Question: So, we have to find $\lim\limits\_{x\to 0}(\frac{\sin(x)}{x})^{\frac{\sin(x)}{x-\sin(x)}}$ with and without L'Hopital's Rule.

**My Work**: Let $\lim\limits\_{x\to 0}(\frac{\sin(x)}{x})^{\frac{\sin(x)}{x-\sin(x)}}=L$   

Taking $\ln$ of both sides and bringing the exponent down.

$\lim\limits\_{x\to 0}{\frac{\sin(x)}{x-\sin(x)}\ln(\frac{\sin(x)}{x})}=\ln(L)$

But it changes to undefined form? The answer (in my textbook) is $$\boxed{L=\frac1{e}}$$

Answer: 
$$
\begin{align}
\log\left(\left(\frac{\sin(x)}{x}\right)^{\frac{\large\sin(x)}{\large x-\sin(x)}}\right)
&=\frac{\sin(x)}{x-\sin(x)}\log\left(\frac{\sin(x)}{x}\right)\\
&=\frac{\sin(x)}x\frac1{\color{#C00000}{1-\frac{\sin(x)}x}}\log\left(1-\left(\color{#C00000}{1-\frac{\sin(x)}x}\right)\right)
\end{align}
$$
Since $\lim\limits\_{x\to0}\frac{\sin(x)}x=1$, we also have $\lim\limits\_{x\to0}\left(1-\frac{\sin(x)}x\right)=0$. All we now need is $\lim\limits\_{u\to0}\frac{\log(1-u)}u=-1$.
Change variable: $\frac{sinx}{x} = u$. Then $u\to 1$ when $x\to 0 $. 

Then $$ln(L) = \lim\_{u\to 1} \frac{ln(u)}{1/u - 1}$$
It will be then much easier to work with/without L'Hospital rule