Evaluate $\lim_{x \to 0} \left(\frac{ \sin x }{x} \right)^{\frac{1}{x^2}}$ $$\lim_{x \to 0} \bigg(\frac{\sin x}{x} \bigg)^{\frac{1}{x^2}}$$
The task should be solved by using Maclaurin series so I did some kind of asymptotic simplification
$$\lim_{x \to 0} \bigg(\frac{\sin x}{x} \bigg)^{\frac{1}{x^2}} \approx \lim_{x \to 0} \bigg(\frac{x - \frac{x^3}{6}}{x} \bigg)^{\frac{1}{x^2}} \approx \lim_{x \to 0} \bigg(1 - \frac{x^2}{6} \bigg)^{\frac{1}{x^2}}$$ How can we say answer that is $e$ in the power of $-\frac{1}{6}$. I want some proving of that fact. 
Thank you.

If you saw something variable shows on the exponent, first take logarithm usually helps. Write 
\begin{align}
& \left(\frac{\sin x}{x}\right)^{1/x^2} \\
= & \exp\left[\frac{1}{x^2}\log\left(\frac{\sin x}{x}\right)\right] \\
= & \exp\left[\frac{1}{x^2}\log\left(\frac{x - \frac{x^3}{6} + o(x^3)}{x}\right)\right] \\
= & \exp\left[\frac{1}{x^2}\log\left(1 - \frac{1}{6}x^2 + o(x^2)\right)\right] \\
= & \exp\left[\frac{1}{x^2}\left(- \frac{1}{6}x^2 + o(x^2)\right)\right] \\
= & \exp \left(-\frac{1}{6} + o(1)\right) \\
\to & e^{-1/6}
\end{align}
as $x \to 0$.
