Prove the limit $\mathop {\lim }\limits_{x \to {0^ + }} {x^{\sin (x)}} = 1$ Prove the following limit:
$$\mathop {\lim }\limits_{x \to {0^ + }} {x^{\sin (x)}} = 1$$
I can use limits arithmatic, squeezing principle, "well-known" limits etc..
We didn't learn Lopital law so I can't use it.
I tried to use the above tools, but didn't reach anything significant.
I'm suspecting it's involving trigonometric/logarithmic magic I'm not able to see.  

EDIT: Note that on $(0,1)$, $x^a < x^b$ whenever $a>b$. The adjustment below reflects this. I give a shout out to Baranovskiy for point out this error.
I have an alternative proof. Note that $1 = x^0 \geq x^{\sin(x)} \geq x^x$ in some neighbourhood $(0,\delta)$. What does this tell us? Then squeeze.
