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 $$\boxed{L=\frac1{e}}$$

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, $L=e^{-1}$
