Given below are two statements: Statement I: x 0lim( ^-1x+ _e(1+x)/(1-x)-2xx^5)=(2)/(5) St…

Question

Given below are two statements:

Statement I: $\underset{x	o 0}{lim}(\frac{{	an }^{-1}x+{\log }_{e}\sqrt{\frac{1+x}{1-x}}-2x}{{x}^{5}})=\frac{2}{5}$

Statement II: $\underset{x	o 1}{lim}({x}^{\frac{2}{1-x}})=\frac{1}{{e}^{2}}$

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

Accepted Answer

$$\begin{aligned}
&\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\tan ^{-1} x+\frac{1}{2}[\ln (1+x)-\ln (1-x)]-2 x}{x^5} \
& =\lim _{x \rightarrow 0} \frac{\left(x-\frac{x^3}{3}+\frac{x^5}{5} \ldots\right)+\frac{1}{2}\left[x-\frac{x^2}{2}+\frac{x^3}{3} \ldots-\left(-x-\frac{x^2}{2}-\frac{x^3}{3} \ldots\right)\right]-2 x}{x^5} \
& =\lim _{x \rightarrow 0} \frac{2 x+\frac{2 x^5}{5} \ldots .-2 x}{x^5}=\frac{2}{5} \
& \lim _{x \rightarrow 1} x^{\frac{2}{(1-x)}}=e^{\lim _{x \rightarrow 1}\left(\frac{2}{1-x}\right)(x-1)}=e^{-2}
\end{aligned}\
\&\Rightarrow \text { Both statements correct }
\end{aligned}$$

Solutions · 1