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.

If $L$ is the desired limit then
\begin{aligned}
\log L &= \log\left\{\lim_{x \to 0}\left(\frac{\sin x}{x}\right)^{1/(1 - \cos x)}\right\}\notag\\
&= \lim_{x \to 0}\log\left(\frac{\sin x}{x}\right)^{1/(1 - \cos x)}\text{ (by continuity of log)}\notag\\
&= \lim_{x \to 0}\frac{1}{1 - \cos x}\cdot\log\left(\frac{\sin x}{x}\right)\notag\\
&= \lim_{x \to 0}\frac{1 + \cos x}{\sin^{2} x}\cdot\log\left(1 + \frac{\sin x}{x} - 1\right)\notag\\
&= \lim_{x \to 0}(1 + \cos x)\cdot\frac{x^{2}}{\sin^{2} x}\dfrac{\dfrac{\sin x}{x} - 1}{x^{2}}\cdot\dfrac{\log\left(1 + \dfrac{\sin x}{x} - 1\right)}{\dfrac{\sin x}{x} - 1}\notag\\
&= \lim_{x \to 0}(1 + 1)\cdot 1\cdot\frac{\sin x - x}{x^{3}}\cdot 1\notag\\
&= 2\lim_{x \to 0}\frac{\sin x - x}{x^{3}}\notag\\
&= 2\lim_{x \to 0}\frac{\cos x - 1}{3x^{2}}\text{ (by L'Hospital Rule)}\notag\\
&= \frac{2}{3}\lim_{x \to 0}\frac{(\cos x - 1)(\cos x + 1)}{x^{2}(\cos x + 1)}\notag\\
&= \frac{2}{3}\lim_{x \to 0}\left(-\frac{\sin^{2}x}{x^{2}}\right)\frac{1}{1 + \cos x}\notag\\
&= -\frac{2}{3}\cdot\frac{1}{1 + 1} = -\frac{1}{3}\notag
\end{aligned}
It follows that $L = e^{-1/3}$.
