​ Compute, calculus help

$\text{Compute}:\displaystyle\lim_{x\to0}(\frac{\sin{x}}{x})^\frac{1}{1-\cos{x}}$

We can approximate $\sin{x} \text{ and } \cos{x}$ by their Taylor series.

After apply substitutions:

$\displaystyle\lim_{x\to0}(\frac{\sin{x}}{x})^\frac{1}{1-\cos{x}}=\lim_{x\to0}(1-\frac{x^2}{6})^\frac{2}{x^2}=\lim_{x\to0}(1+\frac{x^2}{6})^{-\frac{2}{x^2}}$

$=\displaystyle\lim_{x\to\infty}(1+\frac{1}{6x^2})^{-2x^2}=\lim_{x\to\infty}(1+\frac{1}{x})^{-x/3}=\boxed{e^{-1/3}}$

This is because $e=\displaystyle\lim_{x\to\infty}(1+\frac1x)^x$

I hope this helped,

Gavin