calculus - Should L'Hopital's Rule be used for this limit?

Question

$$\lim_{x \to 0} \left(\dfrac{\sin x}{x}\right)^{\dfrac{1}{1 - \cos x}}$$

Attempt

This limit is equal to,

$$\lim_{x \to 0} \exp\left({\dfrac{1}{1 - \cos x}\ln \left(\dfrac{\sin x}{x}\right)}\right).$$

I hope to solve this by focusing on the limit,

$$\lim_{x \to 0} {\dfrac{1}{1 - \cos x}\ln \left(\dfrac{\sin x}{x}\right)}.$$

Which is indeterminate of the form "$\frac{0}{0}$". However, repeated application of L'Hopital's Rule seems to lead to an endless spiral of trigonometric terms, but wolfram Alpha says the limit I am focusing on exists.

Is there another way to solve this?

Answer

It should be used. There may be some little mistakes in your computation.

$$\begin{align*}
\lim_{x\rightarrow0}\frac{1}{1-\cos x}\ln\frac{\sin x}{x}
&=\lim_{x\rightarrow0}\frac{\frac{\cos x}{\sin x}-\frac{1}{x}}{\sin x}\\
&=\lim_{x\rightarrow0}\frac{x\cos x-\sin x}{x\sin^{2}x}\\
&=\lim_{x\rightarrow0}\frac{-x\sin x}{\sin^{2}x+2x\sin x\cos x}\\

&=-\lim_{x\rightarrow0}\frac{1}{\frac{\sin x}{x}+2\cos x}\\
&=-\frac{1}{3}
\end{align*}$$