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 : $\lim _{\mathrm{x} \rightarrow \mathrm{l}}\left(\mathrm{x}^{\frac{2}{1-\mathrm{x}}}\right)=\frac{1}{\mathrm{e}^{2}}$ In the light of the above statements, choose the correct answer from the options given below :

(D) Both Statement I and Statement II are true

(A) Statement I is false but Statement II is true

(B) Statement I is true but Statement II is false

(C) Both Statement I and Statement II are false

JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Limits

Given below are two statements :
Statement I :

lim_{x \to 0} (\frac{t a n ^{- 1} x + l o g _{e} \frac{1 + x}{1 - x} - 2 x}{x ^{5}}) = \frac{2}{5}

Statement II :

lim_{x \to l} (x^{\frac{2}{1 - x}}) = \frac{1}{e ^{2}}

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

Answer: (D)

For Statement I, the series

tan^{- 1} x = x - \frac{x ^{3}}{3} + \frac{x ^{5}}{5} - \dots

and

\frac{1}{2} ln \frac{1 + x}{1 - x} = x + \frac{x ^{3}}{3} + \frac{x ^{5}}{5} + \dots

add to

2 x + \frac{2 x ^{5}}{5} + \dots

; subtracting

2 x

leaves

\frac{2 x ^{5}}{5} + \dots

, so the limit is

\frac{2}{5}

and Statement I is true. For Statement II,

lim_{x \to 1} x^{\frac{2}{1 - x}} = exp (lim_{x \to 1} \frac{2 ln x}{1 - x})

; since

ln x \sim (x - 1)

, the exponent tends to

- 2

, giving

e^{- 2} = \frac{1}{e ^{2}}

, so Statement II is also true. Both statements are true.

JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Single correct option