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?

Another way to think about this: 
$$\lim_{x \to 0} \sin x \ln x = \lim_{x \to 0} x \ln  x$$
(because $(\sin x) /x \to 1$ as $x \to 0$).  Writing $x = 1/y$,
we have
$$\lim_{x \to 0} x \ln x = - \lim_{y \to +\infty}(\ln y)/y.$$ 
Now $\ln y$ grows slower than any positive power of $y$, and so 
$\ln y / y \to 0$ as $y \to +\infty$.
(Or, one can make another subsitution: $y = e^z,$ so that
$(\ln y)/y = z/e^z,$ and then use the fact that $e^z$ grows faster
than any power of $z$, so that $z/e^z \to 0$ as $z \to + \infty$.)
