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.

$$
\lim_{x \to 0} \left( 1 - \frac {x^2}6\right)^{\frac 1{x^2}} = \lim_{x \to 0} \left [ \left( 1 - \frac {x^2}6\right)^{\frac 6{x^2}} \right ]^{\frac 16} = \left [ \left (e^{-1} \right ) \right ]^{\frac 16} = e^{-\frac 16}
$$
Here, I used somewhat modified limit regarding Euler's number
$$
\lim_{t \to 0} \left( 1 - t\right )^{\frac 1t} = e^{-1}
$$
More info, and proofs can be found here.
