\frac{\log a}{\log b}$
D $\frac{\log b}{\log a}$

The correct answer is:
$\frac{\log a}{\log b}$

Solution:
$\begin{aligned} \lim _{x \rightarrow 0} \frac{a^{\sin x}-1}{b^{\sin x}-1} & =\lim _{x \rightarrow 0} \frac{a^{\sin x}-1}{\sin x} \times \frac{\sin x}{b^{\sin x}-1} \\ & =\log _e a \times \frac{1}{\log _e b}=\frac{\log a}{\log b}\end{aligned}$

Topic: Mathematics, Limits, JEE Main
$\lim _{x \rightarrow 0^{+}} \frac{x \sin ^{-1}\left(\frac{2 x}{1+x^2}\right)}{\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right) \tan ^{-1}\left(\frac{3 x-x^3}{1-3 x^2}\right)}$ is equal to

Options:
A $\frac{1}{2}$
B $\frac{1}{3}$
C $\frac{1}{4}$
D $\frac{1}{6}$

The correct answer is:
$\frac{1}{3}$

Solution:
$$
\lim _{x \rightarrow 0^{+}} \frac{x \sin ^{-1}\left(\frac{2 x}{1+x^2}\right)}{\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right) \cdot \tan ^{-1}\left(\frac{3 x-x^3}{1-3 x^2}\right)}
$$
$\because$ We know that $x \rightarrow 0^{+}$
$$
\begin{aligned}
& 2 \tan ^{-1} x=\sin ^{-1} \frac{2 x}{1+x^2}=\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right) \\
& 3 \tan ^{-1} x=\tan ^{-1}\left(\frac{3 x-x^3}{1-3 x^2}\right)
\end{aligned}
$$
$\therefore$ Limit becomes
$$
\lim _{x \rightarrow 0^{+}} \frac{x \cdot\left(2 \tan ^{-1} x\right)}{\left(2 \tan ^{-1} x\right)\left(3 \tan ^{-1} x\right)}
$$
$$
\Rightarrow \lim _{x \rightarrow 0^{+}} \frac{x}{3 \tan ^{-1} x}
$$
Using L Hospital rule
$$
\lim _{x \rightarrow 0^{+}}=\frac{1}{3 \times \frac{1}{1+x^2}}=\frac{1}{3 \times\left(\frac{1}{1+0}\right)}=\frac{1}{3}
$$

Topic: Mathematics, Limits, JEE Main
$\lim _{x \rightarrow 0}(\operatorname{cosec} x)^{1 / \log x}$ is equal to :

Options:
A 0
B 1
C $\frac{1}{e}$
D None of these

The correct answer is:
$\frac{1}{e}$

Solution:
Let $y=\lim _{x \rightarrow 0}(\operatorname{cosec} x)^{1 / \log x}$

Taking $\log$ on both sides, we get

$\log y=\lim _{x \rightarrow 0} \frac{\log \operatorname{cosec} x}{\log x}\left[\frac{\infty}{\infty}\right.$ form $]$

$=\lim _{x \rightarrow 0} \frac{-\cot x}{1 / x} \quad$ (ByL' Hospital rule)

$=-\lim _{x \rightarrow 0} \frac{x}{\tan x} \quad\left(\because \cot x=\frac{1}{\tan x}\right)$

$\Rightarrow \log y=-1$

$\Rightarrow y=e^{-1}=\frac{1}{e}$

Hence, required limit $=\frac{1}{e}$

Topic: Mathematics, Limits, JEE Main
$\lim _{x \rightarrow 0} \frac{e^{1 / x}-1}{e^{1 / x}+1}=$

Options:
A $0$
B $1$
C $-1$
D Does not exist

The correct answer is:
Does not exist

Solution:
$f(x)=\left(\frac{e^{1 / x}-1}{e^{1 / x}+1}\right)$, then
$\lim _{x \rightarrow 0+} f(x)=\lim _{h \rightarrow 0}\left(\frac{e^{1 / h}-1}{e^{1 / h}+1}\right)=\lim _{h \rightarrow 0} \frac{e^{1 / h}\left(1-\frac{1}{e^{1 / h}}\right)}{e^{1 / h}\left(1+\frac{1}{e^{1 / h}}\right)}=1$
Similarly $\lim _{x \rightarrow 0-} f(x)=-1$. Hence limit does not exist.

Topic: Mathematics, Limits, JEE Main
$\lim _{x \rightarrow 0} \frac{a^x-b^x}{e^x-1}=$

Options:
A $\log \left(\frac{a}{b}\right)$
B $\log \left(\frac{b}{a}\right)$
C $\log (a b)$
D $\log (a+b)$

The correct answer is:
$\log \left(\frac{a}{b}\right)$

Solution:
$\begin{aligned}
& \qquad \lim _{x \rightarrow 0} \frac{a^x-b^x}{e^x-1}=\lim _{x \rightarrow 0} \frac{a^x-b^x}{x} \cdot \frac{x}{e^x-1} \\
& =\lim _{x \rightarrow 0}\left[\frac{a^x-1}{x}-\frac{b^x-1}{x}\right] \frac{x}{e^x-1} \\
& =\left(\log _e a-\log _e b\right) \cdot \frac{1}{\log _e e}=\log _e\left(\frac{a}{b}\right)
\end{aligned}$
Trick : Apply L-Hospital's rule.

Topic: Mathematics, Limits, JEE Main
limx→0⁡ex2-cos⁡xsin2⁡ x is equal to

Options:
A 2
B 32
C 54
D 3

The correct answer is:
32

Solution:
limx→0ex2-cosxsin2x=limx→0(ex2−1)+(1−cosx)x2·sin2xx2
=limx→0ex2−1x2+1−cosxx2
=1+12=32

Topic: Mathematics, Limits, JEE Main, JEE Main 2015 (10 Apr Online)
limxâ 0â¡ln1+xx2+x-1x=

Options:
A ∞
B 12
C -12
D 1

The correct answer is:
12

Solution:
limx→ 0⁡ln1+x+x2-xx2

using expansion, we know,
ln1+x=x - x22+x33……
⇒limx→0⁡x-x22+....+x2-xx2
=12

Topic: Mathematics, Limits, JEE Main
$\lim _{x \rightarrow 0} \frac{x \cot 4 x}{\sin ^2 x \cdot \cot ^2(2 x)}$ is equal to

Options:
A $0$
B $1$
C $4$
D $\frac{1}{4}$

The correct answer is:
$1$

Solution:
$\begin{aligned} \lim _{x \rightarrow 0} \frac{x \cot 4 x}{\sin ^2 x \cot ^2(2 x)} & =\lim _{x \rightarrow 0} \frac{x \tan ^2 2 x}{\sin ^2 x \tan 4 x} \\ & =\lim _{x \rightarrow 0} \frac{4\left(\frac{\tan 2 x}{2 x}\right)^2}{4\left(\frac{\sin x}{x}\right)^2\left(\frac{\tan 4 x}{4 x}\right)} \\ & =1\end{aligned}$

Topic: Mathematics, Limits, JEE Main
$\lim _{x \rightarrow 0} \frac{x \cot (4 x)}{\sin ^{2} x \cot ^{2}(2 x)}$ is equal to:

Options:
A 0
B 2
C 4
D 1

The correct answer is:
1

Solution:
$\lim _{x \rightarrow 0} \frac{x \cot 4 x}{\sin ^{2} x \cdot \cot ^{2} 2 x}=\lim _{x \rightarrow 0} \frac{x \cdot \tan ^{2} 2 x}{\sin ^{2} x \cdot \tan 4 x}$

$=\lim _{x \rightarrow 0}\left(\frac{x}{\sin x}\right)^{2} \cdot\left(\frac{\tan 2 x}{2 x}\right)^{2} \cdot\left(\frac{4 x}{\tan 4 x}\right) \cdot \frac{4}{2^{2}}=1$

Topic: Mathematics, Limits, JEE Main, JEE Main 2019 (11 Jan Shift 2)
$\lim _{x \rightarrow 0} \frac{\sin a x}{\sin b x}=$

Options:
A $a / b$
B $b / a$
C $1$
D None of these

The correct answer is:
$a / b$

Solution:
$\lim _{x \rightarrow 0} \frac{\sin a x}{\sin b x}=\lim _{x \rightarrow 0} \frac{a}{b} \frac{\sin a x}{a x} \frac{b x}{\sin b x}=\frac{a}{b}$.

Topic: Mathematics, Limits, JEE Main
limx→0sin2πcos4xx4 is equal to :

Options:
A 2π2
B π2
C 4π2
D 4π

The correct answer is:
4π2

Solution:
Given that

limx→0sin2π1-sin2x2x4
=limx→0sin2π1+sin4x-2sin2xx4

=limx→0sin2π-π2sin2x-sin4xx4
=limx→0sin2π2sin2x-sin4xx4=limx→0sinπ2sin2x-sin4xπ2sin2x-sin4x2·π22sin2x-sin4x2x4
When x → 0 ⇒2sin2x - sin4x → 0

Let k = 2sin2x - sin4x

=limk→0sinπk πk2.limx→0π2. sin4x 2 - sin2x2x4

We know that limx→0sinkxx = k.

=π2π2.π2.limx→0sinx x4.limx→02-sin2x2

=π2 . 1 . 22

=4π2.

Topic: Mathematics, Limits, JEE Main, JEE Main 2021 (31 Aug Shift 1)
limx→0sinπcos2xx2 is equal to

Options:
A - π
B Ï
C π 2
D 1

The correct answer is:
π

Solution:
We have limx→0sinπcos2xx2
=limx→0sinπ-πcos2xx2 ∵ sin π - θ = sin θ
=limx→0sinπ sin2xπ sin2x×π sin2xx2
=limx→0πsinπ sin2xπ sin2x sin x x 2
=π11 As, limx→0sin tt=1
= π .

Topic: Mathematics, Limits, JEE Main, JEE Main 2014 (06 Apr)
$\lim _{x \rightarrow 0} \frac{\sqrt{1+x \sin x}-\sqrt{\cos x}}{\tan ^2 2 x}=$

Options:
A 3
B $\frac{3}{2}$
C $\frac{3}{4}$
D $\frac{3}{16}$

The correct answer is:
$\frac{3}{16}$

Solution:
$\lim _{x \rightarrow 0} \frac{\sqrt{1+x \sin x}-\sqrt{\cos x}}{\tan ^2 2 x}\left(\frac{0}{0}\right.$ form $)$
On rationalising
$$
\begin{aligned}
& =\lim _{x \rightarrow 0} \frac{\sqrt{1+x \sin x}-\sqrt{\cos x}}{\tan ^2 2 x} \times \frac{\sqrt{1+x \sin x}+\sqrt{\cos x}}{\sqrt{1+x \sin x}+\sqrt{\cos x}} \\
& =\lim _{x \rightarrow 0} \frac{1+x \sin x-\cos x}{\tan ^2 2 x} \\
& \times \lim _{x \rightarrow 0} \frac{1}{\sqrt{1+x \sin x}+\sqrt{\cos x}} \\
& =\lim _{x \rightarrow 0} \frac{(1-\cos x)+x \sin x}{\tan ^2 2 x} \times \frac{1}{2} \\
& =\lim _{x \rightarrow 0} \frac{2 \sin ^2 \frac{x}{2}+2 x \sin \frac{x}{2} \cos \frac{x}{2}}{\tan ^2 2 x} \times \frac{1}{2} \\
& =\lim _{x \rightarrow 0} \frac{2 \sin ^2 \frac{x}{2}\left[1+\frac{x}{\tan \frac{x}{2}}\right]}{\tan ^2 2 x} \times \frac{1}{2} \\
& =\lim _{x \rightarrow 0} \frac{\frac{1}{4} \cdot\left(\frac{\sin \frac{x}{2}}{\frac{x}{2}}\right)^2\left[1+\frac{2 \cdot \frac{x}{2}}{\tan \frac{x}{2}}\right]}{4\left[\frac{\tan 2 x}{2 x}\right]^2} \\
&
\end{aligned}
$$
On applying limits, we get
$$
=\frac{\frac{1}{4} \times(1)^2 \times[1+2 \times 1]}{4(1)^2}=\frac{3}{16} \text {. }
$$

Topic: Mathematics, Limits, JEE Main
$\lim _{x \rightarrow 0} \frac{\sin m x}{\tan n x}=$

Options:
A $\frac{n}{m}$
B $\frac{m}{n}$
C $\mathrm{mn}$
D None of these

The correct answer is:
$\frac{m}{n}$

Solution:
m/n (formula).

Topic: Mathematics, Limits, JEE Main
$\lim _{x \rightarrow 0} \frac{e^{\tan x}-e^x}{\tan x-x}=$

Options:
A $1$
B $e$
C $e^{-1}$
D $0$

The correct answer is:
$1$

Solution:
$\begin{aligned} & \lim _{x \rightarrow 0} \frac{e^{\tan x}-e^x}{\tan x-x}=\lim _{x \rightarrow 0} \frac{e^x\left[e^{\tan x-x}-1\right]}{\tan x-x} \\ & =\lim _{x \rightarrow 0} e^x \cdot \lim _{x \rightarrow 0} \frac{e^{\tan x-x}-1}{\tan x-x}=e^0 \times 1=1 .\end{aligned}$

Topic: Mathematics, Limits, JEE Main
limx→0tanπ4+x1/x is equal to

Options:
A e
B 2
C 1
D e2

The correct answer is:
e2

Solution:
limx→0tanπ4+x1/x 1∞form

=elimx→0tanπ4+x-1x limx→afxgx=elimx→agxfx-1, limx→afx=1 & limx→agx=∞

=elimx→01+tanx1-tanx-1x

=elimx→02tanxx1-tanx

=elimx→0tanxx21-tanx

=e121-0=e2

Topic: Mathematics, Limits, JEE Main, JEE Main 2020 (02 Sep Shift 2)
$\lim _{x \rightarrow 0}\left\{\tan \left(\frac{\pi}{4}+x\right)\right\}^{1 / x}=$

Options:
A $1$
B $-1$
C $e^2$
D $e$

The correct answer is:
$e^2$

Solution:
$\begin{aligned} \text { Given limit } & =\lim _{x \rightarrow 0}\left(\frac{1+\tan x}{1-\tan x}\right)^{1 / x} \\ & =\lim _{x \rightarrow 0} \frac{\left\{(1+\tan x)^{1 / \tan x}\right\}^{(\tan x) / x}}{\left\{(1-\tan x)^{1 / \tan x}\right\}^{(\tan x) / x}}=\frac{e}{e^{-1}}=e^2 .\end{aligned}$

Topic: Mathematics, Limits, JEE Main
$\lim _{x \rightarrow 0}\left(\frac{1}{x} \ln \sqrt{\frac{1+x}{1-x}}\right)$ is

Options:
A $1 / 2$
B 0
C 1
D does not exist

The correct answer is:
1

Solution:
$\lim _{x \rightarrow 0} \frac{1 / 2(\ln (1+x)-\ln (1-x))}{x}$
$=\frac{1}{2} \lim _{x \rightarrow 0}\left(\frac{1}{1+x}+\frac{1}{1-x}\right)=1$

Topic: Mathematics, Limits, JEE Main
limx→0x22x-x2sinx-x23x+cosx-3xcosx-1=

Options:
A 1log3log2-1
B 4log31-log2
C 4log3log2-1
D 2log3log2-1

The correct answer is:
2log3log2-1

Solution:
Given that: limx→0x22x-x2sinx-x23x+cosx-3xcosx-1

=limx→0x22x-1-x2sinx3x1-cosx-1-cosx

=limx→0x22x-1-sinx3x1-cosx-1-cosx

=limx→02x-1-sinx3x1-cosxx2-1-cosxx2

=limx→02x-1-sinx(3x-1)1-cosxx2

As we know, limx→01-cosxx2=12

=limx→02(2x-1-sinx)3x-1

Using L'Hospital Rule

=limx→02(2xlog2-cosx)3xlog3

=2log3(log2-1)

Topic: Mathematics, Limits, JEE Main
$\lim _{x \rightarrow 0} \frac{\cos a x-\cos b x}{x^{2}}$ is equal to

Options:
A $\frac{a^{2}-b^{2}}{2}$
B $\frac{b^{2}-a^{2}}{2}$
C $a^{2}-b^{2}$
D $b^{2}-a^{2}$

The correct answer is:
$\frac{b^{2}-a^{2}}{2}$

Solution:
$\lim _{x \rightarrow 0} \frac{\cos a x-\cos b x}{x^{2}}$
$=\lim _{x \rightarrow 0} \frac{-a \sin a x+b \sin b x}{2 x}$ (using L'Hospital's rule)
$=\lim _{x \rightarrow 0} \frac{-a^{2} \cos a x+b^{2} \sin b x}{2}$
$=\frac{b^{2}-a^{2}}{2}$ (using L'Hospital's rule)

Topic: Mathematics, Limits, JEE Main
$\lim _{x \rightarrow 0} \frac{\sin \left(\pi \cos ^2 x\right)}{x^2}$ equals

Options:
A
$-\pi$

B
1

C
$-1$

D
$\pi$

The correct answer is:
$\pi$

Solution:
Consider
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\sin \left(\pi \cos ^2 x\right)}{x^2} \\
& =\lim _{x \rightarrow 0} \frac{\sin \left(\pi-\pi \sin ^2 x\right)}{x^2} \\
& =\lim _{x \rightarrow 0} \frac{\sin \left(\pi \sin ^2 x\right)}{\pi \sin ^2 x} \times \frac{\left(\pi \sin ^2 x\right)}{x^2}=\pi
\end{aligned}
$$

Topic: Mathematics, Limits, JEE Main, JEE Main 2012 (26 May Online)
$\lim _{x \rightarrow 0} \frac{\sin \left(\pi \cos ^2 x\right)}{x^2}$ is equal to

Options:
A $-\pi$
B $\pi$
C $\frac{\pi}{2}$
D 1

The correct answer is:
$\pi$

Solution:
$\begin{aligned} & \lim _{x \rightarrow 0} \frac{\sin \left(\pi \cos ^2 x\right)}{x^2}=\lim _{\substack{x \rightarrow 0}} \frac{\sin \left(\pi\left\{1-\sin ^2 x\right\}\right)}{x^2} \\ & =\lim _{x \rightarrow 0} \frac{\sin \left(\pi-\pi \sin ^2 x\right)}{x^2}=\lim _{\substack{2}} \frac{\sin \left(\pi \sin ^2 x\right)}{x^2} \\ & =\lim _{x \rightarrow 0} \frac{\sin \left(\pi \sin ^2 x\right)}{\pi \sin ^2 x} \cdot \frac{\pi \sin ^2 x}{x^2}=\pi\end{aligned}$

Topic: Mathematics, Limits, JEE Main
$\lim _{x \rightarrow 0} \frac{x e^x-\log (1+x)}{x^2}$ equals

Options:
A $\frac{2}{3}$
B $\frac{1}{3}$
C $\frac{1}{2}$
D $\frac{3}{2}$

The correct answer is:
$\frac{3}{2}$

Solution:
$\text { Let } y=\lim _{x \rightarrow 0} \frac{x e^x-\log (1+x)}{x^2},\left(\frac{0}{0} \text { form }\right)$
Applying L-Hospital's rule,
$\begin{aligned}
& y=\lim _{x \rightarrow 0} \frac{e^x+x e^x-\frac{1}{1+x}}{2 x},\left(\frac{0}{0} \text { form }\right) \\
& y=\lim _{x \rightarrow 0} \frac{1}{2}\left[e^x+e^x+x e^x+\frac{1}{(1+x)^2}\right] \\
& y=\lim _{x \rightarrow 0} \frac{1}{2}[1+1+0+1]=\frac{3}{2}
\end{aligned}$

Topic: Mathematics, Limits, JEE Main
$\lim _{x \rightarrow 0} \frac{\tan ^3 x-\sin ^3 x}{x^5}$ is equal to

Options:
A $\frac{5}{2}$
B $\frac{3}{2}$
C $\frac{3}{5}$
D $\frac{2}{5}$

The correct answer is:
$\frac{3}{2}$

Solution:
$\begin{aligned} & \lim _{x \rightarrow 0} \frac{\tan ^3 x-\sin ^3 x}{x^5} \\ & =\lim _{x \rightarrow 0} \frac{\left[\begin{array}{c}\left(x+\frac{x^3}{3}+\frac{2}{15} x^5+\ldots\right)^3 \\ -\left(x-\frac{x^3}{3 !}+\frac{x^5}{5 !}+\ldots\right)^3\end{array}\right]}{x^5} \\ & =\lim _{x \rightarrow 0} \frac{\left[\begin{array}{c}\left(1+\frac{x^2}{2}+\frac{2}{15} x^4+\ldots\right)^3 \\ -\left(1-\frac{x^2}{3 !}+\frac{x^4}{5 !}+\ldots\right)^3\end{array}\right]}{x^2} \\ & \left\{\left(1+\frac{x^2}{3}+\frac{2}{15} x^4+\ldots\right)-\left(1-\frac{x^2}{3 !}+\frac{x^4}{5 !}+\ldots\right)\right\} \\ & =\lim _{x \rightarrow 0} \frac{\left\{\begin{array}{c}\left(1+\frac{x^2}{3}+\ldots\right)^2+\left(1-\frac{x^2}{3 !}+\ldots\right)^2 \\ +\left(1+\frac{x^2}{3}+\ldots\right)\left(1-\frac{x^2}{3 !}+\ldots\right)\end{array}\right\}}{x^2} \\ & \left.\text { [using }\left(a^3-b^3\right)=(a-b)\left(a^2+b^2+a b\right)\right] \\ & =\lim _{x \rightarrow 0} \frac{\left(\frac{x^2}{2}+\frac{x^4}{8}+\ldots\right)\left\{\begin{array}{c}\left.\left(1+\frac{x^2}{3}+\ldots\right)^2+\left(1-\frac{x^2}{3 !}+\ldots\right)^2\right\} \\ +\left(1+\frac{x^2}{3}+\ldots\right)\left(1-\frac{x^2}{3}+\ldots\right)\end{array}\right\}}{x^2} \\ & =\lim _{x \rightarrow 0}\left(\frac{1}{2}+\frac{x^2}{8}\right)\left\{\begin{array}{c}\left(1+\frac{x^2}{3}+\ldots\right)^2+\left(1-\frac{x^2}{3 !}+\ldots\right)^2 \\ +\left(1+\frac{x^2}{3}+\ldots\right)\left(1-\frac{x^2}{3 !}+\ldots\right)\end{array}\right\} \\ & \end{aligned}$
$\begin{aligned} & =\left(\frac{1}{2}+0\right)\left\{(1+0+\ldots)^2+(1-0+\ldots)^2\right. \\ & =\frac{1}{2}(1+1+1)=\frac{3}{2}\end{aligned}$

Topic: Mathematics, Limits, JEE Main
$\lim _{x \rightarrow 0}\left(\frac{a^x-b^x}{x}\right)=$

Options:
A $\log \left(\frac{b}{a}\right)$
B $\log \left(\frac{a}{b}\right)$
C $\frac{a}{b}$
D $\log a^b$

The correct answer is:
$\log \left(\frac{a}{b}\right)$

Solution:
$\begin{aligned} \lim _{x \rightarrow 0} \frac{a^x-b^x}{x} & =\lim _{x \rightarrow 0}\left(\frac{a^x-1}{x}\right)-\lim _{x \rightarrow 0}\left(\frac{b^x-1}{x}\right) \\ & =\log a-\log b=\log (a / b) .\end{aligned}$

Topic: Mathematics, Limits, JEE Main
$\lim _{x \rightarrow 0} \frac{e-(1+2 x)^{\frac{1}{2 x}}}{x}$ is equal to

Options:
A 0
B $\frac{-2}{e}$
C e
D $e-e^2$

The correct answer is:
e

Solution:
$\begin{aligned} & \operatorname{Lim}_{x \rightarrow 0} \frac{e-e^{\frac{1}{2 x} \ln (1+2 x)}}{x} \\ & =\operatorname{Lim}_{x \rightarrow 0}(-e) \frac{\left(e^{\frac{\ln (1+2 x)}{2 x}-1}-1\right)}{x} \\ & =\operatorname{Lim}_{x \rightarrow 0}(-e) \frac{\ln (1+2 x)-2 x}{2 x^2} \\ & =(-e) \times(-1) \frac{4}{2 \times 2}=e\end{aligned}$

Topic: Mathematics, Limits, JEE Main, JEE Main 2024 (09 Apr Shift 2)
limx→0xe1+x2+x4-1/x-11+x2+x4-1

Options:
A is equal to e
B is equal to 1
C is equal to 0
D does not exist

The correct answer is:
is equal to 1

Solution:
limx→0e1+x2+x4−1x-11+x2+x4−1x

put1+x2+x4−1x=t

clearly x→0⇒t→0

∴ given limit =limt→0et−1t=1

Topic: Mathematics, Limits, JEE Main, JEE Main 2020 (05 Sep Shift 2)
$\lim _{x \rightarrow 0} \frac{\log \cos x}{x}=$

Options:
A $0$
B $1$
C $\infty$
D Does not exist

The correct answer is:
$0$

Solution:
$\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\log \cos x}{x}=\lim _{x \rightarrow 0} \frac{\log (1-2 \sin^2 (\frac{x}{2})) }{x} \\
& =\lim _{x \rightarrow 0} \frac{-\left[2 \sin ^2 \frac{x}{2}+\left(\frac{2 \sin ^2 \frac{x}{2}}{2}\right)^2+\ldots \ldots .\right]}{x}=0
\end{aligned}$
Aliter : Apply L-Hospital's rule,
$\lim _{x \rightarrow 0} \frac{\log \cos x}{x}=\lim _{x \rightarrow 0} \frac{-\tan x}{1}=0 .$

Topic: Mathematics, Limits, JEE Main
$\operatorname{Lim}_{x \rightarrow 0} \frac{\log x^n-[x]}{[x]}, \mathrm{n} \in \mathrm{N}([\mathrm{x}]$ denotes greatest integer less than or equal to $\mathrm{x})$

Options:
A
has value -1

B
has value 0

C
has value 1

D
does not exist

The correct answer is:
does not exist

Solution:
Since $\operatorname{Lim}_{x \rightarrow 0}[x]$ does not exist, hence the required limit does not exist

Topic: Mathematics, Limits, JEE Main, JEE Main 2002
$\lim _{x \rightarrow 0} \frac{\sin |x|}{x}$ is equal to

Options:
A 1
B 0
C positive infinity
D does not exist

The correct answer is:
does not exist

Solution:
Hints: $\operatorname{Lim}_{x \rightarrow 0} \frac{\sin |x|}{x}$
LHL $=-1 \quad$ RHL $=1$
Limit does not exist

Topic: Mathematics, Limits, JEE Main