How to find the limit $\lim_{x \to 0}\sin(x)^{2\sin(x)}$ I can't find the following limit:
$$\lim_{x \to 0}\sin(x)^{2\sin(x)}$$
My attempt :


*

*Let $y =\sin(x)^{2\sin(x)}$. Then we take the natural logarithm of this expression:
$$y=\sin(x)^{2\sin(x)}$$
$$\ln(y) = 2\sin(x)\cdot \ln(\sin(x))$$


$$y=e^{2\sin(x)\cdot \ln(\sin(x))}$$
Then we put this in limit
$$\lim_{x \to 0}\sin(x)^{2\sin(x)}= \lim_{x \to 0}(e^{2\sin(x)\cdot \ln(\sin(x))})=e^{0\cdot (-\infty)}=e^0=1$$
I want to know whether it is correct.
 A: Let's focus on $\sin x\cdot\log\sin x$.  It is well-known that $\sin x$ is asymptotic to $x$ as $x\to0$ (see: Taylor series).  Thus we need only consider $$\lim_{x\to0^+}x\log x$$
(we approach $0$ from $+x$ because $\log$ of negative reals is not very nice to work with). If we substitute $u=\frac1x$:
$$ \lim_{x\to0^+}x\log x=\lim_{u\to\infty}\frac{-\log u}{u} $$
It should be plainly obvious that the latter expression approaches $0$ for large $u$, and if not one may apply the ever-useful L'Hopital's rule to simplify the limit to $\frac1u$.  Thus, we obtain that $$\lim_{x\to0}\sin^{2\sin x} x=\lim_{x\to0}e^{2\sin x\log \sin x}=e^0=1$$
Without this methodology, $0\cdot (-\infty)$ is indeterminate and you cannot assume it to be $0$.  For example, $\lim_{x\to\infty}x\cdot\frac1x$ becomes $\infty\cdot 0$ but evaluates to $1$.
A: You can use that $\lim\limits_{x\to 0}\frac{\sin (x)}{x}=1$ and that $\lim\limits_{x\to 0}x\mathrm{ln}(x)= 0$ to show that $\lim\limits_{x\to 0}\sin(x)\mathrm{ln}(\sin(x))=0$ and then you can use the continuity of the exponential function to get the result.
