Given below are two statements: Statement I: \lim\limits_{x o 0} \left( \frac{ an^{-1} x + \log_e \sqrt{\frac{1+x}{1-x}} - 2x}{x^5} ight) = \frac{2}{5} Statement II: \lim\limits_{x o 1} \left( x^{\frac{2}{1-x}} ight) = \frac{1}{e^2} In the light of the above statements, choose the correct a

Question

Given below are two statements: Statement I: \lim\limits_{x 	o 0} \left( \frac{	an^{-1} x + \log_e \sqrt{\frac{1+x}{1-x}} - 2x}{x^5} ight) = \frac{2}{5} Statement II: \lim\limits_{x 	o 1} \left( x^{\frac{2}{1-x}} ight) = \frac{1}{e^2} In the light of the above statements, choose the correct answer from the options given below:

Accepted Answer

Answer: C. Both Statement I and Statement II are true \begin{aligned} &\begin{aligned} & \lim _{x ightarrow 0} \frac{	an ^{-1} x+\frac{1}{2}[\ln (1+x)-\ln (1-x)]-2 x}{x^5} \ & =\lim _{x ightarrow 0} \frac{\left(x-\frac{x^3}{3}+\frac{x^5}{5} \ldotsight)+\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ight)ight]-2 x}{x^5} \ & =\lim _{x ightarrow 0} \frac{2 x+\frac{2 x^5}{5} \ldots .-2 x}{x^5}=\frac{2}{5} \ & \lim _{x ightarrow 1} x^{\frac{2}{(1-x)}}=e^{\lim _{x ightarrow 1}\left(\frac{2}{1-x}ight)(x-1)}=e^{-2} \end{aligned}\ &\Rightarrow 	ext { Both statements correct } \end{aligned}

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