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.

One may recall that, as $x \to 0$, we have
$$
\cos x =1-\frac {x^2}{2}+\mathcal{O}(x^3), \quad
\sin x =x-\frac{x^3}{6}+\mathcal{O}(x^4), \quad
\ln (1+x)=x+\mathcal{O}(x^2),
$$ giving
$$
\begin{align}
\frac{1}{1-\cos x}&=\frac 2{x^2}+\mathcal{O}(1)\\\\
\ln \left(\frac{\sin x}x\right)& =-\frac{x^2}{6}+\mathcal{O}(x^3).
\end{align}
$$ Then, as $x \to 0$,
$$
\begin{align}
\left(\frac{\sin x}x\right)^{\small \dfrac{1}{1-\cos x}}&=e^{\Large \frac{1}{1-\cos x}\ln \left(\frac{\sin x}x\right)}\\\\
& =e^{\large \left(\frac 2{x^2}+\mathcal{O}(1)\right)\left(-\frac{x^2}{6}+\mathcal{O}(x^3)\right)}\\\\
& =e^{\Large -\frac 13+\mathcal{O}(x)}
\end{align}
$$ and the desired limit is $\displaystyle e^{-\large \frac 13}$.
