Given below are two statements:
Statement I: $ \lim\limits_{x \to 0} \left( \frac{\tan^{-1} x + \log_e \sqrt{\frac{1+x}{1-x}} - 2x}{x^5} \right) = \frac{2}{5} $
Statement II: $ \lim\limits_{x \to 1} \left( x^{\frac{2}{1-x}} \right) = \frac{1}{e^2} $
In the light of the above statements, choose the correct answer from the options given below:
Solution
<p>$$\begin{aligned}
&\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\tan ^{-1} x+\frac{1}{2}[\ln (1+x)-\ln (1-x)]-2 x}{x^5} \\
& =\lim _{x \rightarrow 0} \frac{\left(x-\frac{x^3}{3}+\frac{x^5}{5} \ldots\right)+\frac{1}{2}\left[x-\frac{x^2}{2}+\frac{x^3}{3} \ldots-\left(-x-\frac{x^2}{2}-\frac{x^3}{3} \ldots\right)\right]-2 x}{x^5} \\
& =\lim _{x \rightarrow 0} \frac{2 x+\frac{2 x^5}{5} \ldots .-2 x}{x^5}=\frac{2}{5} \\
& \lim _{x \rightarrow 1} x^{\frac{2}{(1-x)}}=e^{\lim _{x \rightarrow 1}\left(\frac{2}{1-x}\right)(x-1)}=e^{-2}
\end{aligned}\\
&\Rightarrow \text { Both statements correct }
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Limits, Continuity and Differentiability · Topic: Limits and Standard Results
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