without L'Hospital's rule find the $lim_{xto 0+} (csc x)^{sin^2 x}$












1












$begingroup$


$lim_{xto 0+}(csc x)^{sin^{2}x}$



Without using L'Hospital's rule, I've managed to get do:



Let $y=(csc x)^{sin^{2}x}$ now by taking $ln$ of both sides I got



$lim_{xto 0+}y$=$lim_{xto 0+}sin^{2}xln(csc x)$. Now by using the limit rules of a product if $lim_{xto 0+}sin^{2}x=T$ and the $lim_{xto 0+}ln(csc x)=H$ then $lim_{xto 0+}(csc x)^{sin^{2}x}=TH$



$lim_{xto 0+}sin^{2}x=0$ by algebra of limits, so the overall limit should either be indeterminate form or $0$ since $0.H=0$ if H is a non-indeterminate form



Knowing this I used the epsilon-delta definition to prove $lim_{xto 0+}ln(csc x)=0$



Let $epsilon$>$0$ such that



$|ln(csc x)-0|<epsilon$. I can ensure that $ln(csc x)<epsilon$ by requiring $x<sin^{-1}$ $frac{1}{e^{epsilon}}$. Thus:



I let $delta=$ $sin^{-1}$ $frac{1}{e^{epsilon}}$. And I think the conditions of the definition were satisfied. There the $lim_{xto 0+}ln(csc x)=0$



Thus $lim_{xto 0+}(csc x)^{sin^{2}x}=0$



My question is where have I gone wrong because the answer is 1, but I cannot find my own fault.



Thank you in advance,.






real-analysis limits limits-without-lhopital







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You computed$ lim ln(y).$
    $endgroup$
    – hamam_Abdallah
    Dec 23 '18 at 20:43






  • 2




    $begingroup$
    An obvious start would be to switch variable to $u=sin x$ -- then you're looking for $lim_{uto 0^+} u^{-u^2}$ instead and need not trouble yourself with more trigonometry.
    $endgroup$
    – Henning Makholm
    Dec 23 '18 at 20:52












  • $begingroup$
    @ricky If $ln y$ goes to zero, it means that $y$ goes to one.
    $endgroup$
    – hamam_Abdallah
    Dec 23 '18 at 21:04










  • $begingroup$
    could you expand further I'm struggling to see how
    $endgroup$
    – ricky
    Dec 23 '18 at 21:06






  • 1




    $begingroup$
    But $limlimits_{xto 0^+}ln(sin x) = -infty$, so $limlimits_{xto 0^+}ln(csc x) = infty$
    $endgroup$
    – Ted Shifrin
    Dec 23 '18 at 21:21


















1












$begingroup$


$lim_{xto 0+}(csc x)^{sin^{2}x}$



Without using L'Hospital's rule, I've managed to get do:



Let $y=(csc x)^{sin^{2}x}$ now by taking $ln$ of both sides I got



$lim_{xto 0+}y$=$lim_{xto 0+}sin^{2}xln(csc x)$. Now by using the limit rules of a product if $lim_{xto 0+}sin^{2}x=T$ and the $lim_{xto 0+}ln(csc x)=H$ then $lim_{xto 0+}(csc x)^{sin^{2}x}=TH$



$lim_{xto 0+}sin^{2}x=0$ by algebra of limits, so the overall limit should either be indeterminate form or $0$ since $0.H=0$ if H is a non-indeterminate form



Knowing this I used the epsilon-delta definition to prove $lim_{xto 0+}ln(csc x)=0$



Let $epsilon$>$0$ such that



$|ln(csc x)-0|<epsilon$. I can ensure that $ln(csc x)<epsilon$ by requiring $x<sin^{-1}$ $frac{1}{e^{epsilon}}$. Thus:



I let $delta=$ $sin^{-1}$ $frac{1}{e^{epsilon}}$. And I think the conditions of the definition were satisfied. There the $lim_{xto 0+}ln(csc x)=0$



Thus $lim_{xto 0+}(csc x)^{sin^{2}x}=0$



My question is where have I gone wrong because the answer is 1, but I cannot find my own fault.



Thank you in advance,.






real-analysis limits limits-without-lhopital







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You computed$ lim ln(y).$
    $endgroup$
    – hamam_Abdallah
    Dec 23 '18 at 20:43






  • 2




    $begingroup$
    An obvious start would be to switch variable to $u=sin x$ -- then you're looking for $lim_{uto 0^+} u^{-u^2}$ instead and need not trouble yourself with more trigonometry.
    $endgroup$
    – Henning Makholm
    Dec 23 '18 at 20:52












  • $begingroup$
    @ricky If $ln y$ goes to zero, it means that $y$ goes to one.
    $endgroup$
    – hamam_Abdallah
    Dec 23 '18 at 21:04










  • $begingroup$
    could you expand further I'm struggling to see how
    $endgroup$
    – ricky
    Dec 23 '18 at 21:06






  • 1




    $begingroup$
    But $limlimits_{xto 0^+}ln(sin x) = -infty$, so $limlimits_{xto 0^+}ln(csc x) = infty$
    $endgroup$
    – Ted Shifrin
    Dec 23 '18 at 21:21
















1












1








1





$begingroup$


$lim_{xto 0+}(csc x)^{sin^{2}x}$



Without using L'Hospital's rule, I've managed to get do:



Let $y=(csc x)^{sin^{2}x}$ now by taking $ln$ of both sides I got



$lim_{xto 0+}y$=$lim_{xto 0+}sin^{2}xln(csc x)$. Now by using the limit rules of a product if $lim_{xto 0+}sin^{2}x=T$ and the $lim_{xto 0+}ln(csc x)=H$ then $lim_{xto 0+}(csc x)^{sin^{2}x}=TH$



$lim_{xto 0+}sin^{2}x=0$ by algebra of limits, so the overall limit should either be indeterminate form or $0$ since $0.H=0$ if H is a non-indeterminate form



Knowing this I used the epsilon-delta definition to prove $lim_{xto 0+}ln(csc x)=0$



Let $epsilon$>$0$ such that



$|ln(csc x)-0|<epsilon$. I can ensure that $ln(csc x)<epsilon$ by requiring $x<sin^{-1}$ $frac{1}{e^{epsilon}}$. Thus:



I let $delta=$ $sin^{-1}$ $frac{1}{e^{epsilon}}$. And I think the conditions of the definition were satisfied. There the $lim_{xto 0+}ln(csc x)=0$



Thus $lim_{xto 0+}(csc x)^{sin^{2}x}=0$



My question is where have I gone wrong because the answer is 1, but I cannot find my own fault.



Thank you in advance,.






real-analysis limits limits-without-lhopital







share|cite|improve this question











$endgroup$




$lim_{xto 0+}(csc x)^{sin^{2}x}$



Without using L'Hospital's rule, I've managed to get do:



Let $y=(csc x)^{sin^{2}x}$ now by taking $ln$ of both sides I got



$lim_{xto 0+}y$=$lim_{xto 0+}sin^{2}xln(csc x)$. Now by using the limit rules of a product if $lim_{xto 0+}sin^{2}x=T$ and the $lim_{xto 0+}ln(csc x)=H$ then $lim_{xto 0+}(csc x)^{sin^{2}x}=TH$



$lim_{xto 0+}sin^{2}x=0$ by algebra of limits, so the overall limit should either be indeterminate form or $0$ since $0.H=0$ if H is a non-indeterminate form



Knowing this I used the epsilon-delta definition to prove $lim_{xto 0+}ln(csc x)=0$



Let $epsilon$>$0$ such that



$|ln(csc x)-0|<epsilon$. I can ensure that $ln(csc x)<epsilon$ by requiring $x<sin^{-1}$ $frac{1}{e^{epsilon}}$. Thus:



I let $delta=$ $sin^{-1}$ $frac{1}{e^{epsilon}}$. And I think the conditions of the definition were satisfied. There the $lim_{xto 0+}ln(csc x)=0$



Thus $lim_{xto 0+}(csc x)^{sin^{2}x}=0$



My question is where have I gone wrong because the answer is 1, but I cannot find my own fault.



Thank you in advance,.






real-analysis limits limits-without-lhopital




real-analysis limits limits-without-lhopital






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 23 '18 at 20:56









Bernard

124k742117




124k742117










asked Dec 23 '18 at 20:41









rickyricky

1337




1337








  • 1




    $begingroup$
    You computed$ lim ln(y).$
    $endgroup$
    – hamam_Abdallah
    Dec 23 '18 at 20:43






  • 2




    $begingroup$
    An obvious start would be to switch variable to $u=sin x$ -- then you're looking for $lim_{uto 0^+} u^{-u^2}$ instead and need not trouble yourself with more trigonometry.
    $endgroup$
    – Henning Makholm
    Dec 23 '18 at 20:52












  • $begingroup$
    @ricky If $ln y$ goes to zero, it means that $y$ goes to one.
    $endgroup$
    – hamam_Abdallah
    Dec 23 '18 at 21:04










  • $begingroup$
    could you expand further I'm struggling to see how
    $endgroup$
    – ricky
    Dec 23 '18 at 21:06






  • 1




    $begingroup$
    But $limlimits_{xto 0^+}ln(sin x) = -infty$, so $limlimits_{xto 0^+}ln(csc x) = infty$
    $endgroup$
    – Ted Shifrin
    Dec 23 '18 at 21:21
















  • 1




    $begingroup$
    You computed$ lim ln(y).$
    $endgroup$
    – hamam_Abdallah
    Dec 23 '18 at 20:43






  • 2




    $begingroup$
    An obvious start would be to switch variable to $u=sin x$ -- then you're looking for $lim_{uto 0^+} u^{-u^2}$ instead and need not trouble yourself with more trigonometry.
    $endgroup$
    – Henning Makholm
    Dec 23 '18 at 20:52












  • $begingroup$
    @ricky If $ln y$ goes to zero, it means that $y$ goes to one.
    $endgroup$
    – hamam_Abdallah
    Dec 23 '18 at 21:04










  • $begingroup$
    could you expand further I'm struggling to see how
    $endgroup$
    – ricky
    Dec 23 '18 at 21:06






  • 1




    $begingroup$
    But $limlimits_{xto 0^+}ln(sin x) = -infty$, so $limlimits_{xto 0^+}ln(csc x) = infty$
    $endgroup$
    – Ted Shifrin
    Dec 23 '18 at 21:21










1




1




$begingroup$
You computed$ lim ln(y).$
$endgroup$
– hamam_Abdallah
Dec 23 '18 at 20:43




$begingroup$
You computed$ lim ln(y).$
$endgroup$
– hamam_Abdallah
Dec 23 '18 at 20:43




2




2




$begingroup$
An obvious start would be to switch variable to $u=sin x$ -- then you're looking for $lim_{uto 0^+} u^{-u^2}$ instead and need not trouble yourself with more trigonometry.
$endgroup$
– Henning Makholm
Dec 23 '18 at 20:52






$begingroup$
An obvious start would be to switch variable to $u=sin x$ -- then you're looking for $lim_{uto 0^+} u^{-u^2}$ instead and need not trouble yourself with more trigonometry.
$endgroup$
– Henning Makholm
Dec 23 '18 at 20:52














$begingroup$
@ricky If $ln y$ goes to zero, it means that $y$ goes to one.
$endgroup$
– hamam_Abdallah
Dec 23 '18 at 21:04




$begingroup$
@ricky If $ln y$ goes to zero, it means that $y$ goes to one.
$endgroup$
– hamam_Abdallah
Dec 23 '18 at 21:04












$begingroup$
could you expand further I'm struggling to see how
$endgroup$
– ricky
Dec 23 '18 at 21:06




$begingroup$
could you expand further I'm struggling to see how
$endgroup$
– ricky
Dec 23 '18 at 21:06




1




1




$begingroup$
But $limlimits_{xto 0^+}ln(sin x) = -infty$, so $limlimits_{xto 0^+}ln(csc x) = infty$
$endgroup$
– Ted Shifrin
Dec 23 '18 at 21:21






$begingroup$
But $limlimits_{xto 0^+}ln(sin x) = -infty$, so $limlimits_{xto 0^+}ln(csc x) = infty$
$endgroup$
– Ted Shifrin
Dec 23 '18 at 21:21












2 Answers
2






active

oldest

votes


















2












$begingroup$

No need to use $varepsilon$-$delta$. See also at the end for other comments.



Note first that $lncsc x=-lnsin x$ so you have to compute
$$
lim_{xto0}-sin^2x lnsin x
$$
Now use the substitution $x=arcsin t$, so the limit becomes
$$
lim_{tto 0}-t^2ln t
$$
which is known to be $0$. Since
$$
lim_{xto0^+}lnbigl((csc x)^{sin^2x}bigr)=0
$$
you can conclude that
$$
lim_{xto0^+}(csc x)^{sin^2x}=1
$$




Note that
$$
lim_{xto0}lncsc x=infty
$$
and certainly not $0$ as you claim. So, even if you found the correct limit $0$, your method is flawed.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you, made it look so simple haha
    $endgroup$
    – ricky
    Dec 23 '18 at 23:54



















1












$begingroup$

Let's see where you went wrong. We know that
$$lim_{xto 0^+}sin(x)=0$$
$$Longrightarrow lim_{xto 0^+}cosec(x)=+infty$$
$$(because sin(x)>0 forall xin(0,π))$$
$$Longrightarrow lim_{xto 0^+}ln(cosec(x))=+infty$$
And you claimed that to be $0$.



Now, coming to the question, let the required limit be $y$.
$$Longrightarrow y=lim_{xto 0^+}(cosec(x))^{sin^2(x)}$$
$$Longrightarrow ln(y)=lim_{xto 0^+}(sin^2(x))ln(cosec(x))$$
As $xto 0^+,cosec(x)to infty$. Hence,
$$Longrightarrow ln(y)=lim_{uto infty}frac{ln(u)}{u^2}$$
$$Longrightarrow ln(y)=lim_{utoinfty}frac{ln(u)}{u}×lim_{utoinfty}frac{1}{u}$$
both of which limits are clearly $0$.
$$Longrightarrow ln(y)=0×0=0$$
$$Longrightarrow y=1$$
which is the required answer.



Hope it helps






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you, I would of verified your answer but I can only choose one
    $endgroup$
    – ricky
    Dec 23 '18 at 23:55












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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

No need to use $varepsilon$-$delta$. See also at the end for other comments.



Note first that $lncsc x=-lnsin x$ so you have to compute
$$
lim_{xto0}-sin^2x lnsin x
$$
Now use the substitution $x=arcsin t$, so the limit becomes
$$
lim_{tto 0}-t^2ln t
$$
which is known to be $0$. Since
$$
lim_{xto0^+}lnbigl((csc x)^{sin^2x}bigr)=0
$$
you can conclude that
$$
lim_{xto0^+}(csc x)^{sin^2x}=1
$$




Note that
$$
lim_{xto0}lncsc x=infty
$$
and certainly not $0$ as you claim. So, even if you found the correct limit $0$, your method is flawed.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you, made it look so simple haha
    $endgroup$
    – ricky
    Dec 23 '18 at 23:54
















2












$begingroup$

No need to use $varepsilon$-$delta$. See also at the end for other comments.



Note first that $lncsc x=-lnsin x$ so you have to compute
$$
lim_{xto0}-sin^2x lnsin x
$$
Now use the substitution $x=arcsin t$, so the limit becomes
$$
lim_{tto 0}-t^2ln t
$$
which is known to be $0$. Since
$$
lim_{xto0^+}lnbigl((csc x)^{sin^2x}bigr)=0
$$
you can conclude that
$$
lim_{xto0^+}(csc x)^{sin^2x}=1
$$




Note that
$$
lim_{xto0}lncsc x=infty
$$
and certainly not $0$ as you claim. So, even if you found the correct limit $0$, your method is flawed.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you, made it look so simple haha
    $endgroup$
    – ricky
    Dec 23 '18 at 23:54














2












2








2





$begingroup$

No need to use $varepsilon$-$delta$. See also at the end for other comments.



Note first that $lncsc x=-lnsin x$ so you have to compute
$$
lim_{xto0}-sin^2x lnsin x
$$
Now use the substitution $x=arcsin t$, so the limit becomes
$$
lim_{tto 0}-t^2ln t
$$
which is known to be $0$. Since
$$
lim_{xto0^+}lnbigl((csc x)^{sin^2x}bigr)=0
$$
you can conclude that
$$
lim_{xto0^+}(csc x)^{sin^2x}=1
$$




Note that
$$
lim_{xto0}lncsc x=infty
$$
and certainly not $0$ as you claim. So, even if you found the correct limit $0$, your method is flawed.






share|cite|improve this answer









$endgroup$



No need to use $varepsilon$-$delta$. See also at the end for other comments.



Note first that $lncsc x=-lnsin x$ so you have to compute
$$
lim_{xto0}-sin^2x lnsin x
$$
Now use the substitution $x=arcsin t$, so the limit becomes
$$
lim_{tto 0}-t^2ln t
$$
which is known to be $0$. Since
$$
lim_{xto0^+}lnbigl((csc x)^{sin^2x}bigr)=0
$$
you can conclude that
$$
lim_{xto0^+}(csc x)^{sin^2x}=1
$$




Note that
$$
lim_{xto0}lncsc x=infty
$$
and certainly not $0$ as you claim. So, even if you found the correct limit $0$, your method is flawed.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 23 '18 at 23:46









egregegreg

186k1486209




186k1486209












  • $begingroup$
    Thank you, made it look so simple haha
    $endgroup$
    – ricky
    Dec 23 '18 at 23:54


















  • $begingroup$
    Thank you, made it look so simple haha
    $endgroup$
    – ricky
    Dec 23 '18 at 23:54
















$begingroup$
Thank you, made it look so simple haha
$endgroup$
– ricky
Dec 23 '18 at 23:54




$begingroup$
Thank you, made it look so simple haha
$endgroup$
– ricky
Dec 23 '18 at 23:54











1












$begingroup$

Let's see where you went wrong. We know that
$$lim_{xto 0^+}sin(x)=0$$
$$Longrightarrow lim_{xto 0^+}cosec(x)=+infty$$
$$(because sin(x)>0 forall xin(0,π))$$
$$Longrightarrow lim_{xto 0^+}ln(cosec(x))=+infty$$
And you claimed that to be $0$.



Now, coming to the question, let the required limit be $y$.
$$Longrightarrow y=lim_{xto 0^+}(cosec(x))^{sin^2(x)}$$
$$Longrightarrow ln(y)=lim_{xto 0^+}(sin^2(x))ln(cosec(x))$$
As $xto 0^+,cosec(x)to infty$. Hence,
$$Longrightarrow ln(y)=lim_{uto infty}frac{ln(u)}{u^2}$$
$$Longrightarrow ln(y)=lim_{utoinfty}frac{ln(u)}{u}×lim_{utoinfty}frac{1}{u}$$
both of which limits are clearly $0$.
$$Longrightarrow ln(y)=0×0=0$$
$$Longrightarrow y=1$$
which is the required answer.



Hope it helps






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you, I would of verified your answer but I can only choose one
    $endgroup$
    – ricky
    Dec 23 '18 at 23:55
















1












$begingroup$

Let's see where you went wrong. We know that
$$lim_{xto 0^+}sin(x)=0$$
$$Longrightarrow lim_{xto 0^+}cosec(x)=+infty$$
$$(because sin(x)>0 forall xin(0,π))$$
$$Longrightarrow lim_{xto 0^+}ln(cosec(x))=+infty$$
And you claimed that to be $0$.



Now, coming to the question, let the required limit be $y$.
$$Longrightarrow y=lim_{xto 0^+}(cosec(x))^{sin^2(x)}$$
$$Longrightarrow ln(y)=lim_{xto 0^+}(sin^2(x))ln(cosec(x))$$
As $xto 0^+,cosec(x)to infty$. Hence,
$$Longrightarrow ln(y)=lim_{uto infty}frac{ln(u)}{u^2}$$
$$Longrightarrow ln(y)=lim_{utoinfty}frac{ln(u)}{u}×lim_{utoinfty}frac{1}{u}$$
both of which limits are clearly $0$.
$$Longrightarrow ln(y)=0×0=0$$
$$Longrightarrow y=1$$
which is the required answer.



Hope it helps






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you, I would of verified your answer but I can only choose one
    $endgroup$
    – ricky
    Dec 23 '18 at 23:55














1












1








1





$begingroup$

Let's see where you went wrong. We know that
$$lim_{xto 0^+}sin(x)=0$$
$$Longrightarrow lim_{xto 0^+}cosec(x)=+infty$$
$$(because sin(x)>0 forall xin(0,π))$$
$$Longrightarrow lim_{xto 0^+}ln(cosec(x))=+infty$$
And you claimed that to be $0$.



Now, coming to the question, let the required limit be $y$.
$$Longrightarrow y=lim_{xto 0^+}(cosec(x))^{sin^2(x)}$$
$$Longrightarrow ln(y)=lim_{xto 0^+}(sin^2(x))ln(cosec(x))$$
As $xto 0^+,cosec(x)to infty$. Hence,
$$Longrightarrow ln(y)=lim_{uto infty}frac{ln(u)}{u^2}$$
$$Longrightarrow ln(y)=lim_{utoinfty}frac{ln(u)}{u}×lim_{utoinfty}frac{1}{u}$$
both of which limits are clearly $0$.
$$Longrightarrow ln(y)=0×0=0$$
$$Longrightarrow y=1$$
which is the required answer.



Hope it helps






share|cite|improve this answer











$endgroup$



Let's see where you went wrong. We know that
$$lim_{xto 0^+}sin(x)=0$$
$$Longrightarrow lim_{xto 0^+}cosec(x)=+infty$$
$$(because sin(x)>0 forall xin(0,π))$$
$$Longrightarrow lim_{xto 0^+}ln(cosec(x))=+infty$$
And you claimed that to be $0$.



Now, coming to the question, let the required limit be $y$.
$$Longrightarrow y=lim_{xto 0^+}(cosec(x))^{sin^2(x)}$$
$$Longrightarrow ln(y)=lim_{xto 0^+}(sin^2(x))ln(cosec(x))$$
As $xto 0^+,cosec(x)to infty$. Hence,
$$Longrightarrow ln(y)=lim_{uto infty}frac{ln(u)}{u^2}$$
$$Longrightarrow ln(y)=lim_{utoinfty}frac{ln(u)}{u}×lim_{utoinfty}frac{1}{u}$$
both of which limits are clearly $0$.
$$Longrightarrow ln(y)=0×0=0$$
$$Longrightarrow y=1$$
which is the required answer.



Hope it helps







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 23 '18 at 23:57

























answered Dec 23 '18 at 23:30









MartundMartund

2,1451213




2,1451213












  • $begingroup$
    Thank you, I would of verified your answer but I can only choose one
    $endgroup$
    – ricky
    Dec 23 '18 at 23:55


















  • $begingroup$
    Thank you, I would of verified your answer but I can only choose one
    $endgroup$
    – ricky
    Dec 23 '18 at 23:55
















$begingroup$
Thank you, I would of verified your answer but I can only choose one
$endgroup$
– ricky
Dec 23 '18 at 23:55




$begingroup$
Thank you, I would of verified your answer but I can only choose one
$endgroup$
– ricky
Dec 23 '18 at 23:55


















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