Given below are two statements : one is labelled as Statement I and the other is labelled as Statement II: Statement I: lim x rightarrow 0 (frac tan -1 x+log e √(1+x/1-x)-2 x x 5 )=2/5 Statement II: lim x rightarrow 1 (x 2/1-x )=frac 1 e 2 In the light of the above statements, choose the correct answer from the options given below :

Both Statement I and Statement II are true

Statement I is false but Statement II is true

Both Statement I and Statement II are false

Statement I is true but Statement II is false

Mathematics · Limits, Continuity and Differentiability

JEE Main 2025 — 8 April, Evening Shift — Question 27

Given below are two statements : one is labelled as Statement I and the other is labelled as Statement II:

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

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

Option A:
Statement I is false but Statement II is true
Option B:
Both Statement I and Statement II are false
Option C:
Both Statement I and Statement II are true
Correct
Option D:
Statement I is true but Statement II is false

Answer: C

Step-by-step solution

$\text{Statement\; I:\; } \lim_{x \to 0} \frac{\tan^{-1} x + \log \sqrt{\frac{1+x}{1-x}} - 2x}{x^5}$

$\text{Series\; expansions:\; }$ $\tan^{-1}x = x - \frac{x^3}{3} + \frac{x^5}{5} + O(x^7),\quad$ $\log \sqrt{\frac{1+x}{1-x}} = x + \frac{x^3}{3} + \frac{x^5}{5} + O(x^7)$

$\text{Combine:\; } \tan^{-1}x + \log\sqrt{\frac{1+x}{1-x}} - 2x = \frac{2}{5} x^5 + O(x^7)$

$\Rightarrow \lim_{x \to 0} \frac{2 x^5 /5}{x^5} = \frac{2}{5}$

$\text{Statement\; II:\; } \lim_{x \to 1} x^{2/(1-x)} = \lim_{x \to 1} e^{\frac{2 \ln x}{1-x}} = e^{2 \cdot (-1)} = \frac{1}{e^2}$

$\text{Hence\; both\; statements\; are\; correct.}$

Answer key and solution verified before publishing.

Practise Limits, Continuity and Differentiability

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2025
Paper: 8 April, Evening Shift
Subject: Mathematics
Chapter: Limits, Continuity and Differentiability
Topic: Evaluation of Limit of Functions