Why does $\lim_{x\to 0^+} x^{\sin x}=1$? I tried writing $y=x^{\sin x}$, so $\ln y=\sin x\ln x$. I tried to rewrite the function as $\ln(x)/\csc x$ and apply l'Hopital to the last function, but it's a mess. 
Is there a by hand way to do it?

Rewrite the limit of the logarithm as
$$\lim_{x\to 0^+}\frac{\sin x}{x}x\ln x.$$
The behaviour of $\frac{\sin x}{x}$ is known. So we need $\lim_{x\to 0^+} x\ln x$, which is probably familiar, and yields to one application of L'Hospital's Rule. 
