Undetermined form of limits i have to find the solution of 
limit of $\sin(x)^{\arcsin(x)}$ as $x$ approaches $0$.
i have been trying a lot of different ways on how to find the solution of it but none of them helped me.

I tried to turn it into $0/0$ form and use l'hopital's rule since its the form $0^0$ but i am stuck and i can not proceed it anymore.

i did this 
$f(x)=\sin(x)$,
$g(x)=\arcsin(x)$
turned it like this

$\exp\left(\arcsin(x)/(1/\ln(\sin(x)))\right)$ as x approaches $0$ but no result
$$y=\sin x ^ {\arcsin x} $$

$$ \ln y = \arcsin x. \ln \sin x = \frac {\ln \sin x}{\frac {1}{\arcsin x}}$$

Now apply L.Hospital Rule and you will get $$\lim_{x\to 0}  \ln y =0$$ which gives you the $\lim_{x\to 0} y=1$
Hints:

Thanks to the continuity of the arc sine function,

$$\lim_{x\to 0}(\arcsin x)^{\arcsin x}=\lim_{y\to 0}y^y.$$



By direct computation, consider the logarithm

$$\arcsin x\log(\arcsin x)=\frac{\log(\arcsin x)}{\dfrac1{\arcsin x}}$$

and by L'Hospital, this is the limit of

$$-\frac{\dfrac{(\arcsin x)'}{\arcsin x}}{\dfrac{(\arcsin x)'}{\arcsin^2x}}=-\arcsin x.$$