Given below are two statements : Statement I : _ x 0 ( ^ -1 x+ _e… - Vidyadip

MCQ

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 : $\lim _{x \rightarrow 1}\left(x^{\frac{2}{1-x}}\right)=\frac{1}{\mathrm{e}^{2}}$
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
✓
Both Statement I and Statement II are true

✓

Answer

Correct option: D.

Both Statement I and Statement II are true

(D) Both Statement I and Statement II are true
$\lim _{x \rightarrow 0} \frac{\tan ^{-1} x+\frac{1}{2}[ \ell n (1+x)-\ell n (1-x)]-2 x}{x^{5}}$
$=\lim _{x \rightarrow 0} \frac{\left(x-\frac{x^{3}}{3}+\frac{x^{5}}{5} \ldots \right)+\frac{1}{2}\left[x-\frac{x^{2}}{2}+\frac{x^{3}}{3} \ldots-\left(-x-\frac{x^{2}}{2}-\frac{x^{3}}{3} \ldots\right)\right]-2 x}{x^{5}}$
$=\lim _{x \rightarrow 0} \frac{2 x+\frac{2 x^{5}}{5} \ldots-2 x}{x^{5}}=\frac{2}{5}$
$\lim _{x \rightarrow 1} x^{\frac{2}{(1-x)}}=e^{\lim _{x \rightarrow 1}\left(\frac{2}{1-x}\right)(x-1)}=e^{-2}$
$\Rightarrow$ Both statements correct

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