datavorous/entrance-exam-dataset huggingface dataset:
$\lim _{x \rightarrow 0} \frac{|x|}{x}=$

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

The correct answer is:
Does not exist

Solution:
Since $\lim _{x \rightarrow 0-} \frac{|x|}{x}=-1$ and $\lim _{x \rightarrow 0+} \frac{|x|}{x}=1$, hence limit does not exist.

Topic: Mathematics, Limits, JEE Main
$\lim _{x \rightarrow 0} \frac{2 x}{|x|+x^2}=$

Options:
A Limit exists
B Limit does not exists
C $2$
D $-2$

The correct answer is:
Limit does not exists

Solution:
$\lim _{x \rightarrow 0} \frac{2 x}{|x|+x^2}$
L.H.L $\lim _{x \rightarrow 0^{-}} \frac{2 x}{-x+x^2}=\lim _{x \rightarrow 0^{-}} \frac{2}{-1+x}=-\infty$
R.H.L $\lim _{x \rightarrow 0^{+}} \frac{2 x}{x+x^2}=\lim _{x \rightarrow 0^{+}} \frac{2}{1+x}=\infty$
$\because$ L.H.L $\neq$ R.H.L $\Rightarrow$ limit does not exist

Topic: Mathematics, Limits, JEE Main
$\lim _{x \rightarrow 0} \frac{x}{|x|+x^2}=$

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

The correct answer is:
Does not exist

Solution:
$\begin{aligned}
& \lim _{x \rightarrow 0-} f(x)=\lim _{h \rightarrow 0} \frac{0-h}{h+h^2}=\lim _{h \rightarrow 0} \frac{-1}{1+h}=-1 \\
& \text { and } \lim _{x \rightarrow 0+} f(x)=\lim _{h \rightarrow 0} \frac{h}{h+h^2}=\lim _{h \rightarrow 0} \frac{1}{1+h}=1
\end{aligned}$
Hence limit does not exist.

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

Options:
A
equals 1

B
equals 0

C
does not exist

D
equals $-1$

The correct answer is:
equals 0

Solution:
Consider $\lim _{x \rightarrow 0}\left(\frac{x-\sin x}{x}\right) \sin \left(\frac{1}{x}\right)$
$=\lim _{x \rightarrow 0}\left[\frac{x\left(1-\frac{\sin x}{x}\right)}{x}\right] \times \lim _{x \rightarrow 0} \sin \left(\frac{1}{x}\right)$
$=\lim _{x \rightarrow 0}\left[1-\frac{\sin x}{x}\right] \times \lim _{x \rightarrow 0} \sin \left(\frac{1}{x}\right)$
$=\left[1-\lim _{x \rightarrow 0} \frac{\sin x}{x}\right] \times \lim _{x \rightarrow 0} \sin \left(\frac{1}{x}\right)$
$=0 \times \lim _{x \rightarrow 0} \sin \left(\frac{1}{x}\right)=0$

Topic: Mathematics, Limits, JEE Main, JEE Main 2012 (07 May Online)
$\lim _{x \rightarrow 0} x^x=$

Options:
A $0$
B $1$
C $e$
D None of these

The correct answer is:
$1$

Solution:
Let $y=x^x \Rightarrow \log y=x \log x$
$\lim _{y \rightarrow 0} \log y=\lim _{x \rightarrow 0} x \log x=0=\log 1 \Rightarrow \lim _{x \rightarrow 0} x^x=1$

Topic: Mathematics, Limits, JEE Main
$\lim _{x \rightarrow 0} \frac{y^2}{x}=\ldots \ldots \ldots$ where $y^2=a x+b x^2+c x^3$

Options:
A $0$
B $1$
C a
D None of these

The correct answer is:
a

Solution:
$\lim _{x \rightarrow 0} \frac{a x+b x^2+c x^3}{x}=\lim _{x \rightarrow 0} \frac{a+b x+c x^2}{1}=a$

Topic: Mathematics, Limits, JEE Main
limx→1∫0(x-1)2tcost2dt(x-1)sin(x-1)

Options:
A is equal to 12.
B is equal to 1.
C is equal to -12.
D is equal to 0.

The correct answer is:
is equal to 0.

Solution:
Applying L'Hospital's Rule, for numerator we have to apply Newton Leibnitz integral rule

limx→12(x-1)×(x-1)2cosx-14-0(x-1)cos(x-1)+sin(x-1) 00

Divide x-1

=limx→12(x-1)2cos(x-1)4cos(x-1)+sin(x-1)(x-1)

=01+1

=0

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2020 (06 Sep Shift 1)
$\lim _{x \rightarrow 1} \frac{1+\log x-x}{1-2 x+x^2}=$

Options:
A $1$
B $-1$
C $0$
D $-\frac{1}{2}$

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

Solution:
Applying L-Hospital's rule,
$\begin{aligned}
& \lim _{x \rightarrow 1} \frac{1+\log x-x}{1-2 x+x^2}=\lim _{x \rightarrow 1} \frac{\frac{1}{x}-1}{-2+2 x}=\lim _{x \rightarrow 1} \frac{1-x}{2 x(x-1)} \\
& \text { Again applying L-Hospital's rule, } \lim _{x \rightarrow 1} \frac{-1}{4 x-2}=-\frac{1}{2} .
\end{aligned}$

Topic: Mathematics, Limits, JEE Main
$\lim _{x \rightarrow 1} \frac{1}{|1-x|}=$

Options:
A $0$
B $1$
C $2$
D $\infty$

The correct answer is:
$\infty$

Solution:
$\begin{aligned} & \lim _{x \rightarrow 1-} \frac{1}{|1-x|}=\lim _{h \rightarrow 0} \frac{1}{1-(1-h)}=\infty \\ & \text { and } \lim _{x \rightarrow 1+} \frac{1}{|1-x|}=\lim _{h \rightarrow 0} \frac{1}{1+h-1}=\infty \\ & \text { Hence } \lim _{x \rightarrow 1 \mid} \frac{1}{1-x \mid}=\infty .\end{aligned}$

Topic: Mathematics, Limits, JEE Main
limx→1-π-2sin-1x1-x is equal to

Options:
A π
B 2π
C 12π
D π2

The correct answer is:
2π

Solution:
Given limit can be written as

L=limx→1-π-2sin-1x1-x×π+2sin-1xπ+2sin-1x

⇒L=limx→1-π-2sin-1x1-x π+2sin-1x=limx→1-2cos-1x1-x π+2sin-1x

Let K=limx→1-cos-1x1-x and put x=cosθ, we get

K=limθ→0θ2.22.sinθ2=limθ→02sinθ2θ2=2 ∵limx→0sinxx=1

∴ L=222 π=2π.

Topic: Mathematics, Limits, JEE Main, JEE Main 2019 (12 Jan Shift 2)
limx→12sincos-1x-x1-tancos-1x is equal to

Options:
A 12
B -12
C 2
D -1

The correct answer is:
-12

Solution:
Let cos−1x=t i.e. x=cost

So limt→π4sint−cost1−tant =limt→π4−cost=−12

Topic: Mathematics, Limits, JEE Main, JEE Main 2022 (26 Jun Shift 1)
$\lim _{x \rightarrow 1} \frac{x-1}{2 x^2-7 x+5}=$

Options:
A $1 / 3$
B $1 / 11$
C $-1 / 3$
D None of these

The correct answer is:
$-1 / 3$

Solution:
$\lim _{x \rightarrow 1} \frac{x-1}{(x-1)(2 x-5)}=-\frac{1}{3} .$
Aliter : Apply L-Hospital's rule.
So, \(\lim _{x \rightarrow 1} \frac{1}{4 x-7}=-\frac{1}{3} .\)

Topic: Mathematics, Limits, JEE Main
$\lim _{x \rightarrow \infty}\left[1+\frac{1}{m x}\right]^x$ equal to

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

The correct answer is:
$e^{1 / m}$

Solution:
$\begin{aligned} & \text { Let } y=\lim _{x \rightarrow \infty}\left(1+\frac{1}{m x}\right)^x=\lim _{x \rightarrow \infty}\left(1+\frac{1}{m x}\right)^{m x \cdot \frac{1}{m}} \\ & \Rightarrow y=e^{1 / m},\left(\text{as } \lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^x=e\right)\end{aligned}$

Topic: Mathematics, Limits, JEE Main
$\lim _{x \rightarrow \infty}\left(1+\frac{2}{x}\right)^x=$

Options:
A $e$
B $\frac{1}{e}$
C $e^2$
D None of these

The correct answer is:
$e^2$

Solution:
$\left[\lim _{x \rightarrow \infty}\left(1+\frac{1}{x / 2}\right)^{x / 2}\right]^2=e^2$ as \(\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^x=e\)

Topic: Mathematics, Limits, JEE Main
$\lim _{x \rightarrow 1}[x]=$

Options:
A $0$
B $1$
C Does not exist
D None of these

The correct answer is:
Does not exist

Solution:
$\lim _{x \rightarrow 1-}[x]=\lim _{h \rightarrow 0}[1-h]=\lim _{h \rightarrow 0} 0=0$
and $\lim _{x \rightarrow 1+}[x]=\lim _{h \rightarrow 0}[1+h]=\lim _{h \rightarrow 0} 1=1$
Hence limit does not exist.

Topic: Mathematics, Limits, JEE Main
$\lim _{x \rightarrow \frac{\pi}{2}} \frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}} \cdot \frac{1-\sin x}{(\pi-2 x)^3}=$

Options:
A $\frac{1}{32}$
B $0$
C $\frac{1}{16}$
D $\frac{1}{8}$

The correct answer is:
$\frac{1}{32}$

Solution:
$\begin{aligned} & \lim _{x \rightarrow \frac{\pi}{2}} \frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}} \cdot \frac{1-\sin x}{(\pi-2 x)^3} \\ & =\lim _{x \rightarrow \frac{\pi}{2}} \tan \left(\frac{\pi}{4}-\frac{x}{2}\right): \frac{1-\sin x}{(\pi-2 x)^3} \\ & \text { On putting } x=\frac{\pi}{2}-y, y \rightarrow 0 \\ & =\lim _{y \rightarrow 0} \frac{\tan \left(\frac{\pi}{4}-\frac{\frac{\pi}{2}-y}{2}\right) \cdot\left(1-\sin \left(\frac{\pi}{2}-y\right)\right.}{\left(\pi-2\left(\frac{\pi}{2}-y\right)\right)^3} \\ & =\lim _{y \rightarrow 0} \frac{\left(\tan \frac{y}{2}\right) \cdot(1-\cos y)}{8 y^3} \\ & =\lim _{y \rightarrow 0} \frac{\left(\tan \frac{y}{2}\right) \cdot 2 \cdot\left(\sin \frac{y}{2}\right)\left(\sin \frac{y}{2}\right)}{64 \cdot\left(\frac{y}{2}\right)\left(\frac{y}{2}\right)\left(\frac{y}{2}\right)} \\ & =\frac{2}{64}=\frac{1}{32} \\ & \end{aligned}$

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

Options:
A
$\infty$

B
$\frac{1}{8}$

C
0

D
$\frac{1}{32}$

The correct answer is:
$\frac{1}{32}$

Solution:
$\lim _{x \rightarrow \frac{\pi}{2}} \frac{\tan \left(\frac{\pi}{4}-\frac{x}{2}\right) \cdot(1-\sin x)}{(\pi-2 x)^3}$
Let $\mathrm{x}=\frac{\pi}{2}+\mathrm{y} ; \mathrm{y} \rightarrow 0$
$=\lim _{y \rightarrow 0} \frac{-\tan \left(-\frac{y}{2}\right) \cdot(1-\cos y)}{(-2 y)^3}=\lim _{y \rightarrow 0} \frac{-\tan \frac{y}{2} 2 \sin ^2 \frac{y}{2}}{(-8) \cdot \frac{y^3}{8} \cdot 8}$
$=\lim _{y \rightarrow 0} \frac{1}{32} \frac{\tan \frac{y}{2}}{\left(\frac{y}{2}\right)} \cdot\left[\frac{\sin y / 2}{y / 2}\right]^2=\frac{1}{32}$

Topic: Mathematics, Limits, JEE Main, JEE Main 2003
limx→π21x-π22∫x3π23cos1t3dt is equal to

Options:
A 3π8
B 3π24
C 3π28
D 3π4

The correct answer is:
3π28

Solution:
Let, y=limx→π2∫x3π23cost13dtx-π22

⇒y=limh→0∫π2-h3π23cost13dtπ2-h-π22

⇒y=limh→0∫π2-h3π23cost13dth2

Applying L-Hospital's rule and Newton Leibnitz Theorem we get,

⇒y=limh→00+cosπ2-h×3×π2-h22h

⇒y=limh→0sinh×3×π2-h22h

We know that, limh→0sinhh=1

⇒y=limh→03×π2-h22

⇒y=3π28

Topic: Mathematics, Limits, JEE Main, JEE Main 2024 (29 Jan Shift 1)
$\lim _{x \rightarrow \pi / 2} \frac{1+\cos 2 x}{(\pi-2 x)^2}=$

Options:
A $1$
B $2$
C $3$
D $1/2$

The correct answer is:
$1/2$

Solution:
First, let's substitute \(y=\pi-2 x\). As \(x \rightarrow \frac{\pi}{2}, y \rightarrow 0\).
So, the limit becomes:
\(\lim _{y \rightarrow 0} \frac{1+\cos (\pi-y)}{y^2}\)
Using the trigonometric identity \(\cos (\pi-y)=-\cos (y)\), we can rewrite the limit as:
\(\lim _{y \rightarrow 0} \frac{1-\cos (y)}{y^2}\)
Now, we know the standard limit:
\(\lim _{y \rightarrow 0} \frac{1-\cos (y)}{y^2}=\frac{1}{2}\)

Topic: Mathematics, Limits, JEE Main
$\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cot x-\cos x}{(\pi-2 x)^3}$ equals

Options:
A $\frac{1}{24}$
B $\frac{1}{16}$
C $\frac{1}{8}$
D $\frac{1}{4}$

The correct answer is:
$\frac{1}{16}$

Solution:
$\begin{aligned} \text { Let } \mathrm{I} & =\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cot x-\cos x}{(\pi-2 x)^3} \\ & =\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cos x(1-\sin x)}{\sin x(\pi-2 x)^3} \\ \text { Put } x & =\frac{\pi}{2}-\mathrm{h} \\ \therefore \quad \pi-2 x & =2 \mathrm{~h} \\ \text { As } x & \rightarrow \frac{\pi}{2}, \mathrm{~h} \rightarrow 0\end{aligned}$
$\begin{aligned} \therefore \quad I & =\lim _{h \rightarrow 0} \frac{\cos \left(\frac{\pi}{2}-h\right)\left(1-\sin \left(\frac{\pi}{2}-h\right)\right)}{\sin \left(\frac{\pi}{2}-h\right)(2 h)^3} \\ & =\lim _{h \rightarrow 0} \frac{\sin h \cdot 2 \sin ^2\left(\frac{h}{2}\right)}{\cosh \cdot 8 h^3} \\ & =\frac{2}{8} \lim _{h \rightarrow 0} \frac{\sin h}{h} \times \lim _{h \rightarrow 0} \frac{\sin ^2 \frac{h}{2}}{\frac{h^2}{4} \cdot 4} \\ & =\frac{2}{8}(1) \times(1) \times \frac{1}{4} \\ & =\frac{1}{16}\end{aligned}$

Topic: Mathematics, Limits, JEE Main
$\lim _{x \rightarrow 2}\left(\frac{5 x-8}{8-3 x}\right)^{\frac{3}{2 x-4}}=$

Options:
A $e^{5 / 2}$
B $\mathrm{e}^{3 / 2}$
C $\mathrm{e}^2$
D $\mathrm{e}^6$

The correct answer is:
$\mathrm{e}^6$

Solution:
$\begin{aligned} & \lim _{x \rightarrow 2}\left(\frac{5 x-8}{8-3 x}\right) \frac{3}{2 x-4}=e^{\lim _{x \rightarrow 2}\left(\frac{5 x-8}{8-3 x}-1\right) \times \frac{3}{2 x-4}} \\ & =e^{\lim _{x \rightarrow 2} \frac{8(x-2) \times 3}{(8-3 x) 2(x-2)}}=e^6 \\ & \end{aligned}$

Topic: Mathematics, Limits, JEE Main
limx→π2cotx-cosxπ-2x3 equals

Options:
A 124
B 116
C 18
D 14

The correct answer is:
116

Solution:
Put, x=π2-h in the given limit, we get
=limh→0cotπ2-h-cosπ2-hπ-2×π2-h3
=limh→0tanh-sinhπ-2×π2+2h3
=limh→0tanh-sinh8h3
= 18 limh→0sinhhsech-1h2
=18 limh→0sinhhlimh→01-coshcosh h2

=18 ×1×limh→01-coshcosh h2

=18 limh→02sin2h2cosh h2

=18 limh→02sin2h2h24×4×cosh

=116.

Topic: Mathematics, Limits, JEE Main, JEE Main 2017 (02 Apr)
$\lim _{x \rightarrow \pi / 2}(\sec \theta-\tan \theta)=$

Options:
A $0$
B $1/2$
C $2$
D $\infty$

The correct answer is:
$0$

Solution:
$\lim _{\theta \rightarrow \pi / 2} \frac{1-\sin \theta}{\cos \theta}=\lim _{\theta \rightarrow \pi / 2} \frac{\left(\cos \frac{\theta}{2}-\sin \frac{\theta}{2}\right)^2}{\left(\cos \frac{\theta}{2}-\sin \frac{\theta}{2}\right)\left(\cos \frac{\theta}{2}+\sin \frac{\theta}{2}\right)}=0$.

Topic: Mathematics, Limits, JEE Main
$\lim _{x \rightarrow \pi / 2}(\sec x-\tan x)$ is equal to

Options:
A 2
B $-1$
C 1
D 0

The correct answer is:
0

Solution:
$\lim _{x \rightarrow \frac{\pi}{2}}(\sec x-\tan x)$
$\Rightarrow \quad \lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{1-\sin x}{\cos x}\right) \quad\left(\frac{0}{0}\right.$ form $)$
Using 'L' hospital rule, $\Rightarrow \quad \lim _{x \rightarrow \frac{\pi}{2}} \frac{-\cos x}{-\sin x}$
$=\lim _{x \rightarrow \frac{\pi}{2}} \cot x=\cot \frac{\pi}{2}=0$

Topic: Mathematics, Limits, JEE Main
limx→π2tan2x2sin2x+3sinx+412-sin2x+6sinx+212 is equal to

Options:
A 112
B -118
C -112
D 16

The correct answer is:
112

Solution:
Given, limx→π2tan2x2sin2x+3sinx+4-sin2x+6sinx+2

=limx→π2tan2x2sin2x+3sinx+42-sin2x+6sinx+222sin2x+3sinx+4+sin2x+6sinx+2

=limx→π2tan2xsin2x-3sinx+29+9

=limx→π2tan2xsinx-1sinx-26

=16limx→π2tan2x1-sinx

=16limx→π2sin2x1-sinx1-sinx1+sinx=112

Topic: Mathematics, Limits, JEE Main, JEE Main 2022 (25 Jun Shift 2)
$\lim _{x \rightarrow \pi / 2} \tan x \log \sin x=$

Options:
A $0$
B $1$
C $-1$
D None of these

The correct answer is:
$0$

Solution:
$\begin{aligned} & \lim _{x \rightarrow \frac{\pi}{2}} \tan x \log \sin x=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\log \sin x}{\cot x} \\ & =\lim _{x \rightarrow \frac{\pi}{2}} \frac{\frac{1}{\sin x} \cos x}{-\operatorname{cosec}^2 x}=0 \quad \text {(Applying L-Hospital's)}\end{aligned}$

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

Options:
A $\frac{5 \pi^2}{9}$
B $\frac{9 \pi^2}{8}$
C $\frac{11 \pi^2}{10}$
D $\frac{3 \pi^2}{2}$

The correct answer is:
$\frac{9 \pi^2}{8}$

Solution:
$\begin{aligned} & \lim _{x \rightarrow \frac{\pi}{2}} \frac{0-\{\sin (2 x)+\cos (x)\} \cdot 3 x^2}{2\left(x-\frac{\pi}{2}\right)} \\ & =\lim _{x \rightarrow \frac{\pi}{2}} \frac{-\{2 \sin x \cos x+\cos x\} 3 x^2}{2\left(x-\frac{\pi}{2}\right)} \\ & =\lim _{x \rightarrow \frac{\pi}{2}}\left\{\frac{2 \sin x \sin \left(\frac{\pi}{2}-x\right)}{2\left(x-\frac{\pi}{2}\right)}+\frac{\sin \left(\frac{\pi}{2}-x\right)}{2\left(\frac{\pi}{2}-x\right)}\right\} 3 x^2 \\ & =\left(1(1)+\frac{1}{2}\right) 3\left(\frac{\pi}{2}\right)^2 \\ & =\frac{9 \pi^2}{8}\end{aligned}$

Topic: Mathematics, Limits, JEE Main, JEE Main 2024 (09 Apr Shift 2)
$\lim _{x \rightarrow 2} \frac{|x-2|}{x-2}=$

Options:
A $1$
B $-1$
C Does not exist
D None of these

The correct answer is:
Does not exist

Solution:
$\lim _{x \rightarrow 2^{-}} \frac{|x-2|}{x-2}=\lim _{h \rightarrow 0} \frac{|2-h-2|}{2-h-2}=-1$
and $\lim _{x \rightarrow 2+} \frac{|x-2|}{x-2}=\lim _{h \rightarrow 0} \frac{|2+h-2|}{2+h-2}=1$
Hence limit does not exist.

Topic: Mathematics, Limits, JEE Main
$\lim _{x \rightarrow \infty} \frac{(2 x+1)^{40}(4 x-1)^5}{(2 x+3)^{45}}=$

Options:
A $16$
B $24$
C $32$
D $8$

The correct answer is:
$32$

Solution:
$\lim _{x \rightarrow \infty} \frac{\left(2+\frac{1}{x}\right)^{40}\left(4-\frac{1}{x}\right)^5}{\left(2+\frac{3}{x}\right)^{45}}=\frac{2^{40} \cdot 4^5}{2^{45}}=2^5=32$.

Topic: Mathematics, Limits, JEE Main
limx→3⁡3x-32x-4- 2 is equal to

Options:
A 12
B 122
C 32
D 3

The correct answer is:
12

Solution:
Given,

limx→3⁡3x-32x-4- 2
On rationalization we get,
limx→3⁡3x-9×2x-4+ 22x-4-2×3x+3
= limx→3⁡3x-32x-3×2x-4+ 23x+3
=32×226=12.

Topic: Mathematics, Limits, JEE Main, JEE Main 2017 (08 Apr Online)
limx→π482-cosx+sinx72-2sin2x is equal to

Options:
A 14
B 7
C 142
D 72

The correct answer is:
14

Solution:
limx→π482-cosx+sinx72-2sin2x 00form

=limx→π4-7cosx+sinx6-sinx+cosx-22cos2x (using L'Hospital Rule)

=limx→π456cosx-sinx22cos2x 00 form

=limx→π4-56sinx+cosx-42sin2x (using L'Hospital Rule)

=56242=14

Topic: Mathematics, Limits, JEE Main, JEE Main 2022 (25 Jul Shift 2)
$\lim _{x \rightarrow 4} \frac{\cos 7 x^{\circ}-\cos 2 x^{\circ}}{x^2}$ is

Options:
A $\frac{-45}{2} \pi^2$
B $\frac{-45}{2} \pi$
C $\frac{-\pi^2}{1440}$
D $\frac{-\pi^2}{2880}$

The correct answer is:
$\frac{-\pi^2}{1440}$

Solution:
$\begin{aligned} & \lim _{x \rightarrow 0} \frac{\cos 7 x^{\circ}-\cos 2 x^{\circ}}{x^2} \\ & =\lim _{x \rightarrow 0} \frac{\cos \left(\frac{7 \pi}{180}\right) x-\cos \left(\frac{2 \pi}{180}\right) x}{x^2} \\ & =\frac{\left(\frac{2 \pi}{180}\right)^2-\left(\frac{7 \pi}{180}\right)^2}{2} \\ & \quad \ldots\left[\because \lim _{x \rightarrow 0} \frac{\cos m x-\cos n x}{x^2}=\frac{n^2-m^2}{2}\right] \\ & =\frac{-\pi^2}{1440}\end{aligned}$

Topic: Mathematics, Limits, JEE Main
$\lim _{x \rightarrow \pi / 6} \frac{\cot ^2 \theta-3}{\operatorname{cose} \theta-2}=$

Options:
A $2$
B $4$
C $6$
D $0$

The correct answer is:
$4$

Solution:
$\lim _{x \rightarrow \pi / 6} \frac{\operatorname{cosec}^2 \theta-4}{\operatorname{cosec} \theta-2}=\lim _{x \rightarrow \pi / 6} \operatorname{cosec} \theta+2=4$ \(\left( \text{as, }1+\cot ^2 \theta=\operatorname{cosec}^2 \theta\right)\)

Topic: Mathematics, Limits, JEE Main
$\lim _{x \rightarrow 8} \frac{\sqrt{1+\sqrt{1+x}}-2}{x-8}$ is equal to

Options:
A $\frac{3}{2}$
B $\frac{1}{4}$
C $\frac{1}{24}$
D $\frac{1}{12}$

The correct answer is:
$\frac{1}{24}$

Solution:
$\begin{aligned} & \lim _{x \rightarrow 8} \frac{\sqrt{1+\sqrt{1+x}}-2}{x-8} \times \frac{\sqrt{1+\sqrt{1+x}}+2}{\sqrt{1+\sqrt{1+x}}+2} \\ & =\lim _{x \rightarrow 8} \frac{1+\sqrt{1+x}-4}{(\sqrt{1+\sqrt{1+x}}+2)(x-8)} \\ & =\lim _{x \rightarrow 8} \frac{\sqrt{1+x}-3}{(\sqrt{1+\sqrt{1+x}}+2)(x-8)} \times \frac{\sqrt{1+x}+3}{(\sqrt{1+x}+3)} \\ & =\lim _{x \rightarrow 8} \frac{1+x-9}{(\sqrt{1+x}+3)(\sqrt{1+\sqrt{1+x}}+2)(x-8)} \\ & =\lim _{x \rightarrow 8} \frac{1}{(\sqrt{1+x}+3)(\sqrt{1+\sqrt{1+x}}+2)} \\ & =\frac{1}{(\sqrt{1+8}+3)(\sqrt{1+\sqrt{1+8}}+2)} \\ & =\frac{1}{(3+3)(2+2)}=\frac{1}{24}\end{aligned}$

Topic: Mathematics, Limits, JEE Main
limx→aa+2x13-3x133a+x13-4x13 a≠0 is equal to:

Options:
A 292313
B 2343
C 2943
D 232913

The correct answer is:
232913

Solution:
At x=a, limit is of 00 form.

So, applying L'Hospital rule, we get

limx→aa+2x13-3x133a+x13-4x13=limx→a13a+2x-23.2-13.3x-23.3133a+x-23.-13.4x-23.4

=133a-23.2-3134a-23.1-4=3-234-23.13

=243913.13=232913

Topic: Mathematics, Limits, JEE Main, JEE Main 2020 (03 Sep Shift 2)
$\lim _{x \rightarrow a} f(x) . g(x)$ exists, if

Options:
A $\lim _{x \rightarrow a} f(x)$ and $\lim _{x \rightarrow a} g(x)$ exist
B $\lim _{x \rightarrow a} f(x)^{g(x)}$ exists
C $\lim _{x \rightarrow a} \frac{f(x)}{g(x)}$ exists
D $\lim _{x \rightarrow a} f(x) g\left(\frac{1}{x}\right)$ exists

The correct answer is:
$\lim _{x \rightarrow a} f(x)$ and $\lim _{x \rightarrow a} g(x)$ exist

Solution:
If both \(\lim _{x \rightarrow a} f(x)\) and \(\lim _{x \rightarrow a} g(x)\) exist, then \(\lim _{x \rightarrow a} f(x) \cdot g(x)\) also exists by the product limit rule.

Topic: Mathematics, Limits, JEE Main
$\lim _{x \rightarrow a} \frac{\sqrt{3 x-a}-\sqrt{x+a}}{x-a}=$

Options:
A $\sqrt{2} a$
B $1 / \sqrt{2 a}$
C $2 a$
D $1 / 2 a$

The correct answer is:
$1 / \sqrt{2 a}$

Solution:
$\begin{aligned}
& \lim _{x \rightarrow a} \frac{\sqrt{3 x-a}-\sqrt{x+a}}{x-a} \\
& \quad=\lim _{x \rightarrow a} \frac{\sqrt{3 x-a}-\sqrt{x+a}}{(x-a)} \times \frac{\sqrt{3 x-a}+\sqrt{x+a}}{\sqrt{3 x-a}+\sqrt{x+a}} \\
& \quad=\frac{2}{2 \sqrt{2 a}}=\frac{1}{\sqrt{2 a}}
\end{aligned}$
Aliter : Apply L-Hospital's rule
$\begin{aligned}
\lim _{x \rightarrow a} \frac{\sqrt{3 x-a}-\sqrt{x+a}}{x-a} & =\lim _{x \rightarrow a} \frac{3}{2 \sqrt{3 x-a}}-\frac{1}{2 \sqrt{x+a}} \\
& =\frac{3}{2 \sqrt{2 a}}-\frac{1}{2 \sqrt{2 a}}=\frac{1}{\sqrt{2 a}} .
\end{aligned}$

Topic: Mathematics, Limits, JEE Main
$\lim _{x \rightarrow a} \frac{(x+2)^{5 / 3}-(a+2)^{5 / 3}}{x-a}=$

Options:
A $\frac{5}{3}(a+2)^{2 / 3}$
B $\frac{5}{3}(a+2)^{5 / 3}$
C $\frac{5}{3} a^{2 / 3}$
D $\frac{5}{3} a^{5 / 3}$

The correct answer is:
$\frac{5}{3}(a+2)^{2 / 3}$

Solution:
Apply the L-Hospital's rule, $\lim _{x \rightarrow a} \frac{f(x)}{g(x)}=\lim _{x \rightarrow a} \frac{f^{\prime}(x)}{g^{\prime}(x)}$.
\(\begin{aligned}
& =\lim _{x \rightarrow a} \frac{\frac{5}{3}(x+2)^{2 / 3}}{1 \frac{1}{3}} \\
& =\frac{5}{3}(a+2)^{\frac{2}{3}}
\end{aligned}\)

Topic: Mathematics, Limits, JEE Main
$\lim _{x \rightarrow a} \frac{x^9+a^9}{x+a}=9$, then $a=$

Options:
A $9^{1 / 8}$
B $\pm 2$
C $\pm 3$
D None of these

The correct answer is:
$9^{1 / 8}$

Solution:
$\lim _{x \rightarrow a} \frac{x^9+a^9}{x+a}=9 \Rightarrow \frac{2 a^9}{2 a}=9 \Rightarrow a^8=9 \Rightarrow a=9^{1 / 8}$

Topic: Mathematics, Limits, JEE Main
$\lim _{x \rightarrow \infty}\left[\frac{1^3+2^3+3^3+\ldots \ldots .+n^3}{n^4}\right]=$

Options:
A $1$
B $\frac{1}{3}$
C $\frac{1}{4}$
D None of these

The correct answer is:
$\frac{1}{4}$

Solution:
\(\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{1^3+2^3+3^3+\ldots+n^3}{n^4} \\
& =\lim _{n \rightarrow \infty} \frac{\left[\frac{1}{2} n(n+1)\right]^2}{n^4} \\
& {\left[1^3+2^3+3^3+\ldots+n^3=\left(\frac{1}{2} n(n+1)\right)^2\right]} \\
& =\lim _{n \rightarrow \infty} \frac{\frac{1}{4} n^2(n+1)^2}{n^4} \\
& =\lim _{n \rightarrow \infty} \frac{\frac{1}{4} n^2\left(n^2+1+2 n\right)}{n^4} \\
& =\lim _{4 n \rightarrow \infty} \frac{n^4\left(1+\frac{1}{n^2}+\frac{2}{n}\right)}{n^4} \quad \quad\left[\frac{\infty}{\infty} \text { form }\right] \\
& =\frac{1}{4} \lim _{n \rightarrow \infty} \frac{\left(1+\frac{1}{n^2}+\frac{2}{n}\right)}{1}
\end{aligned}\)
When \(\mathrm{n} \rightarrow \infty\), then \(\frac{1}{\mathrm{n}} \rightarrow 0=\frac{1}{4}\)

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

Options:
A \( -2 \)
B \( 0 \)
C \( 2 \)
D \( \infty \)

The correct answer is:
\( 2 \)

Solution:
Put x=tan⁡θ ⇒ θ=tan-1⁡x
As x→0⇒θ→0
∴ limθ→0⁡1tan⁡θsin-1⁡2tan⁡θ1+tan2⁡θ
=limθ→0 1tan⁡θsin-1⁡(sin⁡2θ)
=limθ→02θtan⁡θ=2

Topic: Mathematics, Limits, JEE Main
\(\lim _{x \rightarrow 0} \frac{8}{\sin ^8 x}\) \(\left\{1-\cos \left(\frac{x^2}{2}\right)-\cos \left(\frac{x^2}{4}\right)+\cos \left(\frac{x^2}{2}\right) \cos \left(\frac{x^2}{4}\right)\right\}=\)

Options:
A \(\frac{1}{16}\)
B \(\frac{1}{32}\)
C \(\frac{1}{64}\)
D \(\frac{1}{8}\)

The correct answer is:
\(\frac{1}{32}\)

Solution:
\(\begin{aligned}
\lim _{x \rightarrow 0} \frac{8}{\sin ^8 x}\left\{1-\cos \left(\frac{x^2}{2}\right)-\right. & \cos \left(\frac{x^2}{4}\right) \\
& \left.+\cos \left(\frac{x^2}{2}\right) \cdot \cos \left(\frac{x^2}{4}\right)\right\}
\end{aligned}\)
\(\begin{gathered}\Rightarrow \lim _{x \rightarrow 0} \frac{8}{\sin ^8 x}\left\{\left[1-\cos \left(\frac{x^2}{2}\right)\right]\left[-\cos \left(\frac{x^2}{4}\right)\right]\left[1-\cos ^2\left(\frac{x^2}{2}\right)\right]\right\} \\ \Rightarrow \quad \lim _{x \rightarrow 0} \frac{8}{\sin ^8 x}\left[1-\cos \left(\frac{x^2}{2}\right)\right]\left(1-\cos \left(\frac{x^2}{4}\right)\right) \\ \Rightarrow \quad \lim _{x \rightarrow 0} \frac{8}{\sin ^8 x}\left(2 \sin ^2\left(\frac{x^2}{4}\right)\right)\left(2 \sin ^2\left(\frac{x^2}{8}\right)\right) \\ \Rightarrow \quad \lim _{x \rightarrow 0} \frac{32 \sin ^2\left(\frac{x^2}{4}\right) \cdot \sin ^2\left(\frac{x^2}{8}\right)}{\sin ^8 x} \\ \Rightarrow \quad \lim _{x \rightarrow 0} \frac{32 \cdot \sin ^2\left(\frac{x^2}{4}\right) \cdot \sin ^2\left(\frac{x^2}{8}\right) \times x^8}{\left(\frac{x^2}{4} \times \frac{x^2}{8}\right)^2 \cdot \sin ^8 x \times(4 \times 8)^2}=\frac{1}{32}\end{gathered}\)

Topic: Mathematics, Limits, JEE Main
\( \lim _{x \rightarrow 0}\left\{(1+x)^{\frac{2}{x}}\right\} \) (where \( \{.\} \) denotes the fractional part of \( x \) ) is equal to

Options:
A \( e^{2}-7 \)
B \( e^{2}-8 \)
C \( e^{2}-6 \)
D None of these

The correct answer is:
\( e^{2}-7 \)

Solution:
We know, 1+x2/x=1+x2/x-1+x2/x

Also, limx→01+x2/x=e2

⇒limx→01+x2/x=limx→01+x2/x-limx→01+x2/x

=e2-e2=e2-7

Topic: Mathematics, Limits, JEE Main
\( \lim _{x \rightarrow 2} \frac{3^{x}+3^{3-x}-12}{3^{-\frac{x}{2}}-3^{1-x}} \) is equal to

Options:

The correct answer is:
36.00

Solution:
Let, 3x2=t, xâ2âtâ3

limtâ3t227t2-121t-3t2=limtâ3t4+27-12t2t-3

limtâ3t2-3t+3t-3t-3=6Ã6=36

Topic: Mathematics, Limits, JEE Main