How to find limit of $\lim_{x\to 0}\sin x^x$?

Question

$\displaystyle\lim_{x\to 0}^{}\sin{x}^{x}$. I tried this $ P =\displaystyle \lim_{x\to 0}(\sin{x})^x$ taking the natural log on both side we get $\log_{e}P= \displaystyle \lim_{x\to 0}\log_{e}(\sin{x})^x$

$=\displaystyle \lim_{x\to 0}x\ln(\sin{x})$

$=\displaystyle \lim_{x\to 0}\frac{\ln(\sin{x})}{1\over x}$

this is $\frac{\infty}{\infty}$

I think we should use L'Hospital's rule, but I can't find the answer.

Answer 1

$$\lim_{x\to0}(\sin x)^x=\lim_{x\to0}e^{x\ln(\sin x)}=\lim_{x\to0}e^{\displaystyle \frac{\ln(\sin x)}{1/x}}\stackrel{L'H}=\lim_{x\to0}e^{\displaystyle \frac{\cot x}{-1/x^2}}=\lim_{x\to0}e^{\displaystyle -x\times\frac{x}{\tan x}}=e^{0\times1}=1$$

Answer 2

For the right-hand side limit:

Suppose $0<x<\pi$. Then: $$x\ln(\sin x)=\frac x{\sin x}\cdot \sin x \ln(\sin x)$$ Now $\dfrac x{\sin x}\to 1$ as $x\to 0$, and $\,u\ln u\to 0$ as $u\to 0_+$, so
$$x\ln(\sin x)\to 0\enspace\text{and}\quad\lim_{x\to 0_+} (\sin x)^x = 1$$