Find
$$L=\lim_{x \to 0}\left(\frac{\arcsin x}{x}\right)^{\frac1{x^2}}$$
Not much experienced with limits of this kind, I have tried to reduce this limit to a form without exponent:
$$\ln L=\lim_{x \to 0}\ln\left(\frac{\arcsin x}{x}\right)^{\frac1{x^2}}$$
$$\ln L=\lim_{x \to 0}\frac{\ln\arcsin x – \ln x}{x^2}$$
…and then I tried to apply L'Hospital to numerator and denominator. However, the solution becomes a complete mess and you can repeat derivation as many times as you want without ever reaching a conclusion.
I also tried to introduce substitution $x=\sin y$ which leads to:
$$L=\lim_{y \to 0}\left(\frac{y}{\sin y}\right)^{\frac1{\sin^2 y}}$$
This makes manipulation slightly easier but the same approach (L'Hospital after log) is still unable to generate a solution.
Best Answer
Notice that $\frac{\arcsin x}{x} \to 1$ as $x \to 0$, then
$$\ln L = \lim_{x \to 0} \frac{\ln\left(1+\frac{\arcsin x}{x}-1\right)}{x^2} = \lim_{x \to 0} \frac{\frac{\arcsin x}{x}-1}{x^2} = \lim_{x \to 0} \frac{\arcsin x-x}{x^3}.$$
Taking $x = \sin t$, the limit becomes
$$\ln L = \lim_{t \to 0} \frac{t - \sin t}{\sin^3 t} = \lim_{t \to 0} \frac{t - \sin t}{t^3} = \lim_{t \to 0} \frac{1 - \cos t}{3t^2} = \frac{1}{6},$$
and hence $L = e^{1/6}$.
Note: Such method is useful to most of $1^{\infty}$-form limits.