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







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












$begingroup$


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






share|cite|improve this question















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











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3027754%2fwhy-is-lim-x-to-%25e2%2588%259e-fracxx-1x-e%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3027754%2fwhy-is-lim-x-to-%25e2%2588%259e-fracxx-1x-e%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Berounka

Image
Berounka.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Caractéristiques Longueur 139  km Bassin 8 861  km 2 Bassin collecteur Elbe Débit moyen ? Cours Source Confluence de la Mže et de la Radbuza · Localisation Pilsen · Altitude 305  m · Coordonnées 49° 45′ 13″ N, 13° 23′ 24″ E Confluence Vltava · Localisation Prague · Altitude 187  m · Coordonnées 49° 59′ 43″ N, 14° 24′ 04″ E Géographie Pays traversés   République tchèque modifier   La Berounka est une rivière de la République tchèque. Elle porte le nom de Mže depuis sa source en Allemagne près de la frontière jusqu'à sa confluence avec la Radbuza à Plzeň. Elle continue ensuite sous le nom de Berounka jusqu'à sa confluence avec la Vltava, près de Prague. La rivière est une destination prisée pour les amateurs de canoë, qui appréc
Read more

Fiat S.p.A.

Image
Cet article court présente un sujet plus développé dans : Fiat et Fiat Chrysler Automobiles. Fiat S.p.A. Fiat S.p.A. ( Fabbrica Italiana Automobili Torino , acronyme FIAT) est un groupe, notamment connu comme société-mère du constructeur automobile italien, Fiat. Portail des entreprises Portail de Turin Portail de l’automobile This page is only for reference, If you need detailed information, please check here
Read more

Type 'String' is not a subtype of type 'int' of 'index'

Image
up vote 0 down vote favorite My App can authenticate successfully when debugging against the Android emulator, but if I try to authenticate using debugging against the physical device (with same OS version), an error message appears after more than one minute waiting: Exception has occurred. _TypeError (type 'String' is not a subtype of type 'int' of 'index') The error message points to the following code: if (responseJson[AuthUtils.authTokenKey] != null) { AuthUtils.insertDetails( userNameController.text, passwordController.text, responseJson); ... } else { ... }; And in the DEBUG CONSOLE I get the following: I/flutter ( 9531): Auth: SocketException: OS Error: Connection timed out, errno = 110, address = example.com, port = 3897
Read more