Finding $\lim\limits_{x\to 0}(\frac{\sin(x)}{x})^{\frac{\sin(x)}{x-\sin(x)}}$ with and without L'Hopital's Rule.

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 1

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 $L = e^{-1} = 1/e$.

Answer 2

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, $L=e^{-1}$

Answer 3

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

Answer 4

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.

Answer 5

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