JEE Main 2025 Apr 8 Shift 2, Mathematics Q20: Methods to Evaluate Limits

JEE Main2025Apr 8, Shift 2Mathematics

Given below are two statements:

Statement I: $x \to 0 lim \frac{tan ^{- 1} x + lo g _{e} \frac{1 + x}{1 - x} - 2 x}{x ^{5}} = \frac{2}{5}$

Statement II: $x \to 1 lim x^{\frac{2}{1 - x}} = \frac{1}{e ^{2}}$

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

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
D
Both Statement I and Statement II are true

Solution

For Statement I, use the Maclaurin series: $tan^{- 1} x = x - \frac{x ^{3}}{3} + \frac{x ^{5}}{5} - \dots$ and $\frac{1}{2} ln \frac{1 + x}{1 - x} = x + \frac{x ^{3}}{3} + \frac{x ^{5}}{5} + \dots$ .

Adding these: $tan^{- 1} x + \frac{1}{2} ln \frac{1 + x}{1 - x} = 2 x + \frac{2 x ^{5}}{5} + O (x^{7})$ .

Subtracting $2 x$ : the numerator becomes $\frac{2 x ^{5}}{5} + O (x^{7})$ , so dividing by $x^{5}$ and taking the limit gives $\frac{2}{5}$ ; Statement I is true.

For Statement II, write $x^{2/ (1 - x)} = exp (\frac{2 ln x}{1 - x})$ . As $x \to 1$ , let $u = x - 1 \to 0$ ; then $ln x = ln (1 + u) \approx u - \frac{u ^{2}}{2} + \dots$ and $1 - x = - u$ .

So $\frac{2 ln x}{1 - x} \to \frac{2 u}{- u} = - 2$ , giving the limit $e^{- 2} = \frac{1}{e ^{2}}$ ; Statement II is true.

Both statements are correct, matching option (4).

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JEE Main 2025 Apr 8 Shift 2, Mathematics Q20: Methods to Evaluate Limits

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