Find the limit of the expression $\lim_{x\to 0}\left(\frac{\sin x}{\arcsin x}\right)^{1/\ln(1+x^2)}$

Question

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 $o\left(\frac{-\frac{1}{3}x^2}{1+\frac{1}{6}x^2+o(x^2)}\right)$.

Answer 1

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?

Answer 2

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.