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 $o\left(\frac{-\frac{1}{3}x^2}{1+\frac{1}{6}x^2+o(x^2)}\right)$.
Find the limit of the expression $\lim_{x\to 0}\left(\frac{\sin x}{\arcsin x}\right)^{1/\ln(1+x^2)}$
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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 07:57
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1Why 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 07:59
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@BrevanEllefsen We can also give an help for homework as hint and suggestion without give a full solution. – user Nov 29 '18 at 07:59
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@J_G Let follow the given hint, what do you obtain? – user Nov 29 '18 at 08:00
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@BrevanEllefsen Yes, I agree with you, indeed the point is not whether it is a homework or not, the point is that the asker must be pushed to do the main work on that. – user Nov 29 '18 at 08:04
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1Now 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 08:10
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@J_G I think that here Taylor's series is the most direct way. Are you aware about that? – user Nov 29 '18 at 08:15
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1If 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 08:16
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@ParamanandSingh Do you argree to proceed by Taylor's series or you would suggest also different ways? – user Nov 29 '18 at 08:16
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@gimusi: I only want the asker to do some simplification via standard limits to reach a point after which both Taylor and L'Hospital's Rule can be used with almost equal ease (Taylor being slightly easier). – Paramanand Singh Nov 29 '18 at 08:17
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@J_G It would be helpful if you can also shown your work here with the full resolution. We'll take a look to that. – user Nov 29 '18 at 08:17
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@ParamanandSingh Yes I agree indieed the initial set up is fully correct, maybe he wasn't aware how to deal with the two expansions in order to obtain one single expansion for the $\sin x/\arcsin x$ term. – user Nov 29 '18 at 08:19
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@J_G There is somethign wrong in your derivation. Start from here $$\frac{\sin x}{\arcsin x}=\left(1-\frac16x^2+o(x^2)\right)\left(1+\frac16x^2+o(x^2)\right)^{-1}$$ and apply binomial series expansion to the term $$\left(1+\frac16x^2+o(x^2)\right)^{-1}=1-\frac16x^2+o(x^2)$$ then proceed with the product. – user Nov 29 '18 at 08:27
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@BrevanEllefsen What do you think about the way followed here to deal with such kind of question? I think the asker is fully motivated to proceed in the solution by him/herself. – user Nov 29 '18 at 08:28
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1@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 08:30
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@BrevanEllefsen That's fine from your side thanks. Often people here are fast to downvote but also after an answer has been improved they usually don't remove the downvotes. Now there are also 4 votes for close the OP. I think that up to now the question does not deserve that. Thanks again, Bye. – user Nov 29 '18 at 08:33
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@J_G As a reference to proceed after you did the simplification according to the last comment, take also a look HERE. Let me know where you are stucked at. – user Nov 29 '18 at 08:42
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@J_G You should not accept an answer if you have not solved your problem. Let us know if you can now proceed to obtain a solution. Thanks – user Nov 29 '18 at 09:17
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@gimusi Please stop spamming the chat. It is quite annoying. If you would like to discuss something, open a chat room. – Brevan Ellefsen Nov 29 '18 at 09:23
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@BrevanEllefsen Yes sorry, you are right! I’ll follow your suggestion! Bye – user Nov 29 '18 at 09:25
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@J_G Do not give up! Show your work again, we'll take a look to that. – user Nov 29 '18 at 10:32
2 Answers
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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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 08:01
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@ParamanandSingh Yes it was just a hint to start with. Do you think it too small? – user Nov 29 '18 at 08:05
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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 08:07
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@ParamanandSingh After the editing, I've added a different hint to start with using Taylor's series. – user Nov 29 '18 at 08:12
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.
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+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 02:45