Q37 / 75 · JEE Main 2025 (8 Apr Shift 2) · Mathematics
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 :
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 :
- 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
- Correct answerDBoth Statement I and Statement II are true
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
$\Rightarrow$ Both statements correct
Hence, the correct answer is option: 'D'.