Question

$\underset{x\to 0}{\lim }{\left\{\frac{e^x+e^{-x}-2}{x^2}\right\}}^{1/x^2}$

Answer 1

$$\begin{gathered} \underset{x\to 0}{\lim }{\left[\frac{e^x+e^{-x}-2}{x^2}\right]}^{\left(\frac{1}{x^2}\right)} \\ =\underset{x\to 0}{\lim }{\left[1+\frac{e^x+e^{-x}-2}{x^2}-1\right]}^{\left(\frac{1}{x^2}\right)} \\ =e{}^{\underset{x\to 0}{\lim }\left(\frac{e^x+e^{-x}-2}{x^2}-1\right)\times \left(\frac{1}{x^2}\right)} \\ \because e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+......\propto \\ {e}^{-x}=1-\frac{x}{1!}+\frac{x^2}{2!}-\frac{x^3}{3!}+......\propto \\ \Rightarrow e^x+e^{-x}=2+\frac{2x^2}{2!}+\frac{2x^4}{4!}+.....\propto \\ =e^{\underset{x\to 0}{\lim }\left(\frac{2+{\displaystyle \frac{2x^2}{2!}}+{\displaystyle \frac{2x^4}{4!}}...\propto -2}{x^2}-1\right)\times \left(\frac{1}{x^2}\right)} \\ =e^{\underset{x\to 0}{\lim }\left(\frac{{\displaystyle \frac{2x^2}{2!}}+{\displaystyle \frac{2x^4}{4!}}+......\propto }{x^4}-\frac{1}{x^2}\right)} \\ =e^{\underset{x\to 0}{\lim }\left(\frac{x^2+{\displaystyle \frac{x^4}{12}}+.....\propto -x^2}{x^4}\right)} \\ =e^{\frac{1}{12}} \end{gathered}$$