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 $\lim\limits\_{u\to0}\frac{\log(1-u)}u=-1$.
Let $y=\dfrac{\sin x}{x},$ then $x\to 0\iff y\to 1.$
$$\lim\limits\_{x\to 0}\left(\dfrac{\sin x}{x}\right)^{\dfrac{\sin x}{x-\sin x}}
=\lim\_{y\to1}y^{\left(\dfrac{y}{1-y}\right)}$$ continue from here.