Given below are two statements : Statement I : Statement II : In the light of the above statements, choose the correct answer from the potions given below. [2025]

Question

Given below are two statements :

Statement I : $\lim_{x \to 0} (\frac{\tan^{- 1} x + \log_{e} \sqrt{\frac{1 + x}{1 - x}} - 2 x}{x^{5}}) = \frac{2}{5}$

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

In the light of the above statements, choose the correct answer from the potions given below. [2025]

Smart Achievers · Accepted Answer

(2)

Let $L = \lim_{x \to 0} (\frac{\tan^{- 1} x + \log_{e} \sqrt{\frac{1 + x}{1 - x}} - 2 x}{x^{5}})$

$= \lim_{x \to 0} \frac{(x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - . . .) + \frac{1}{2} [\log_{e} (1 + x) - \log_{e} (1 - x)] - 2 x}{x^{5}}$

$= \lim_{x \to 0} \frac{(x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - . . .) + \frac{1}{2} [x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - . . . . . - (- x - \frac{x^{2}}{2} - \frac{x^{3}}{3} - . . . . .)] - 2 x}{x^{5}}$

$= \lim_{x \to 0} \frac{x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - . . . + (x + \frac{x^{3}}{3} + \frac{x^{5}}{5} + . . .) - 2 x}{x^{5}}$

$= \lim_{x \to 0} \frac{(2 x + \frac{2 x^{5}}{5} + . . .) - 2 x}{x^{5}} = \frac{2}{5}$

$∴$ Statement I is true.

Let $I = \lim_{x \to 1} x^{(\frac{2}{1 - x})} = \lim_{x \to 1} e^{\ln (x^{2 / 1 - x})} = \lim_{x \to 1} e^{2 \ln x / (1 - x)}$

Let us evaluate $\lim_{x \to 1} \frac{2 \ln x}{1 - x}$ $(\frac{0}{0} form)$

$= \lim_{x \to 1} \frac{2 / x}{- 1}$ [Using L'Hospital rule]

= – 2

$∴ I = \lim_{x \to 1} e^{- 2} = e^{- 2}$

$∴$ Statement II is also true.

Solution

1 Statement I is true but Statement II is false

2 Both Statement I and Statement II are true

3 Both Statement I and Statement II are false

4 Statement I is false but Statement II is true

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