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 $$\boxed{L=\frac1{e}}$$
 A: 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 $L = e^{-1} = 1/e$.
A: $$
\begin{align}
\log\left(\left(\frac{\sin(x)}{x}\right)^{\frac{\large\sin(x)}{\large x-\sin(x)}}\right)
&=\frac{\sin(x)}{x-\sin(x)}\log\left(\frac{\sin(x)}{x}\right)\\
&=\frac{\sin(x)}x\frac1{\color{#C00000}{1-\frac{\sin(x)}x}}\log\left(1-\left(\color{#C00000}{1-\frac{\sin(x)}x}\right)\right)
\end{align}
$$
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 $\lim\limits_{u\to0}\frac{\log(1-u)}u=-1$.
A: 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, $L=e^{-1}$
A: 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
A: 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.
