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
On
Folowing your idea
$$\sin x ^ {\arcsin x}=e^{\arcsin x\log (\sin x)}\to e^0=1$$
indeed
$$\arcsin x\log (\sin x)=\frac{\arcsin x}{x}\frac x{\sin x}\cdot \sin x\cdot\log (\sin x)\to 1\cdot 1 \cdot 0=0$$
As an alternative we have that
$$\sin x ^ {\arcsin x}=\left[(\sin x)^{\sin x}\right] ^ {\frac{\arcsin x}{x}\frac{x}{\sin x}}\to 1^1=1$$
indeed by $t=\sin x \to 0$ we have that $(\sin x)^{\sin x}=t^t \to 1$.
$$y=\sin x ^ {\arcsin x} $$
$$ \ln y = \arcsin x. \ln \sin x = \frac {\ln \sin x}{\frac {1}{\arcsin x}}$$
Now apply L.Hospital Rule and you will get $$\lim_{x\to 0} \ln y =0$$ which gives you the $\lim_{x\to 0} y=1$