Find $\lim_{x\to 0} \left(\frac{x}{\arcsin x}\right)^{\frac{1}{x^2}}$. Find $\lim_{x\to 0} \left(\dfrac{x}{\arcsin x}\right)^{\frac{1}{x^2}}$.
$\lim_{x\to 0}f(x)=e^a$ with $a=\lim_{X\to 0}\dfrac{1}{x^2}\left(\dfrac{x}{\arcsin x}-1\right)=\lim_{X\to 0}\dfrac{x-\arcsin x}{x^3}$
Continue i used formular L'hospotal .

My hint:
Put $$L=\lim_{x\to 0}\left(\frac{x}{\arcsin x}\right)^{\frac{1}{x^2}}=\exp\left\{\lim_{x\to 0}\frac{1}{x^2}\ln\frac{x}{\arcsin x}\right\}=\exp\left\{-\lim_{x\to 0}\frac{1}{x^2}\ln\frac{\arcsin x}{x}\right\}$$
We use Taylor' expansion, we get:
$$\arcsin x=x+\frac{x^3}{6}+o(x^3)\to \frac{\arcsin x}{x}=1+\frac{x^2}{6}+o(x^2)$$
Hence:
$$L=\exp\left\{-\lim_{x\to 0}\frac{1}{x^2}\ln\left(1+\frac{x^2}{6}+o(x^2)\right)\right\}=\exp\left\{-\frac{1}{6}\right\}=\frac{1}{\sqrt[6]{e}}$$
