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












3












$begingroup$


Limit: $lim_{xto 0}left(dfrac{sin x}{arcsin x}right)^{1/ln(1+x^2)}$ I have tried to do this: it is equal to $e^{limfrac{log{frac{sin x}{arcsin x}}}{log(1+x^2)}}$, 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.
$$lim_{xrightarrow 0}{frac{log{frac{sin x}{arcsin x}}}{log(1+x^2)}} = lim_{xrightarrow 0}frac{log1 + frac{-frac{1}{3}x^2}{1+frac{1}{6}x^2+o(x^2)}}{log(1+x^2)} = lim_{xrightarrow0}frac{frac{-frac{1}{3}x^2}{1+frac{1}{6}x^2+o(x^2)} + o(frac{-frac{1}{3}x^2}{1+frac{1}{6}x^2+o(x^2)})}{x^2+o(x^2)}$$ using Taylor series. Now I think that it's not clear for me how to simplify $oleft(frac{-frac{1}{3}x^2}{1+frac{1}{6}x^2+o(x^2)}right)$.






real-analysis limits taylor-expansion







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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






  • 1




    $begingroup$
    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






  • 1




    $begingroup$
    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








  • 1




    $begingroup$
    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








  • 1




    $begingroup$
    @J_G that first step seems wrong$$ lim_{xrightarrow 0}{frac{log{frac{sin x}{arcsin x}}}{log(1+x^2)}} = lim_{xrightarrow 0}frac{log1 + frac{-frac{1}{3}x^2}{1+frac{1}{6}x^2+o(x^2)}}{log(1+x^2)}...$$
    $endgroup$
    – gimusi
    Nov 29 '18 at 8:30
















3












$begingroup$


Limit: $lim_{xto 0}left(dfrac{sin x}{arcsin x}right)^{1/ln(1+x^2)}$ I have tried to do this: it is equal to $e^{limfrac{log{frac{sin x}{arcsin x}}}{log(1+x^2)}}$, 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.
$$lim_{xrightarrow 0}{frac{log{frac{sin x}{arcsin x}}}{log(1+x^2)}} = lim_{xrightarrow 0}frac{log1 + frac{-frac{1}{3}x^2}{1+frac{1}{6}x^2+o(x^2)}}{log(1+x^2)} = lim_{xrightarrow0}frac{frac{-frac{1}{3}x^2}{1+frac{1}{6}x^2+o(x^2)} + o(frac{-frac{1}{3}x^2}{1+frac{1}{6}x^2+o(x^2)})}{x^2+o(x^2)}$$ using Taylor series. Now I think that it's not clear for me how to simplify $oleft(frac{-frac{1}{3}x^2}{1+frac{1}{6}x^2+o(x^2)}right)$.






real-analysis limits taylor-expansion







share|cite|improve this question











$endgroup$












  • $begingroup$
    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






  • 1




    $begingroup$
    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






  • 1




    $begingroup$
    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








  • 1




    $begingroup$
    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








  • 1




    $begingroup$
    @J_G that first step seems wrong$$ lim_{xrightarrow 0}{frac{log{frac{sin x}{arcsin x}}}{log(1+x^2)}} = lim_{xrightarrow 0}frac{log1 + frac{-frac{1}{3}x^2}{1+frac{1}{6}x^2+o(x^2)}}{log(1+x^2)}...$$
    $endgroup$
    – gimusi
    Nov 29 '18 at 8:30














3












3








3


1



$begingroup$


Limit: $lim_{xto 0}left(dfrac{sin x}{arcsin x}right)^{1/ln(1+x^2)}$ I have tried to do this: it is equal to $e^{limfrac{log{frac{sin x}{arcsin x}}}{log(1+x^2)}}$, 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.
$$lim_{xrightarrow 0}{frac{log{frac{sin x}{arcsin x}}}{log(1+x^2)}} = lim_{xrightarrow 0}frac{log1 + frac{-frac{1}{3}x^2}{1+frac{1}{6}x^2+o(x^2)}}{log(1+x^2)} = lim_{xrightarrow0}frac{frac{-frac{1}{3}x^2}{1+frac{1}{6}x^2+o(x^2)} + o(frac{-frac{1}{3}x^2}{1+frac{1}{6}x^2+o(x^2)})}{x^2+o(x^2)}$$ using Taylor series. Now I think that it's not clear for me how to simplify $oleft(frac{-frac{1}{3}x^2}{1+frac{1}{6}x^2+o(x^2)}right)$.






real-analysis limits taylor-expansion







share|cite|improve this question











$endgroup$




Limit: $lim_{xto 0}left(dfrac{sin x}{arcsin x}right)^{1/ln(1+x^2)}$ I have tried to do this: it is equal to $e^{limfrac{log{frac{sin x}{arcsin x}}}{log(1+x^2)}}$, 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.
$$lim_{xrightarrow 0}{frac{log{frac{sin x}{arcsin x}}}{log(1+x^2)}} = lim_{xrightarrow 0}frac{log1 + frac{-frac{1}{3}x^2}{1+frac{1}{6}x^2+o(x^2)}}{log(1+x^2)} = lim_{xrightarrow0}frac{frac{-frac{1}{3}x^2}{1+frac{1}{6}x^2+o(x^2)} + o(frac{-frac{1}{3}x^2}{1+frac{1}{6}x^2+o(x^2)})}{x^2+o(x^2)}$$ using Taylor series. Now I think that it's not clear for me how to simplify $oleft(frac{-frac{1}{3}x^2}{1+frac{1}{6}x^2+o(x^2)}right)$.






real-analysis limits taylor-expansion




real-analysis limits taylor-expansion






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share|cite|improve this question













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edited Nov 29 '18 at 12:27









Martin Sleziak

44.8k9118272




44.8k9118272










asked Nov 29 '18 at 7:54







user596269



















  • $begingroup$
    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






  • 1




    $begingroup$
    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






  • 1




    $begingroup$
    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








  • 1




    $begingroup$
    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








  • 1




    $begingroup$
    @J_G that first step seems wrong$$ lim_{xrightarrow 0}{frac{log{frac{sin x}{arcsin x}}}{log(1+x^2)}} = lim_{xrightarrow 0}frac{log1 + frac{-frac{1}{3}x^2}{1+frac{1}{6}x^2+o(x^2)}}{log(1+x^2)}...$$
    $endgroup$
    – gimusi
    Nov 29 '18 at 8:30


















  • $begingroup$
    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






  • 1




    $begingroup$
    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






  • 1




    $begingroup$
    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








  • 1




    $begingroup$
    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








  • 1




    $begingroup$
    @J_G that first step seems wrong$$ lim_{xrightarrow 0}{frac{log{frac{sin x}{arcsin x}}}{log(1+x^2)}} = lim_{xrightarrow 0}frac{log1 + frac{-frac{1}{3}x^2}{1+frac{1}{6}x^2+o(x^2)}}{log(1+x^2)}...$$
    $endgroup$
    – gimusi
    Nov 29 '18 at 8:30
















$begingroup$
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




$begingroup$
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




1




1




$begingroup$
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




$begingroup$
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




1




1




$begingroup$
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






$begingroup$
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






1




1




$begingroup$
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






$begingroup$
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






1




1




$begingroup$
@J_G that first step seems wrong$$ lim_{xrightarrow 0}{frac{log{frac{sin x}{arcsin x}}}{log(1+x^2)}} = lim_{xrightarrow 0}frac{log1 + frac{-frac{1}{3}x^2}{1+frac{1}{6}x^2+o(x^2)}}{log(1+x^2)}...$$
$endgroup$
– gimusi
Nov 29 '18 at 8:30




$begingroup$
@J_G that first step seems wrong$$ lim_{xrightarrow 0}{frac{log{frac{sin x}{arcsin x}}}{log(1+x^2)}} = lim_{xrightarrow 0}frac{log1 + frac{-frac{1}{3}x^2}{1+frac{1}{6}x^2+o(x^2)}}{log(1+x^2)}...$$
$endgroup$
– gimusi
Nov 29 '18 at 8:30










2 Answers
2






active

oldest

votes


















3












$begingroup$

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?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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?
    $endgroup$
    – gimusi
    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.
    $endgroup$
    – gimusi
    Nov 29 '18 at 8:12



















2












$begingroup$

So you want to find the limit
$$L=limlimits_{xto0} frac{lnfrac{sin x}{arcsin x}}{ln(1+x^2)}.$$
Perhaps a reasonable strategy would be to split this into calculating several simpler limits.
We know that
begin{gather*}
limlimits_{xto0} frac{lnfrac{sin x}{arcsin x}}{frac{sin x}{arcsin x}-1}=1\
limlimits_{xto0} frac{ln(1+x^2)}{x^2}=1
end{gather*}
so we eventually get to the limit
$$L=limlimits_{xto0} frac{frac{sin x}{arcsin x}-1}{x^2} =limlimits_{xto0} frac{sin x-arcsin x}{x^2arcsin x}.$$
If we also use that $limlimits_{xto0} frac{arcsin x}x=1$, we get that
$$L=limlimits_{xto0} frac{sin x-arcsin x}{x^3}.$$
And now we can try to calculate separately the two limits
begin{align*}
L_1&=limlimits_{xto0} frac{sin x-x}{x^3}\
L_2&=limlimits_{xto0} 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 $lim_{xto 0} frac{sin x-x}{x^3}$,
Find the limit $lim_{xto0}frac{arcsin x -x}{x^2}$,
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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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

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?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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?
    $endgroup$
    – gimusi
    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.
    $endgroup$
    – gimusi
    Nov 29 '18 at 8:12
















3












$begingroup$

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?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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?
    $endgroup$
    – gimusi
    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.
    $endgroup$
    – gimusi
    Nov 29 '18 at 8:12














3












3








3





$begingroup$

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?






share|cite|improve this answer











$endgroup$



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?







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 29 '18 at 8:11

























answered Nov 29 '18 at 7:58









gimusigimusi

92.8k84494




92.8k84494












  • $begingroup$
    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?
    $endgroup$
    – gimusi
    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.
    $endgroup$
    – gimusi
    Nov 29 '18 at 8:12


















  • $begingroup$
    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?
    $endgroup$
    – gimusi
    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.
    $endgroup$
    – gimusi
    Nov 29 '18 at 8:12
















$begingroup$
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




$begingroup$
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?
$endgroup$
– gimusi
Nov 29 '18 at 8:05




$begingroup$
@ParamanandSingh Yes it was just a hint to start with. Do you think it too small?
$endgroup$
– gimusi
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




$begingroup$
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.
$endgroup$
– gimusi
Nov 29 '18 at 8:12




$begingroup$
@ParamanandSingh After the editing, I've added a different hint to start with using Taylor's series.
$endgroup$
– gimusi
Nov 29 '18 at 8:12











2












$begingroup$

So you want to find the limit
$$L=limlimits_{xto0} frac{lnfrac{sin x}{arcsin x}}{ln(1+x^2)}.$$
Perhaps a reasonable strategy would be to split this into calculating several simpler limits.
We know that
begin{gather*}
limlimits_{xto0} frac{lnfrac{sin x}{arcsin x}}{frac{sin x}{arcsin x}-1}=1\
limlimits_{xto0} frac{ln(1+x^2)}{x^2}=1
end{gather*}
so we eventually get to the limit
$$L=limlimits_{xto0} frac{frac{sin x}{arcsin x}-1}{x^2} =limlimits_{xto0} frac{sin x-arcsin x}{x^2arcsin x}.$$
If we also use that $limlimits_{xto0} frac{arcsin x}x=1$, we get that
$$L=limlimits_{xto0} frac{sin x-arcsin x}{x^3}.$$
And now we can try to calculate separately the two limits
begin{align*}
L_1&=limlimits_{xto0} frac{sin x-x}{x^3}\
L_2&=limlimits_{xto0} 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 $lim_{xto 0} frac{sin x-x}{x^3}$,
Find the limit $lim_{xto0}frac{arcsin x -x}{x^2}$,
Are all limits solvable without L'Hôpital Rule or Series Expansion.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    +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
















2












$begingroup$

So you want to find the limit
$$L=limlimits_{xto0} frac{lnfrac{sin x}{arcsin x}}{ln(1+x^2)}.$$
Perhaps a reasonable strategy would be to split this into calculating several simpler limits.
We know that
begin{gather*}
limlimits_{xto0} frac{lnfrac{sin x}{arcsin x}}{frac{sin x}{arcsin x}-1}=1\
limlimits_{xto0} frac{ln(1+x^2)}{x^2}=1
end{gather*}
so we eventually get to the limit
$$L=limlimits_{xto0} frac{frac{sin x}{arcsin x}-1}{x^2} =limlimits_{xto0} frac{sin x-arcsin x}{x^2arcsin x}.$$
If we also use that $limlimits_{xto0} frac{arcsin x}x=1$, we get that
$$L=limlimits_{xto0} frac{sin x-arcsin x}{x^3}.$$
And now we can try to calculate separately the two limits
begin{align*}
L_1&=limlimits_{xto0} frac{sin x-x}{x^3}\
L_2&=limlimits_{xto0} 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 $lim_{xto 0} frac{sin x-x}{x^3}$,
Find the limit $lim_{xto0}frac{arcsin x -x}{x^2}$,
Are all limits solvable without L'Hôpital Rule or Series Expansion.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    +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














2












2








2





$begingroup$

So you want to find the limit
$$L=limlimits_{xto0} frac{lnfrac{sin x}{arcsin x}}{ln(1+x^2)}.$$
Perhaps a reasonable strategy would be to split this into calculating several simpler limits.
We know that
begin{gather*}
limlimits_{xto0} frac{lnfrac{sin x}{arcsin x}}{frac{sin x}{arcsin x}-1}=1\
limlimits_{xto0} frac{ln(1+x^2)}{x^2}=1
end{gather*}
so we eventually get to the limit
$$L=limlimits_{xto0} frac{frac{sin x}{arcsin x}-1}{x^2} =limlimits_{xto0} frac{sin x-arcsin x}{x^2arcsin x}.$$
If we also use that $limlimits_{xto0} frac{arcsin x}x=1$, we get that
$$L=limlimits_{xto0} frac{sin x-arcsin x}{x^3}.$$
And now we can try to calculate separately the two limits
begin{align*}
L_1&=limlimits_{xto0} frac{sin x-x}{x^3}\
L_2&=limlimits_{xto0} 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 $lim_{xto 0} frac{sin x-x}{x^3}$,
Find the limit $lim_{xto0}frac{arcsin x -x}{x^2}$,
Are all limits solvable without L'Hôpital Rule or Series Expansion.






share|cite|improve this answer











$endgroup$



So you want to find the limit
$$L=limlimits_{xto0} frac{lnfrac{sin x}{arcsin x}}{ln(1+x^2)}.$$
Perhaps a reasonable strategy would be to split this into calculating several simpler limits.
We know that
begin{gather*}
limlimits_{xto0} frac{lnfrac{sin x}{arcsin x}}{frac{sin x}{arcsin x}-1}=1\
limlimits_{xto0} frac{ln(1+x^2)}{x^2}=1
end{gather*}
so we eventually get to the limit
$$L=limlimits_{xto0} frac{frac{sin x}{arcsin x}-1}{x^2} =limlimits_{xto0} frac{sin x-arcsin x}{x^2arcsin x}.$$
If we also use that $limlimits_{xto0} frac{arcsin x}x=1$, we get that
$$L=limlimits_{xto0} frac{sin x-arcsin x}{x^3}.$$
And now we can try to calculate separately the two limits
begin{align*}
L_1&=limlimits_{xto0} frac{sin x-x}{x^3}\
L_2&=limlimits_{xto0} 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 $lim_{xto 0} frac{sin x-x}{x^3}$,
Find the limit $lim_{xto0}frac{arcsin x -x}{x^2}$,
Are all limits solvable without L'Hôpital Rule or Series Expansion.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 29 '18 at 17:11

























answered Nov 29 '18 at 15:08









Martin SleziakMartin Sleziak

44.8k9118272




44.8k9118272












  • $begingroup$
    +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


















  • $begingroup$
    +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
















$begingroup$
+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




$begingroup$
+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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