Finding $\lim_{x \to 0^+} x^{\sin x}$ 
Find $\displaystyle \lim_{x \to 0^+} x^{\sin x}$

This is how I started but I get to a dead end fast: 
$\displaystyle\lim_{x \to 0^+} e^{\ln x^{\sin x}}=\lim_{x \to 0^+} e^{\sin x \ln x}$
I guess I can bound it below with $e^0=1$ but I have no idea with what from above. 
Any advice on how to continue ?
PS: we can't use integration/derivation nor Taylor's theorem because we haven't covered that.

Hint: Note that 
$$(\sin x)(\ln x)=\frac{\sin x}{x}(x\ln x).$$ 
