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?

One can also simplify as follows:
$\displaystyle \begin{aligned}L &= \lim_{x \to 0}\dfrac{1}{1 - \cos x}\log\left(\dfrac{\sin x}{x}\right)\\
&= \lim_{x \to 0}\dfrac{1}{1 - \cos x}\log\left(1 + \dfrac{\sin x}{x} - 1\right)\\
&= \lim_{x \to 0}\dfrac{1}{1 - \cos x}\left(\dfrac{\sin x}{x} - 1\right)\dfrac{\log\left(1 + \dfrac{\sin x}{x} - 1\right)}{\left(\dfrac{\sin x}{x} - 1\right)}\\
&= \lim_{x \to 0}\dfrac{1}{1 - \cos x}\left(\dfrac{\sin x}{x} - 1\right)\cdot 1\\
&= \lim_{x \to 0}\dfrac{x^{2}}{1 - \cos x}\cdot \dfrac{\sin x - x}{x^{3}}\\
&= \lim_{x \to 0}\dfrac{x^{2}}{1 - \cos x}\cdot\lim_{x \to 0}\dfrac{\sin x - x}{x^{3}}\\
&= \lim_{x \to 0}\dfrac{x^{2}}{1 - \cos x}\cdot\lim_{x \to 0}\dfrac{\cos x  - 1}{3x^{2}}\text{ (by L'Hospital's Rule)}\\
&= \lim_{x \to 0}\dfrac{x^{2}}{1 - \cos x}\cdot\dfrac{\cos x  - 1}{3x^{2}}\\
&= -\frac{1}{3}\end{aligned}$
This solves the problem with minimal use of L'Hospital Rule. We have made use of the standard limit $\lim_{y \to 0}\dfrac{\log (1 + y)}{y} = 1$ where $y = \dfrac{\sin x}{x} - 1$ which tends to $0$ as $x \to 0$.
