Evaluating $\lim\limits_{x \to 0}\left(\frac{\sin x}{x}\right)^{\frac{1}{1-\cos x}}$ How do I evaluate
$$\lim_{x \to 0}\left(\frac{\sin x}{x}\right)^{\dfrac{1}{1-\cos x}}\ ?$$
I tried using the fact that $\left(\frac{\sin x}{x}\right)^{\frac{1}{1-\cos x}} = \exp\left(\ln\bigg(\frac{\sin x}{x}\right)\frac{1}{1-\cos x}\bigg)$ and then I am already stuck.

With l'Hospital directly:
\begin{align*}
\lim_{x\to 0}\frac{\log\frac{\sin x}x}{1-\cos x} & \stackrel{\text{l'H}}=\lim_{x\to 0}\frac{\frac 1{\sin x}\cdot\frac{x\cos x-\sin x}x}{\sin x} \\
& =\lim_{x\to 0}\frac{x\cos x-\sin x}{x\sin^2x}\\
& \stackrel{\text{l'H}}=\lim_{x\to 0}\frac{-x\sin x}{\sin^2x+2x\sin x\cos x}\\
&=-\lim_{x\to 0}\frac{x}{\sin x+2x\cos x}\\&=-\lim_{x\to 0}\frac1{\frac{\sin x}x+2\cos x}\\
&=-\frac13
\end{align*}
