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?
 A: Sometimes there can be a lot of repetitions of l'Hopital needed.
And here we even have some "nested" l'Hopital because of $\frac{\sin x}x$.
A nice way to avoid getting lost in l'Hopital bookkeeping is to work with power series expansions. This leads to 
$$\tag1\ln\left(\frac{\sin x}x\right)= -\frac16 x^2+O(x^4)$$
and 
$$1-\cos x= \frac12 x^2+O(x^4),$$
hence the quotient is just $-\frac13+O(x^2)$ and has limit $-\frac13$.

How does this work before one learns about Taylor series or if one does not even want to worry about convergence?
Well, you may simply consider "$f(x)=a_0+a_1x+\ldots +a_{n-1}x^{n-1}+O(x^n)$" simply as a friendly encoding of "$f$ is at least $n-1$ times continuously differentiable at $x=0$ and $f^{(k)}(0)=k!a_k$ for $0\le k<n$". It follow sthat the most obvious rules apply for combining such results.
Especially, $\sin x=x-\frac16x^3+O(x^5)$ implies $\frac{\sin x}x=1-\frac16x^2+O(x^4)$. Together with $\ln(1+x)=x+O(x^2)$ this produces $(1)$ above. $(2)$ follows simply from $\cos x=1-\frac12x^2+O(x^4)$. Cancellation of $x^2$ and dropping the $O$ terms then is equivalent to applying l*Hopital to times.
A: You can try to expand
$$
\log \frac{\sin x}{x} = -\frac{x^2}{6}-\frac{x^4}{180}+O(x^6)
$$
and reduce to the limit of
$$
\frac{-\frac{x^2}{6}-\frac{x^4}{180}+O(x^6)}{1-\cos x} = \frac{-\frac{x^2}{6}-\frac{x^4}{180}+O(x^6)}{\frac{x^2}{2}-\frac{x^4}{24}+O(x^6)},
$$
which is now easy to solve.
A: It should be used. There may be some little mistakes in your computation.
$$\begin{align*}
\lim_{x\rightarrow0}\frac{1}{1-\cos x}\ln\frac{\sin x}{x}
&=\lim_{x\rightarrow0}\frac{\frac{\cos x}{\sin x}-\frac{1}{x}}{\sin x}\\
&=\lim_{x\rightarrow0}\frac{x\cos x-\sin x}{x\sin^{2}x}\\
&=\lim_{x\rightarrow0}\frac{-x\sin x}{\sin^{2}x+2x\sin x\cos x}\\
&=-\lim_{x\rightarrow0}\frac{1}{\frac{\sin x}{x}+2\cos x}\\
&=-\frac{1}{3}
\end{align*}$$
A: Use $\ln(\frac{\sin x}{x})=\ln(\sin x)-\ln x$
Then, $$\frac{d(\ln(\frac{\sin x}{x}))}{dx}=\cot x-\frac{1}{x}=\frac{x\cos x-\sin x}{x\sin x}$$
and $$\frac{d(1-\cos x)}{dx}=\sin x$$ which gives $$\lim_{x \to 0} {(\frac{1}{1 - \cos x})\ln (\frac{\sin x}{x})}=\lim_{x \to 0} \frac{(x\cos x-\sin x)}{x\sin^2 x}$$ which again applying L'Hopital's rule gives $$\lim_{x \to 0} \frac{(x\cos x-\sin x)}{x\sin^2 x}=\lim_{x \to 0} \frac{-x\sin x}{2x\sin x\cos x+\sin^2 x}=\lim_{x \to 0} \frac{-1}{2\cos x+\frac{\sin x}{x}}=-\frac{1}{3}$$ since $\lim_{x\to 0}\frac{\sin x}{x}=1$
A: 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 $x \to 0$.
