Why is $lim_{x to ∞} {(frac{x}{x-1})^x}=e$












0












$begingroup$


Hey guys so I had a quiz a couple days ago and lost partial marks for the following limit:



$$lim_{x to ∞} f(x) = left(frac{x}{x-1}right)^x$$



The way I solved it was by applying $ln$ to get that $x$ down, and after some algebra got the limit $=dfrac{0}{0}$; I applied L'hospital's Rule and got $1$; why my TA says that the answer is $e$?






calculus limits







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$endgroup$








  • 3




    $begingroup$
    If $lim_{xtoinfty} ln f(x) = 1$, then $lim_{xtoinfty} f(x) = e^1 = e$.
    $endgroup$
    – Clement C.
    Dec 5 '18 at 22:42










  • $begingroup$
    Well that's true, I still think I shouldn't have lost marks for it but can't argue that; gotta go back to read the book then
    $endgroup$
    – S..
    Dec 5 '18 at 22:44






  • 2




    $begingroup$
    Well, the loss of marks is beyond what I can do here.
    $endgroup$
    – Clement C.
    Dec 5 '18 at 22:49












  • $begingroup$
    why the massive downvotes? The question is clear and appropriate.
    $endgroup$
    – Masacroso
    Dec 5 '18 at 22:57










  • $begingroup$
    @Masacroso ¯_(ツ)_/¯
    $endgroup$
    – S..
    Dec 5 '18 at 23:03


















0












$begingroup$


Hey guys so I had a quiz a couple days ago and lost partial marks for the following limit:



$$lim_{x to ∞} f(x) = left(frac{x}{x-1}right)^x$$



The way I solved it was by applying $ln$ to get that $x$ down, and after some algebra got the limit $=dfrac{0}{0}$; I applied L'hospital's Rule and got $1$; why my TA says that the answer is $e$?






calculus limits







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    If $lim_{xtoinfty} ln f(x) = 1$, then $lim_{xtoinfty} f(x) = e^1 = e$.
    $endgroup$
    – Clement C.
    Dec 5 '18 at 22:42










  • $begingroup$
    Well that's true, I still think I shouldn't have lost marks for it but can't argue that; gotta go back to read the book then
    $endgroup$
    – S..
    Dec 5 '18 at 22:44






  • 2




    $begingroup$
    Well, the loss of marks is beyond what I can do here.
    $endgroup$
    – Clement C.
    Dec 5 '18 at 22:49












  • $begingroup$
    why the massive downvotes? The question is clear and appropriate.
    $endgroup$
    – Masacroso
    Dec 5 '18 at 22:57










  • $begingroup$
    @Masacroso ¯_(ツ)_/¯
    $endgroup$
    – S..
    Dec 5 '18 at 23:03
















0












0








0





$begingroup$


Hey guys so I had a quiz a couple days ago and lost partial marks for the following limit:



$$lim_{x to ∞} f(x) = left(frac{x}{x-1}right)^x$$



The way I solved it was by applying $ln$ to get that $x$ down, and after some algebra got the limit $=dfrac{0}{0}$; I applied L'hospital's Rule and got $1$; why my TA says that the answer is $e$?






calculus limits







share|cite|improve this question











$endgroup$




Hey guys so I had a quiz a couple days ago and lost partial marks for the following limit:



$$lim_{x to ∞} f(x) = left(frac{x}{x-1}right)^x$$



The way I solved it was by applying $ln$ to get that $x$ down, and after some algebra got the limit $=dfrac{0}{0}$; I applied L'hospital's Rule and got $1$; why my TA says that the answer is $e$?






calculus limits




calculus limits






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 5 '18 at 22:53









amWhy

192k28225439




192k28225439










asked Dec 5 '18 at 22:38









S..S..

545




545








  • 3




    $begingroup$
    If $lim_{xtoinfty} ln f(x) = 1$, then $lim_{xtoinfty} f(x) = e^1 = e$.
    $endgroup$
    – Clement C.
    Dec 5 '18 at 22:42










  • $begingroup$
    Well that's true, I still think I shouldn't have lost marks for it but can't argue that; gotta go back to read the book then
    $endgroup$
    – S..
    Dec 5 '18 at 22:44






  • 2




    $begingroup$
    Well, the loss of marks is beyond what I can do here.
    $endgroup$
    – Clement C.
    Dec 5 '18 at 22:49












  • $begingroup$
    why the massive downvotes? The question is clear and appropriate.
    $endgroup$
    – Masacroso
    Dec 5 '18 at 22:57










  • $begingroup$
    @Masacroso ¯_(ツ)_/¯
    $endgroup$
    – S..
    Dec 5 '18 at 23:03
















  • 3




    $begingroup$
    If $lim_{xtoinfty} ln f(x) = 1$, then $lim_{xtoinfty} f(x) = e^1 = e$.
    $endgroup$
    – Clement C.
    Dec 5 '18 at 22:42










  • $begingroup$
    Well that's true, I still think I shouldn't have lost marks for it but can't argue that; gotta go back to read the book then
    $endgroup$
    – S..
    Dec 5 '18 at 22:44






  • 2




    $begingroup$
    Well, the loss of marks is beyond what I can do here.
    $endgroup$
    – Clement C.
    Dec 5 '18 at 22:49












  • $begingroup$
    why the massive downvotes? The question is clear and appropriate.
    $endgroup$
    – Masacroso
    Dec 5 '18 at 22:57










  • $begingroup$
    @Masacroso ¯_(ツ)_/¯
    $endgroup$
    – S..
    Dec 5 '18 at 23:03










3




3




$begingroup$
If $lim_{xtoinfty} ln f(x) = 1$, then $lim_{xtoinfty} f(x) = e^1 = e$.
$endgroup$
– Clement C.
Dec 5 '18 at 22:42




$begingroup$
If $lim_{xtoinfty} ln f(x) = 1$, then $lim_{xtoinfty} f(x) = e^1 = e$.
$endgroup$
– Clement C.
Dec 5 '18 at 22:42












$begingroup$
Well that's true, I still think I shouldn't have lost marks for it but can't argue that; gotta go back to read the book then
$endgroup$
– S..
Dec 5 '18 at 22:44




$begingroup$
Well that's true, I still think I shouldn't have lost marks for it but can't argue that; gotta go back to read the book then
$endgroup$
– S..
Dec 5 '18 at 22:44




2




2




$begingroup$
Well, the loss of marks is beyond what I can do here.
$endgroup$
– Clement C.
Dec 5 '18 at 22:49






$begingroup$
Well, the loss of marks is beyond what I can do here.
$endgroup$
– Clement C.
Dec 5 '18 at 22:49














$begingroup$
why the massive downvotes? The question is clear and appropriate.
$endgroup$
– Masacroso
Dec 5 '18 at 22:57




$begingroup$
why the massive downvotes? The question is clear and appropriate.
$endgroup$
– Masacroso
Dec 5 '18 at 22:57












$begingroup$
@Masacroso ¯_(ツ)_/¯
$endgroup$
– S..
Dec 5 '18 at 23:03






$begingroup$
@Masacroso ¯_(ツ)_/¯
$endgroup$
– S..
Dec 5 '18 at 23:03












3 Answers
3






active

oldest

votes


















5












$begingroup$

Consider the reciprocal. Then
$$
left(frac{x-1}{x}right)^x=left(
1-frac{1}{x}
right)^xto e^{-1}
$$
as $xto infty$ by a well-known characterization of the exponential function. In particular the original limit is then $e$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Very nice trick!
    $endgroup$
    – gimusi
    Dec 5 '18 at 22:47



















3












$begingroup$

HINT



We have



$${left(frac{x}{x-1}right)^x}={left(1+frac{1}{x-1}right)^x}=frac{{left(1+frac{1}{x-1}right)^x}}{{left(1+frac{1}{x-1}right)}}{left(1+frac{1}{x-1}right)}=ldots$$



Edit: to complete the answer



$$ldots={left(1+frac{1}{x-1}right)^{x-1}}{left(1+frac{1}{x-1}right)}to e cdot 1 =e$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Yeah doing the limit wasn't the problem though, but rather my final answer which was $1$ and correct, why the answer has to be exactly $e$. But it got answered in one of the comments.
    $endgroup$
    – S..
    Dec 5 '18 at 23:00












  • $begingroup$
    @S.. As you can see the expression is equal to $${left(1+frac{1}{x-1}right)^{x-1}}{left(1+frac{1}{x-1}right)}$$ and by$ x-1=y to infty$ by the definition of $e$ $$ {left(1+frac{1}{y}right)^{y}}{left(1+frac{1}{y}right)} to e cdot 1=e $$
    $endgroup$
    – gimusi
    Dec 5 '18 at 23:01












  • $begingroup$
    @S.. Please reselect and accept the other answer, I didn't edited the question for that! Ionly want to be sure You got the method. It seems not correct to me gain an answer after editing. Thanks
    $endgroup$
    – gimusi
    Dec 5 '18 at 23:05





















3












$begingroup$

The other $2$ answers have shown you how you can approach the problem but haven't addressed why your approach was faulty. If you did take the log of both sides, you'd get:



$$begin{aligned}L&=limleft(frac{x}{x-1}right)^x\
ln L&=lim left[xlnleft(frac{x}{x-1}right)right]\
&=lim left[xlnleft(xright)-xlnleft(x-1right)right]
end{aligned}$$



As far as I can see, this doesn't lead you to anything of the form $frac{f}{g}$, with $lim f=lim g=0$ or $pminfty$. Hence, we couldn't apply L'Hopital's rule. My guess is that you erroneously distributed the log with $lnleft(frac{x}{x-1}right)neqfrac{ln(x)}{ln(x-1)}$. Or possibly, you didn't take the log of the base, $frac{x}{x-1}$, when you took logs of both sides of the limit. In any case, the other answers give the correct ways to proceed with the problem.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    For posterity, you can manipulate that into the form $frac{f}{g}$ if you use the substitution $xmapsto frac1u$. So then $frac{f(u)}{g(u)}=frac{lnleft(frac{1}{u}right)-lnleft(frac{1}{u}-1right)}{u}$ and $frac{f'(u)}{g'(u)}=frac{1}{left(1-uright)}$ so $lim_{xtoinfty}frac{f'(u)}{g'(u)}=1$, which would prove the limit but I don't know if that's what you did.
    $endgroup$
    – Jam
    Dec 5 '18 at 23:21











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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









5












$begingroup$

Consider the reciprocal. Then
$$
left(frac{x-1}{x}right)^x=left(
1-frac{1}{x}
right)^xto e^{-1}
$$
as $xto infty$ by a well-known characterization of the exponential function. In particular the original limit is then $e$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Very nice trick!
    $endgroup$
    – gimusi
    Dec 5 '18 at 22:47
















5












$begingroup$

Consider the reciprocal. Then
$$
left(frac{x-1}{x}right)^x=left(
1-frac{1}{x}
right)^xto e^{-1}
$$
as $xto infty$ by a well-known characterization of the exponential function. In particular the original limit is then $e$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Very nice trick!
    $endgroup$
    – gimusi
    Dec 5 '18 at 22:47














5












5








5





$begingroup$

Consider the reciprocal. Then
$$
left(frac{x-1}{x}right)^x=left(
1-frac{1}{x}
right)^xto e^{-1}
$$
as $xto infty$ by a well-known characterization of the exponential function. In particular the original limit is then $e$.






share|cite|improve this answer









$endgroup$



Consider the reciprocal. Then
$$
left(frac{x-1}{x}right)^x=left(
1-frac{1}{x}
right)^xto e^{-1}
$$
as $xto infty$ by a well-known characterization of the exponential function. In particular the original limit is then $e$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 5 '18 at 22:46









Foobaz JohnFoobaz John

21.5k41351




21.5k41351












  • $begingroup$
    Very nice trick!
    $endgroup$
    – gimusi
    Dec 5 '18 at 22:47


















  • $begingroup$
    Very nice trick!
    $endgroup$
    – gimusi
    Dec 5 '18 at 22:47
















$begingroup$
Very nice trick!
$endgroup$
– gimusi
Dec 5 '18 at 22:47




$begingroup$
Very nice trick!
$endgroup$
– gimusi
Dec 5 '18 at 22:47











3












$begingroup$

HINT



We have



$${left(frac{x}{x-1}right)^x}={left(1+frac{1}{x-1}right)^x}=frac{{left(1+frac{1}{x-1}right)^x}}{{left(1+frac{1}{x-1}right)}}{left(1+frac{1}{x-1}right)}=ldots$$



Edit: to complete the answer



$$ldots={left(1+frac{1}{x-1}right)^{x-1}}{left(1+frac{1}{x-1}right)}to e cdot 1 =e$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Yeah doing the limit wasn't the problem though, but rather my final answer which was $1$ and correct, why the answer has to be exactly $e$. But it got answered in one of the comments.
    $endgroup$
    – S..
    Dec 5 '18 at 23:00












  • $begingroup$
    @S.. As you can see the expression is equal to $${left(1+frac{1}{x-1}right)^{x-1}}{left(1+frac{1}{x-1}right)}$$ and by$ x-1=y to infty$ by the definition of $e$ $$ {left(1+frac{1}{y}right)^{y}}{left(1+frac{1}{y}right)} to e cdot 1=e $$
    $endgroup$
    – gimusi
    Dec 5 '18 at 23:01












  • $begingroup$
    @S.. Please reselect and accept the other answer, I didn't edited the question for that! Ionly want to be sure You got the method. It seems not correct to me gain an answer after editing. Thanks
    $endgroup$
    – gimusi
    Dec 5 '18 at 23:05


















3












$begingroup$

HINT



We have



$${left(frac{x}{x-1}right)^x}={left(1+frac{1}{x-1}right)^x}=frac{{left(1+frac{1}{x-1}right)^x}}{{left(1+frac{1}{x-1}right)}}{left(1+frac{1}{x-1}right)}=ldots$$



Edit: to complete the answer



$$ldots={left(1+frac{1}{x-1}right)^{x-1}}{left(1+frac{1}{x-1}right)}to e cdot 1 =e$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Yeah doing the limit wasn't the problem though, but rather my final answer which was $1$ and correct, why the answer has to be exactly $e$. But it got answered in one of the comments.
    $endgroup$
    – S..
    Dec 5 '18 at 23:00












  • $begingroup$
    @S.. As you can see the expression is equal to $${left(1+frac{1}{x-1}right)^{x-1}}{left(1+frac{1}{x-1}right)}$$ and by$ x-1=y to infty$ by the definition of $e$ $$ {left(1+frac{1}{y}right)^{y}}{left(1+frac{1}{y}right)} to e cdot 1=e $$
    $endgroup$
    – gimusi
    Dec 5 '18 at 23:01












  • $begingroup$
    @S.. Please reselect and accept the other answer, I didn't edited the question for that! Ionly want to be sure You got the method. It seems not correct to me gain an answer after editing. Thanks
    $endgroup$
    – gimusi
    Dec 5 '18 at 23:05
















3












3








3





$begingroup$

HINT



We have



$${left(frac{x}{x-1}right)^x}={left(1+frac{1}{x-1}right)^x}=frac{{left(1+frac{1}{x-1}right)^x}}{{left(1+frac{1}{x-1}right)}}{left(1+frac{1}{x-1}right)}=ldots$$



Edit: to complete the answer



$$ldots={left(1+frac{1}{x-1}right)^{x-1}}{left(1+frac{1}{x-1}right)}to e cdot 1 =e$$






share|cite|improve this answer











$endgroup$



HINT



We have



$${left(frac{x}{x-1}right)^x}={left(1+frac{1}{x-1}right)^x}=frac{{left(1+frac{1}{x-1}right)^x}}{{left(1+frac{1}{x-1}right)}}{left(1+frac{1}{x-1}right)}=ldots$$



Edit: to complete the answer



$$ldots={left(1+frac{1}{x-1}right)^{x-1}}{left(1+frac{1}{x-1}right)}to e cdot 1 =e$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 5 '18 at 23:00

























answered Dec 5 '18 at 22:44









gimusigimusi

1




1












  • $begingroup$
    Yeah doing the limit wasn't the problem though, but rather my final answer which was $1$ and correct, why the answer has to be exactly $e$. But it got answered in one of the comments.
    $endgroup$
    – S..
    Dec 5 '18 at 23:00












  • $begingroup$
    @S.. As you can see the expression is equal to $${left(1+frac{1}{x-1}right)^{x-1}}{left(1+frac{1}{x-1}right)}$$ and by$ x-1=y to infty$ by the definition of $e$ $$ {left(1+frac{1}{y}right)^{y}}{left(1+frac{1}{y}right)} to e cdot 1=e $$
    $endgroup$
    – gimusi
    Dec 5 '18 at 23:01












  • $begingroup$
    @S.. Please reselect and accept the other answer, I didn't edited the question for that! Ionly want to be sure You got the method. It seems not correct to me gain an answer after editing. Thanks
    $endgroup$
    – gimusi
    Dec 5 '18 at 23:05




















  • $begingroup$
    Yeah doing the limit wasn't the problem though, but rather my final answer which was $1$ and correct, why the answer has to be exactly $e$. But it got answered in one of the comments.
    $endgroup$
    – S..
    Dec 5 '18 at 23:00












  • $begingroup$
    @S.. As you can see the expression is equal to $${left(1+frac{1}{x-1}right)^{x-1}}{left(1+frac{1}{x-1}right)}$$ and by$ x-1=y to infty$ by the definition of $e$ $$ {left(1+frac{1}{y}right)^{y}}{left(1+frac{1}{y}right)} to e cdot 1=e $$
    $endgroup$
    – gimusi
    Dec 5 '18 at 23:01












  • $begingroup$
    @S.. Please reselect and accept the other answer, I didn't edited the question for that! Ionly want to be sure You got the method. It seems not correct to me gain an answer after editing. Thanks
    $endgroup$
    – gimusi
    Dec 5 '18 at 23:05


















$begingroup$
Yeah doing the limit wasn't the problem though, but rather my final answer which was $1$ and correct, why the answer has to be exactly $e$. But it got answered in one of the comments.
$endgroup$
– S..
Dec 5 '18 at 23:00






$begingroup$
Yeah doing the limit wasn't the problem though, but rather my final answer which was $1$ and correct, why the answer has to be exactly $e$. But it got answered in one of the comments.
$endgroup$
– S..
Dec 5 '18 at 23:00














$begingroup$
@S.. As you can see the expression is equal to $${left(1+frac{1}{x-1}right)^{x-1}}{left(1+frac{1}{x-1}right)}$$ and by$ x-1=y to infty$ by the definition of $e$ $$ {left(1+frac{1}{y}right)^{y}}{left(1+frac{1}{y}right)} to e cdot 1=e $$
$endgroup$
– gimusi
Dec 5 '18 at 23:01






$begingroup$
@S.. As you can see the expression is equal to $${left(1+frac{1}{x-1}right)^{x-1}}{left(1+frac{1}{x-1}right)}$$ and by$ x-1=y to infty$ by the definition of $e$ $$ {left(1+frac{1}{y}right)^{y}}{left(1+frac{1}{y}right)} to e cdot 1=e $$
$endgroup$
– gimusi
Dec 5 '18 at 23:01














$begingroup$
@S.. Please reselect and accept the other answer, I didn't edited the question for that! Ionly want to be sure You got the method. It seems not correct to me gain an answer after editing. Thanks
$endgroup$
– gimusi
Dec 5 '18 at 23:05






$begingroup$
@S.. Please reselect and accept the other answer, I didn't edited the question for that! Ionly want to be sure You got the method. It seems not correct to me gain an answer after editing. Thanks
$endgroup$
– gimusi
Dec 5 '18 at 23:05













3












$begingroup$

The other $2$ answers have shown you how you can approach the problem but haven't addressed why your approach was faulty. If you did take the log of both sides, you'd get:



$$begin{aligned}L&=limleft(frac{x}{x-1}right)^x\
ln L&=lim left[xlnleft(frac{x}{x-1}right)right]\
&=lim left[xlnleft(xright)-xlnleft(x-1right)right]
end{aligned}$$



As far as I can see, this doesn't lead you to anything of the form $frac{f}{g}$, with $lim f=lim g=0$ or $pminfty$. Hence, we couldn't apply L'Hopital's rule. My guess is that you erroneously distributed the log with $lnleft(frac{x}{x-1}right)neqfrac{ln(x)}{ln(x-1)}$. Or possibly, you didn't take the log of the base, $frac{x}{x-1}$, when you took logs of both sides of the limit. In any case, the other answers give the correct ways to proceed with the problem.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    For posterity, you can manipulate that into the form $frac{f}{g}$ if you use the substitution $xmapsto frac1u$. So then $frac{f(u)}{g(u)}=frac{lnleft(frac{1}{u}right)-lnleft(frac{1}{u}-1right)}{u}$ and $frac{f'(u)}{g'(u)}=frac{1}{left(1-uright)}$ so $lim_{xtoinfty}frac{f'(u)}{g'(u)}=1$, which would prove the limit but I don't know if that's what you did.
    $endgroup$
    – Jam
    Dec 5 '18 at 23:21
















3












$begingroup$

The other $2$ answers have shown you how you can approach the problem but haven't addressed why your approach was faulty. If you did take the log of both sides, you'd get:



$$begin{aligned}L&=limleft(frac{x}{x-1}right)^x\
ln L&=lim left[xlnleft(frac{x}{x-1}right)right]\
&=lim left[xlnleft(xright)-xlnleft(x-1right)right]
end{aligned}$$



As far as I can see, this doesn't lead you to anything of the form $frac{f}{g}$, with $lim f=lim g=0$ or $pminfty$. Hence, we couldn't apply L'Hopital's rule. My guess is that you erroneously distributed the log with $lnleft(frac{x}{x-1}right)neqfrac{ln(x)}{ln(x-1)}$. Or possibly, you didn't take the log of the base, $frac{x}{x-1}$, when you took logs of both sides of the limit. In any case, the other answers give the correct ways to proceed with the problem.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    For posterity, you can manipulate that into the form $frac{f}{g}$ if you use the substitution $xmapsto frac1u$. So then $frac{f(u)}{g(u)}=frac{lnleft(frac{1}{u}right)-lnleft(frac{1}{u}-1right)}{u}$ and $frac{f'(u)}{g'(u)}=frac{1}{left(1-uright)}$ so $lim_{xtoinfty}frac{f'(u)}{g'(u)}=1$, which would prove the limit but I don't know if that's what you did.
    $endgroup$
    – Jam
    Dec 5 '18 at 23:21














3












3








3





$begingroup$

The other $2$ answers have shown you how you can approach the problem but haven't addressed why your approach was faulty. If you did take the log of both sides, you'd get:



$$begin{aligned}L&=limleft(frac{x}{x-1}right)^x\
ln L&=lim left[xlnleft(frac{x}{x-1}right)right]\
&=lim left[xlnleft(xright)-xlnleft(x-1right)right]
end{aligned}$$



As far as I can see, this doesn't lead you to anything of the form $frac{f}{g}$, with $lim f=lim g=0$ or $pminfty$. Hence, we couldn't apply L'Hopital's rule. My guess is that you erroneously distributed the log with $lnleft(frac{x}{x-1}right)neqfrac{ln(x)}{ln(x-1)}$. Or possibly, you didn't take the log of the base, $frac{x}{x-1}$, when you took logs of both sides of the limit. In any case, the other answers give the correct ways to proceed with the problem.






share|cite|improve this answer









$endgroup$



The other $2$ answers have shown you how you can approach the problem but haven't addressed why your approach was faulty. If you did take the log of both sides, you'd get:



$$begin{aligned}L&=limleft(frac{x}{x-1}right)^x\
ln L&=lim left[xlnleft(frac{x}{x-1}right)right]\
&=lim left[xlnleft(xright)-xlnleft(x-1right)right]
end{aligned}$$



As far as I can see, this doesn't lead you to anything of the form $frac{f}{g}$, with $lim f=lim g=0$ or $pminfty$. Hence, we couldn't apply L'Hopital's rule. My guess is that you erroneously distributed the log with $lnleft(frac{x}{x-1}right)neqfrac{ln(x)}{ln(x-1)}$. Or possibly, you didn't take the log of the base, $frac{x}{x-1}$, when you took logs of both sides of the limit. In any case, the other answers give the correct ways to proceed with the problem.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 5 '18 at 23:02









JamJam

4,96011431




4,96011431












  • $begingroup$
    For posterity, you can manipulate that into the form $frac{f}{g}$ if you use the substitution $xmapsto frac1u$. So then $frac{f(u)}{g(u)}=frac{lnleft(frac{1}{u}right)-lnleft(frac{1}{u}-1right)}{u}$ and $frac{f'(u)}{g'(u)}=frac{1}{left(1-uright)}$ so $lim_{xtoinfty}frac{f'(u)}{g'(u)}=1$, which would prove the limit but I don't know if that's what you did.
    $endgroup$
    – Jam
    Dec 5 '18 at 23:21


















  • $begingroup$
    For posterity, you can manipulate that into the form $frac{f}{g}$ if you use the substitution $xmapsto frac1u$. So then $frac{f(u)}{g(u)}=frac{lnleft(frac{1}{u}right)-lnleft(frac{1}{u}-1right)}{u}$ and $frac{f'(u)}{g'(u)}=frac{1}{left(1-uright)}$ so $lim_{xtoinfty}frac{f'(u)}{g'(u)}=1$, which would prove the limit but I don't know if that's what you did.
    $endgroup$
    – Jam
    Dec 5 '18 at 23:21
















$begingroup$
For posterity, you can manipulate that into the form $frac{f}{g}$ if you use the substitution $xmapsto frac1u$. So then $frac{f(u)}{g(u)}=frac{lnleft(frac{1}{u}right)-lnleft(frac{1}{u}-1right)}{u}$ and $frac{f'(u)}{g'(u)}=frac{1}{left(1-uright)}$ so $lim_{xtoinfty}frac{f'(u)}{g'(u)}=1$, which would prove the limit but I don't know if that's what you did.
$endgroup$
– Jam
Dec 5 '18 at 23:21




$begingroup$
For posterity, you can manipulate that into the form $frac{f}{g}$ if you use the substitution $xmapsto frac1u$. So then $frac{f(u)}{g(u)}=frac{lnleft(frac{1}{u}right)-lnleft(frac{1}{u}-1right)}{u}$ and $frac{f'(u)}{g'(u)}=frac{1}{left(1-uright)}$ so $lim_{xtoinfty}frac{f'(u)}{g'(u)}=1$, which would prove the limit but I don't know if that's what you did.
$endgroup$
– Jam
Dec 5 '18 at 23:21


















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