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: 
You need to work a bit more. Let's continue it as follows
\begin{align}
\log L &= \lim\_{x \to 0}\frac{\sin x}{x - \sin x}\log\left(\frac{\sin x}{x}\right)\notag\\
&= \lim\_{x \to 0}\frac{\sin x}{x - \sin x}\log\left(1 + \frac{\sin x}{x} - 1\right)\notag\\
&= \lim\_{x \to 0}\frac{\sin x}{x - \sin x}\cdot\left(\dfrac{\sin x}{x} - 1\right)\cdot\dfrac{\log\left(1 + \dfrac{\sin x}{x} - 1\right)}{\dfrac{\sin x}{x} - 1}\notag\\
&= \lim\_{x \to 0}\frac{-\sin x}{x}\cdot\dfrac{\log\left(1 + \dfrac{\sin x}{x} - 1\right)}{\dfrac{\sin x}{x} - 1}\notag\\
&= -1\cdot\lim\_{t \to 0}\frac{\log(1 + t)}{t}\text{ (putting }t = \frac{\sin x}{x} - 1)\notag\\
&= -1\cdot 1 = -1
\end{align}
Hence we have $L = e^{-1} = 1/e$.
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.