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

• This is not a homework solving site. Please see the help center. In particular, show effort and expect to do the work – Brevan Ellefsen Nov 29 '18 at 7:57
• Why don't you try the usual approach of taking logarithm and let us know if you face any problems? Just giving a problem statement is not encouraged here. – Paramanand Singh Nov 29 '18 at 7:59
• Now that you have effectively taken logs and also used $e$ just focus on the exponent itself. Do you recall any standard limits by looking at the denominator $\log(1+x^2)$? – Paramanand Singh Nov 29 '18 at 8:10
• If you look carefully both numerator and denominator are log expressions which tend to $0$ and hence they can be simplified by the use of the same standard limit and you should try to proceed in that manner. – Paramanand Singh Nov 29 '18 at 8:16
• @J_G that first step seems wrong$$\lim_{x\rightarrow 0}{\frac{\log{\frac{\sin x}{\arcsin x}}}{\log(1+x^2)}} = \lim_{x\rightarrow 0}\frac{\log1 + \frac{-\frac{1}{3}x^2}{1+\frac{1}{6}x^2+o(x^2)}}{\log(1+x^2)}...$$ – user Nov 29 '18 at 8:30

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?

• This won't help as it leads to indeterminate form $1^{\infty}$. Or may be you wanted to emphasize that it is a particular indeterminate form. – Paramanand Singh Nov 29 '18 at 8:01
• @ParamanandSingh Yes it was just a hint to start with. Do you think it too small? – user Nov 29 '18 at 8:05
• It will help if the asker knows how to deal with $1^{\infty}$ and unless the asker says anything we can't be sure. – Paramanand Singh Nov 29 '18 at 8:07
• @ParamanandSingh After the editing, I've added a different hint to start with using Taylor's series. – user Nov 29 '18 at 8:12

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.

• +1 this is what I was suggesting in comments. A little algebraic manipulation combined with standard limits always helps. – Paramanand Singh Nov 30 '18 at 2:45