Given below are two statements : Statement I : _ x arrow 0 ( ^ -1 x+ _e (1+x)/(1-x) -2 x x^5 )=(2)/(5) Statement II : _ x arrow 1 (x^ 2 1-x )= 1 e^2 In the light of the above statements, choose the correct answer from

Correct answer: D. Both Statement I and Statement II are true aligned & _ x arrow 0 ^ -1 x+(1)/(2)[ (1+x)- (1-x)]-2 x x^5 & = _ x arrow 0 ((x-(x^3)/(3)+(x^5)/(5) )+(1)/(2)[x-(x^2)/(2)+(x^3)/(3) -(-x-(x^2)/(2)-(x^3)/(3) )]-2 x)/(x^5) & = _ x arrow 0 (2 x+(2 x^5)/(5) -2 x)/(x^5)=(2)/(5) & _ x arrow 1 x^ (2)/((1-x)) =e^ _ x arrow l ((2)/((1-x)))(x-1) =e^ -2 aligned Both statements correct

JEE Main 20258 AprilEvening Shift

Exam: JEE Main
Year: 2025
Date: 8 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Limits

JEE Main 2025 · 8 April · Evening Shift · Mathematics · Limits

Given below are two statements : Statement I : _ x arrow 0 ( ^ -1...

Given below are two statements :
Statement I :

\lim _{x \rightarrow 0}\left(\frac{\tan ^{-1} x+\log _e \sqrt{\frac{1+x}{1-x}}-2 x}{x^5}\right)=\frac{2}{5}

Statement II :

\lim _{\mathrm{x} \rightarrow 1}\left(\mathrm{x}^{\frac{2}{1-\mathrm{x}}}\right)=\frac{1}{\mathrm{e}^2}

In the light of the above statements, choose the correct answer from the options given below :

Options

AStatement I is false but Statement II is true
BStatement I is true but Statement II is false
CBoth Statement I and Statement II are false
DBoth Statement I and Statement II are true

Correct answer

D. Both Statement I and Statement II are true

Step-by-step solution

\begin{aligned}<br/>& \lim _{x \rightarrow 0} \frac{\tan ^{-1} x+\frac{1}{2}[\ln (1+x)-\ln (1-x)]-2 x}{x^5} \\ & =\lim _{x \rightarrow 0} \frac{\left(x-\frac{x^3}{3}+\frac{x^5}{5} \ldots\right)+\frac{1}{2}\left[x-\frac{x^2}{2}+\frac{x^3}{3} \ldots-\left(-x-\frac{x^2}{2}-\frac{x^3}{3} \ldots\right)\right]-2 x}{x^5} \\ & =\lim _{x \rightarrow 0} \frac{2 x+\frac{2 x^5}{5} \ldots-2 x}{x^5}=\frac{2}{5} \\ & \lim _{x \rightarrow 1} x^{\frac{2}{(1-x)}}=e^{\lim _{x \rightarrow l}\left(\frac{2}{(1-x)}\right)(x-1)}=e^{-2}<br/>\end{aligned}

\Rightarrow

Both statements correct

Practice Limits on Quantrex Academy →

More from Limits

Let _ x → 2 (( (x-2))(rx^2 + (p-2)x - 2p))/((x-2)^2) = 5 for some r, p ∈ ℝ . If the set… JEE Main 2026 The value of _ x → 0 ((x^2 ^2 x)/(x^2 - ^2 x)) is: JEE Main 2026 Let f(x) = _ y → 0 ((1 - (xy)) (xy))/(y^3) . Then the number of solutions of the equation… JEE Main 2026 The product of all possible values of α , for which _ x → 0 ((1 - (α x) ((α+1)x)… JEE Main 2026 If _ x → 2 (x^3 - 5x^2 + ax + b) ( x-1 - 1) _e(x-1) = m , then a + b + m is equal to : JEE Main 2026 The value of _ x arrow 0 _ e ( (e x) · (e^ 2 x) · … · (e^ 10 x)) e^ 2 -e^ 2 x is equal to JEE Main 2026 If _ x arrow 0 e^ (a-1) x +2 ~b x+(c-2) e^ -x x x- _ e (1+x) =2 , then a^ 2 +b^ 2 +c^ 2… JEE Main 2026 Let [·] denote the greatest integer function and f(x)= _ n arrow ∞ 1 n^ 3 ∑_ k=1 ^ n [ k^… JEE Main 2026 Let f: R arrow(0, ∞) be a twice differentiable function such that f(3)=18, f^ (3)=0 and… JEE Main 2026 Let α and β be the real numbers such that _ x arrow 0 (1)/(x^3)((α)/(2) ∫_0^x (1)/(1-t^2)… JEE Advanced 2025 Let x_0 be the real number such that e^ x_0 +x_0=0 . For a given real number α , define… JEE Advanced 2025 If the function f(x)=( ( x)- ( x))/( x- x) is continuous at x=0 , then f(0) is equal to… JEE Main 2025 _ x arrow 0^ + (5(x)^ (1)/(3) ) _e(1+3 x^2) ( ^ -1 3 x )^2(e^ 5(x)^ (4)/(3) -1) is equal… JEE Main 2025 If _ x arrow 1^ + ((x-1)(6+λ (x-1))+μ (1-x))/((x-1)^3)=-1 , where λ, μ ∈ ℝ , then λ+μ is… JEE Main 2025 If Lim_ x arrow 0 (( x)/(x))^ (1)/(x^2) =p , then 96 _e p is equal to ______ JEE Main 2025 If _ x arrow 0 ( (2 x)+a (4 x)-b)/(x^4) is finite, then (a+b) is equal to : JEE Main 2025 Full Limits list Limits JEE Maths PYQs JEE Mathematics PYQs All JEE Main PYQs