Find the limit of the expression $\lim_{x\to 0}\left(\frac{\sin x}{\arcsin x}\right)^{1/\ln(1+x^2)}$ Limit: $\lim_{x\to 0}\left(\dfrac{\sin x}{\arcsin x}\right)^{1/\ln(1+x^2)}$ I have tried to do this: it is equal to $e^{\lim\frac{\log{\frac{\sin x}{\arcsin x}}}{\log(1+x^2)}}$, but I can't calculate this with the help of l'Hopital rule or using Taylor series, because there is very complex and big derivatives, so I wish to find more easier way. 
$$\lim_{x\rightarrow 0}{\frac{\log{\frac{\sin x}{\arcsin x}}}{\log(1+x^2)}} = \lim_{x\rightarrow 0}\frac{\log1 + \frac{-\frac{1}{3}x^2}{1+\frac{1}{6}x^2+o(x^2)}}{\log(1+x^2)} = \lim_{x\rightarrow0}\frac{\frac{-\frac{1}{3}x^2}{1+\frac{1}{6}x^2+o(x^2)} + o(\frac{-\frac{1}{3}x^2}{1+\frac{1}{6}x^2+o(x^2)})}{x^2+o(x^2)}$$ using Taylor series. Now I think that it's not clear for me how to simplify $o\left(\frac{-\frac{1}{3}x^2}{1+\frac{1}{6}x^2+o(x^2)}\right)$.
 A: HINT
By Taylor's series
$$\frac{\sin x}{\arcsin x}=\frac{x-\frac16x^3+o(x^3)}{x+\frac16x^3+o(x^3)}=\frac{1-\frac16x^2+o(x^2)}{1+\frac16x^2+o(x^2)}=$$$$=\left(1-\frac16x^2+o(x^2)\right)\left(1+\frac16x^2+o(x^2)\right)^{-1}$$
Can you continue form here using binomial series for the last term?
A: So you want to find the limit
$$L=\lim\limits_{x\to0} \frac{\ln\frac{\sin x}{\arcsin x}}{\ln(1+x^2)}.$$
Perhaps a reasonable strategy would be to split this into calculating several simpler limits.
We know that 
\begin{gather*}
\lim\limits_{x\to0} \frac{\ln\frac{\sin x}{\arcsin x}}{\frac{\sin x}{\arcsin x}-1}=1\\
\lim\limits_{x\to0} \frac{\ln(1+x^2)}{x^2}=1
\end{gather*}
so we eventually get to the limit
$$L=\lim\limits_{x\to0} \frac{\frac{\sin x}{\arcsin x}-1}{x^2} =\lim\limits_{x\to0} \frac{\sin x-\arcsin x}{x^2\arcsin x}.$$
If we also use that $\lim\limits_{x\to0} \frac{\arcsin x}x=1$, we get that
$$L=\lim\limits_{x\to0} \frac{\sin x-\arcsin x}{x^3}.$$
And now we can try to calculate separately the two limits
\begin{align*}
L_1&=\lim\limits_{x\to0} \frac{\sin x-x}{x^3}\\
L_2&=\lim\limits_{x\to0} \frac{x-\arcsin x}{x^3}
\end{align*}
Both $L_1$ and $L_2$ seem as limits where L'Hospital's rule or Taylor expansion should lead to result. In fact, substitution $y=\sin x$ transforms $L_2$ to a limit very similar to $L_1$.
You can probably find also some posts on this site at least for $L_1$ (and maybe also for $L_2$). For example:
Solve $\lim_{x\to 0} \frac{\sin x-x}{x^3}$, 
Find the limit $\lim_{x\to0}\frac{\arcsin x -x}{x^2}$,
Are all limits solvable without L'Hôpital Rule or Series Expansion. 
