<s>[INST] Finding $\lim\limits_{x\to 0}(\frac{\sin(x)}{x})^{\frac{\sin(x)}{x-\sin(x)}}$ with and without L'Hopital's Rule. 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 $$\boxed{L=\frac1{e}}$$
 [/INST] 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 $L = e^{-1} = 1/e$.
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