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.

When $x\to 0$
$$
\bigg(\frac{\sin x}{x} \bigg)^{\frac{1}{x^2}} 
= \exp\left(\frac{\log\frac{x - x^3/6 + o(x^6)}{x}}{x^2}\right)
= \exp\left(\frac{-x^2/6 + o(x^3)}{x^2}\right) 
= \exp(-1/6 + o(1)) \to \exp(-1/6)
$$
