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

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  • $\begingroup$ This is not a homework solving site. Please see the help center. In particular, show effort and expect to do the work $\endgroup$ – Brevan Ellefsen Nov 29 '18 at 7:57
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    $\begingroup$ 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. $\endgroup$ – Paramanand Singh Nov 29 '18 at 7:59
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    $\begingroup$ 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 $\log(1+x^2)$? $\endgroup$ – Paramanand Singh Nov 29 '18 at 8:10
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    $\begingroup$ 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. $\endgroup$ – Paramanand Singh Nov 29 '18 at 8:16
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    $\begingroup$ @J_G that first step seems wrong$$ \lim_{x\rightarrow 0}{\frac{\log{\frac{\sin x}{\arcsin x}}}{\log(1+x^2)}} = \lim_{x\rightarrow 0}\frac{\log1 + \frac{-\frac{1}{3}x^2}{1+\frac{1}{6}x^2+o(x^2)}}{\log(1+x^2)}...$$ $\endgroup$ – user Nov 29 '18 at 8:30
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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?

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

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

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