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: 
Take logarithm as you did then write what follows
\begin{equation\*}
\frac{\sin x}{x-\sin x}\ln \left( \frac{\sin x}{x}\right) =\frac{\ln \left( 
\frac{\sin x}{x}\right) }{\frac{x-\sin x}{\sin x}}=\frac{\ln \left( 1+\left[ 
\frac{\sin x}{x}-1\right] \right) }{\frac{x-\sin x}{x}\times \frac{x}{\sin x}%
}=-\frac{\ln \left( 1+\left[ \frac{\sin x}{x}-1\right] \right) }{\left[ 
\frac{\sin x}{x}-1\right] }\times \frac{\sin x}{x}
\end{equation\*}
Now we use classic limits
\begin{eqnarray\*}
\lim\_{x\rightarrow 0}\frac{\sin x}{x} &=&1 \\
\lim\_{u\rightarrow 0}\frac{\ln (1+u)}{u} &=&1
\end{eqnarray\*}
it follows that 
\begin{equation\*}
\lim\_{x\rightarrow 0}\frac{\sin x}{x-\sin x}\ln \left( \frac{\sin x}{x}%
\right) =-\lim\_{x\rightarrow 0}\frac{\ln \left( 1+\left[ \frac{\sin x}{x}-1%
\right] \right) }{\left[ \frac{\sin x}{x}-1\right] }\times \frac{\sin x}{x}%
=-1\times 1=-1.
\end{equation\*}
Therefore, $L=e^{-1}$
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