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?

Write $x^{\sin{x}} = \exp{\frac{\log{x}}{\frac{1}{\sin{x}}}}$ and apply rule of l'hopital:
$$\lim_{x\rightarrow 0}x^{\sin{x}} = \exp{\left(\lim_{x\rightarrow 0}\frac{\log{x}}{\frac{1}{\sin{x}}}\right)}$$
Now (by 2$\times$ l'hopital)$$\lim_{x\rightarrow 0}\frac{\log{x}}{\frac{1}{\sin{x}}} = \lim_{x\rightarrow 0}\frac{\sin{2x}}{ \cos{x}-x\sin{x}} = 0$$
Thus 
$$\lim_{x\rightarrow 0}x^{\sin{x}} = \exp{\left(\lim_{x\rightarrow 0}\frac{\sin{2x}}{ \cos{x}-x\sin{x}}\right)} = \exp{0} = 1$$
