Find the value of $\lim_{x \to 0}\left (\frac {\sin x}{x}\right)^{\frac {1}{x^2}}$. I am stuck with the following problem :

Find the value of $$\lim_{x \to 0}\left (\frac {\sin x}{x}\right)^{\frac {1}{x^2}}$$

My try : Let $$p=\lim_{x \to 0} \left(\frac {\sin x}{x}\right)^{\frac {1}{x^2}}\implies \log p=\lim_{x \to 0}{\frac {1}{x^2}}\log\left(\frac {\sin x}{x}\right)$$...After applying l'hospitals rule few times  I get 
$$\log p= \lim_{x \to 0}\frac{1-\sec^2x}{4x \tan x+2x^2\sec^2x}$$.. we can again apply l'hospitals rule ,but the calculations get bigger and bigger..Is there any other easier way around or I am missing something?

From the Taylor expansion of $\sin x$ around $0$, namely $\sin u = u -\frac{u^3}{6} + o(u^3)$; and that of $\ln(1+u)$, specifically $\ln(1+u)=u+o(u),$ we get
$$\begin{align}
\left( \frac{\sin x}{x}\right)^{1/x^2}
&= \left(1-\frac{x^2}{6} + o(x^2)\right)^{1/x^2}
= e^{\frac{1}{x^2}\ln\left(1-\frac{x^2}{6} + o(x^2)\right) }
= e^{\frac{1}{x^2}\left(-\frac{x^2}{6} + o(x^2)\right) }\\
&= e^{-\frac{1}{6} + o(1)}
\xrightarrow[x\to0]{}\boxed{e^{-\frac{1}{6}}}
\end{align}$$
