The value of $\underset{x\to 0}{lim} {\left(\frac{\sin x}{x}\right)}^{\frac{1}{x^2}},$ is

Solution:

We have,

$\underset{x\to 0}{lim} {\left(\frac{\sin x}{x}\right)}^{\frac{1}{x^2}}$

$=\underset{x\to 0}{lim} {\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\dots .\right)}^{1/x^2}$

$=\underset{x\to 0}{lim} {\left(1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+\dots \right)}^{\frac{1}{x^2}}$

$$\begin{gathered} =e^{\underset{x\to 0}{lim}\frac{1}{x^2}\left(\frac{-x^2}{3!}+\frac{x^4}{5!}\frac{x^6}{7!}+\dots \right)} \\ =e^{-1/2!}=e^{-1/6} \end{gathered}$$