Evaluating $lim_{xto infty}(frac{x}{x-1})^x$ [duplicate]
$begingroup$
This question already has an answer here:
Prove that $lim_{x to infty}big(frac{x}{x-1}big)^x$ is also $e$.
4 answers
I am going over a solution given to solving the follow limit,
$$lim_{xto infty}(frac{x}{x-1})^x$$
The solution continues as follows,
Consider raising the function to $e^{lncdots}$
We can find the limit as follows,
$$lim_{xto infty} x ln(frac{x}{x-1}) = lim_{xto infty} frac{ln(frac{x}{x-1})}{frac{1}{x}}$$
The solution argues this is just $frac{0}{0}$ and as such we can apply L'Hospital's rule. It continues on to find the limit equals 1, so the limit of the function is $e$.
However, I don't understand how that expression evaluates to $frac{0}{0}$, in fact it seems to express
$$frac{ln(frac{infty}{infty})}{0}$$
I assume the argument is that $frac{infty}{infty}$ equals 1, and $ln(1) = 0$, so we have $frac{0}{0}$. But I thought we cannot evaluate $frac{infty}{infty}$?
calculus limits
asked Dec 7 '18 at 0:45
DanielleDanielle
2179
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marked as duplicate by Jyrki Lahtonen, amWhy calculus
Users with the calculus badge can single-handedly close calculus questions as duplicates and reopen them as needed.
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Dec 7 '18 at 14:12
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Prove that $lim_{x to infty}big(frac{x}{x-1}big)^x$ is also $e$.
4 answers
I am going over a solution given to solving the follow limit,
$$lim_{xto infty}(frac{x}{x-1})^x$$
The solution continues as follows,
Consider raising the function to $e^{lncdots}$
We can find the limit as follows,
$$lim_{xto infty} x ln(frac{x}{x-1}) = lim_{xto infty} frac{ln(frac{x}{x-1})}{frac{1}{x}}$$
The solution argues this is just $frac{0}{0}$ and as such we can apply L'Hospital's rule. It continues on to find the limit equals 1, so the limit of the function is $e$.
However, I don't understand how that expression evaluates to $frac{0}{0}$, in fact it seems to express
$$frac{ln(frac{infty}{infty})}{0}$$
I assume the argument is that $frac{infty}{infty}$ equals 1, and $ln(1) = 0$, so we have $frac{0}{0}$. But I thought we cannot evaluate $frac{infty}{infty}$?
calculus limits
asked Dec 7 '18 at 0:45
DanielleDanielle
2179
$endgroup$
marked as duplicate by Jyrki Lahtonen, amWhy calculus
Users with the calculus badge can single-handedly close calculus questions as duplicates and reopen them as needed.
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Dec 7 '18 at 14:12
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
$ln left(frac{x-1}{x}right)=ln left(1-frac{1}{x}right)$. As $x to infty$, the expression approaches $ln 1=0$.
$endgroup$
– Anurag A
Dec 7 '18 at 0:47
$begingroup$
Do the answerers seriously think that this limit would not have been handled multiple times here already? Shame on you!
$endgroup$
– Jyrki Lahtonen
Dec 7 '18 at 10:27
add a comment |
$begingroup$
This question already has an answer here:
Prove that $lim_{x to infty}big(frac{x}{x-1}big)^x$ is also $e$.
4 answers
I am going over a solution given to solving the follow limit,
$$lim_{xto infty}(frac{x}{x-1})^x$$
The solution continues as follows,
Consider raising the function to $e^{lncdots}$
We can find the limit as follows,
$$lim_{xto infty} x ln(frac{x}{x-1}) = lim_{xto infty} frac{ln(frac{x}{x-1})}{frac{1}{x}}$$
The solution argues this is just $frac{0}{0}$ and as such we can apply L'Hospital's rule. It continues on to find the limit equals 1, so the limit of the function is $e$.
However, I don't understand how that expression evaluates to $frac{0}{0}$, in fact it seems to express
$$frac{ln(frac{infty}{infty})}{0}$$
I assume the argument is that $frac{infty}{infty}$ equals 1, and $ln(1) = 0$, so we have $frac{0}{0}$. But I thought we cannot evaluate $frac{infty}{infty}$?
calculus limits
asked Dec 7 '18 at 0:45
DanielleDanielle
2179
$endgroup$
This question already has an answer here:
Prove that $lim_{x to infty}big(frac{x}{x-1}big)^x$ is also $e$.
4 answers
I am going over a solution given to solving the follow limit,
$$lim_{xto infty}(frac{x}{x-1})^x$$
The solution continues as follows,
Consider raising the function to $e^{lncdots}$
We can find the limit as follows,
$$lim_{xto infty} x ln(frac{x}{x-1}) = lim_{xto infty} frac{ln(frac{x}{x-1})}{frac{1}{x}}$$
The solution argues this is just $frac{0}{0}$ and as such we can apply L'Hospital's rule. It continues on to find the limit equals 1, so the limit of the function is $e$.
However, I don't understand how that expression evaluates to $frac{0}{0}$, in fact it seems to express
$$frac{ln(frac{infty}{infty})}{0}$$
I assume the argument is that $frac{infty}{infty}$ equals 1, and $ln(1) = 0$, so we have $frac{0}{0}$. But I thought we cannot evaluate $frac{infty}{infty}$?
This question already has an answer here:
Prove that $lim_{x to infty}big(frac{x}{x-1}big)^x$ is also $e$.
4 answers
calculus limits
calculus limits
asked Dec 7 '18 at 0:45
DanielleDanielle
2179
asked Dec 7 '18 at 0:45
DanielleDanielle
2179
asked Dec 7 '18 at 0:45
DanielleDanielle
2179
asked Dec 7 '18 at 0:45
DanielleDanielle
2179
asked Dec 7 '18 at 0:45
DanielleDanielle
2179
2179
marked as duplicate by Jyrki Lahtonen, amWhy calculus
Users with the calculus badge can single-handedly close calculus questions as duplicates and reopen them as needed.
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Dec 7 '18 at 14:12
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Jyrki Lahtonen, amWhy calculus
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Dec 7 '18 at 14:12
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
$ln left(frac{x-1}{x}right)=ln left(1-frac{1}{x}right)$. As $x to infty$, the expression approaches $ln 1=0$.
$endgroup$
– Anurag A
Dec 7 '18 at 0:47
$begingroup$
Do the answerers seriously think that this limit would not have been handled multiple times here already? Shame on you!
$endgroup$
– Jyrki Lahtonen
Dec 7 '18 at 10:27
add a comment |
$begingroup$
$ln left(frac{x-1}{x}right)=ln left(1-frac{1}{x}right)$. As $x to infty$, the expression approaches $ln 1=0$.
$endgroup$
– Anurag A
Dec 7 '18 at 0:47
$begingroup$
Do the answerers seriously think that this limit would not have been handled multiple times here already? Shame on you!
$endgroup$
– Jyrki Lahtonen
Dec 7 '18 at 10:27
$begingroup$
$ln left(frac{x-1}{x}right)=ln left(1-frac{1}{x}right)$. As $x to infty$, the expression approaches $ln 1=0$.
$endgroup$
– Anurag A
Dec 7 '18 at 0:47
$begingroup$
$ln left(frac{x-1}{x}right)=ln left(1-frac{1}{x}right)$. As $x to infty$, the expression approaches $ln 1=0$.
$endgroup$
– Anurag A
Dec 7 '18 at 0:47
$begingroup$
Do the answerers seriously think that this limit would not have been handled multiple times here already? Shame on you!
$endgroup$
– Jyrki Lahtonen
Dec 7 '18 at 10:27
$begingroup$
Do the answerers seriously think that this limit would not have been handled multiple times here already? Shame on you!
$endgroup$
– Jyrki Lahtonen
Dec 7 '18 at 10:27
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
No, you do not need to have any conclusions about $ln(infty/infty)$ here (and those wouldn't work anyways!). To see that this is indeed a $0/0$ form, go a bit more carefully through the logarithm: We have
$$lim_{x to infty} ln left(frac{x - 1}{x}right) = lim_{x to infty} ln left(1 - frac 1 xright) = ln 1 = 0$$
which is what you need.
answered Dec 7 '18 at 0:48
T. BongersT. Bongers
23.1k54662
$endgroup$
add a comment |
$begingroup$
HINT
Note that the inverse
$$left(frac{x-1}{x}right)^x=left(1-frac{1}{x}right)^x$$
answered Dec 7 '18 at 0:49
gimusigimusi
92.8k84494
$endgroup$
add a comment |
$begingroup$
L'Hopital is rarely the method of choice. In this case, let $y = x-1$. Then
$$
left(frac{x }{x-1}right)^x
=
left(frac{y+1}{y}right)^{y+1}
=
left(1 + frac{ 1}{y}right)^{y }left(1 + frac{ 1}{y}right)^{ 1}.
$$
Now you can recognize the limit as $y to infty$ as $e times 1 = e$.
answered Dec 7 '18 at 0:58
Ethan BolkerEthan Bolker
42.4k549112
$endgroup$
add a comment |
$begingroup$
$$limlimits_{xto infty}left(frac{x}{x-1}right)^x = limlimits_{yto infty}left(frac{y+1}{y}right)^{y+1} = limlimits_{yto infty}left(1 +frac{1}{y}right)limlimits_{yto infty}left(1+frac{1}{y}right)^{y} = 1 times e = e$$
answered Dec 7 '18 at 0:57
HenryHenry
99.6k479165
$endgroup$
1
$begingroup$
though this does not use $e^{log_eleft(left(frac{x}{x-1}right)^x right)} = e^{xlog_eleft(frac{x}{x-1}right)}$
$endgroup$
– Henry
Dec 7 '18 at 1:01
1
$begingroup$
+1 You beat me by a minute while I was writing the same answer.
$endgroup$
– Ethan Bolker
Dec 7 '18 at 1:01
add a comment |
$begingroup$
Remember, for continuous function $f$, we can say, when we compose the function with some other function $g$,
$$lim_{x to c} f(g(x)) = f left( lim_{x to c} g(x) right)$$
In your case,
$$lim_{x to infty} ln left( frac{x}{x-1} right) = ln left( lim_{x to infty} left( frac{x}{x-1} right) right) $$
Alternatively (or in conjunction with this), in line with what other people have suggested, you could also observe
$$frac{x}{x-1} = frac{x-1+1}{x-1} = 1 + frac{1}{x-1}$$
answered Dec 7 '18 at 0:49
Eevee TrainerEevee Trainer
5,7961936
$endgroup$
add a comment |
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No, you do not need to have any conclusions about $ln(infty/infty)$ here (and those wouldn't work anyways!). To see that this is indeed a $0/0$ form, go a bit more carefully through the logarithm: We have
$$lim_{x to infty} ln left(frac{x - 1}{x}right) = lim_{x to infty} ln left(1 - frac 1 xright) = ln 1 = 0$$
which is what you need.
answered Dec 7 '18 at 0:48
T. BongersT. Bongers
23.1k54662
$endgroup$
add a comment |
$begingroup$
No, you do not need to have any conclusions about $ln(infty/infty)$ here (and those wouldn't work anyways!). To see that this is indeed a $0/0$ form, go a bit more carefully through the logarithm: We have
$$lim_{x to infty} ln left(frac{x - 1}{x}right) = lim_{x to infty} ln left(1 - frac 1 xright) = ln 1 = 0$$
which is what you need.
answered Dec 7 '18 at 0:48
T. BongersT. Bongers
23.1k54662
$endgroup$
add a comment |
$begingroup$
No, you do not need to have any conclusions about $ln(infty/infty)$ here (and those wouldn't work anyways!). To see that this is indeed a $0/0$ form, go a bit more carefully through the logarithm: We have
$$lim_{x to infty} ln left(frac{x - 1}{x}right) = lim_{x to infty} ln left(1 - frac 1 xright) = ln 1 = 0$$
which is what you need.
answered Dec 7 '18 at 0:48
T. BongersT. Bongers
23.1k54662
$endgroup$
No, you do not need to have any conclusions about $ln(infty/infty)$ here (and those wouldn't work anyways!). To see that this is indeed a $0/0$ form, go a bit more carefully through the logarithm: We have
$$lim_{x to infty} ln left(frac{x - 1}{x}right) = lim_{x to infty} ln left(1 - frac 1 xright) = ln 1 = 0$$
which is what you need.
answered Dec 7 '18 at 0:48
T. BongersT. Bongers
23.1k54662
answered Dec 7 '18 at 0:48
T. BongersT. Bongers
23.1k54662
answered Dec 7 '18 at 0:48
T. BongersT. Bongers
23.1k54662
answered Dec 7 '18 at 0:48
T. BongersT. Bongers
23.1k54662
23.1k54662
add a comment |
add a comment |
$begingroup$
HINT
Note that the inverse
$$left(frac{x-1}{x}right)^x=left(1-frac{1}{x}right)^x$$
answered Dec 7 '18 at 0:49
gimusigimusi
92.8k84494
$endgroup$
add a comment |
$begingroup$
HINT
Note that the inverse
$$left(frac{x-1}{x}right)^x=left(1-frac{1}{x}right)^x$$
answered Dec 7 '18 at 0:49
gimusigimusi
92.8k84494
$endgroup$
add a comment |
$begingroup$
HINT
Note that the inverse
$$left(frac{x-1}{x}right)^x=left(1-frac{1}{x}right)^x$$
answered Dec 7 '18 at 0:49
gimusigimusi
92.8k84494
$endgroup$
HINT
Note that the inverse
$$left(frac{x-1}{x}right)^x=left(1-frac{1}{x}right)^x$$
answered Dec 7 '18 at 0:49
gimusigimusi
92.8k84494
answered Dec 7 '18 at 0:49
gimusigimusi
92.8k84494
answered Dec 7 '18 at 0:49
gimusigimusi
92.8k84494
answered Dec 7 '18 at 0:49
gimusigimusi
92.8k84494
92.8k84494
add a comment |
add a comment |
$begingroup$
L'Hopital is rarely the method of choice. In this case, let $y = x-1$. Then
$$
left(frac{x }{x-1}right)^x
=
left(frac{y+1}{y}right)^{y+1}
=
left(1 + frac{ 1}{y}right)^{y }left(1 + frac{ 1}{y}right)^{ 1}.
$$
Now you can recognize the limit as $y to infty$ as $e times 1 = e$.
answered Dec 7 '18 at 0:58
Ethan BolkerEthan Bolker
42.4k549112
$endgroup$
add a comment |
$begingroup$
L'Hopital is rarely the method of choice. In this case, let $y = x-1$. Then
$$
left(frac{x }{x-1}right)^x
=
left(frac{y+1}{y}right)^{y+1}
=
left(1 + frac{ 1}{y}right)^{y }left(1 + frac{ 1}{y}right)^{ 1}.
$$
Now you can recognize the limit as $y to infty$ as $e times 1 = e$.
answered Dec 7 '18 at 0:58
Ethan BolkerEthan Bolker
42.4k549112
$endgroup$
add a comment |
$begingroup$
L'Hopital is rarely the method of choice. In this case, let $y = x-1$. Then
$$
left(frac{x }{x-1}right)^x
=
left(frac{y+1}{y}right)^{y+1}
=
left(1 + frac{ 1}{y}right)^{y }left(1 + frac{ 1}{y}right)^{ 1}.
$$
Now you can recognize the limit as $y to infty$ as $e times 1 = e$.
answered Dec 7 '18 at 0:58
Ethan BolkerEthan Bolker
42.4k549112
$endgroup$
L'Hopital is rarely the method of choice. In this case, let $y = x-1$. Then
$$
left(frac{x }{x-1}right)^x
=
left(frac{y+1}{y}right)^{y+1}
=
left(1 + frac{ 1}{y}right)^{y }left(1 + frac{ 1}{y}right)^{ 1}.
$$
Now you can recognize the limit as $y to infty$ as $e times 1 = e$.
answered Dec 7 '18 at 0:58
Ethan BolkerEthan Bolker
42.4k549112
answered Dec 7 '18 at 0:58
Ethan BolkerEthan Bolker
42.4k549112
answered Dec 7 '18 at 0:58
Ethan BolkerEthan Bolker
42.4k549112
answered Dec 7 '18 at 0:58
Ethan BolkerEthan Bolker
42.4k549112
42.4k549112
add a comment |
add a comment |
$begingroup$
$$limlimits_{xto infty}left(frac{x}{x-1}right)^x = limlimits_{yto infty}left(frac{y+1}{y}right)^{y+1} = limlimits_{yto infty}left(1 +frac{1}{y}right)limlimits_{yto infty}left(1+frac{1}{y}right)^{y} = 1 times e = e$$
answered Dec 7 '18 at 0:57
HenryHenry
99.6k479165
$endgroup$
1
$begingroup$
though this does not use $e^{log_eleft(left(frac{x}{x-1}right)^x right)} = e^{xlog_eleft(frac{x}{x-1}right)}$
$endgroup$
– Henry
Dec 7 '18 at 1:01
1
$begingroup$
+1 You beat me by a minute while I was writing the same answer.
$endgroup$
– Ethan Bolker
Dec 7 '18 at 1:01
add a comment |
$begingroup$
$$limlimits_{xto infty}left(frac{x}{x-1}right)^x = limlimits_{yto infty}left(frac{y+1}{y}right)^{y+1} = limlimits_{yto infty}left(1 +frac{1}{y}right)limlimits_{yto infty}left(1+frac{1}{y}right)^{y} = 1 times e = e$$
answered Dec 7 '18 at 0:57
HenryHenry
99.6k479165
$endgroup$
1
$begingroup$
though this does not use $e^{log_eleft(left(frac{x}{x-1}right)^x right)} = e^{xlog_eleft(frac{x}{x-1}right)}$
$endgroup$
– Henry
Dec 7 '18 at 1:01
1
$begingroup$
+1 You beat me by a minute while I was writing the same answer.
$endgroup$
– Ethan Bolker
Dec 7 '18 at 1:01
add a comment |
$begingroup$
$$limlimits_{xto infty}left(frac{x}{x-1}right)^x = limlimits_{yto infty}left(frac{y+1}{y}right)^{y+1} = limlimits_{yto infty}left(1 +frac{1}{y}right)limlimits_{yto infty}left(1+frac{1}{y}right)^{y} = 1 times e = e$$
answered Dec 7 '18 at 0:57
HenryHenry
99.6k479165
$endgroup$
$$limlimits_{xto infty}left(frac{x}{x-1}right)^x = limlimits_{yto infty}left(frac{y+1}{y}right)^{y+1} = limlimits_{yto infty}left(1 +frac{1}{y}right)limlimits_{yto infty}left(1+frac{1}{y}right)^{y} = 1 times e = e$$
answered Dec 7 '18 at 0:57
HenryHenry
99.6k479165
answered Dec 7 '18 at 0:57
HenryHenry
99.6k479165
answered Dec 7 '18 at 0:57
HenryHenry
99.6k479165
answered Dec 7 '18 at 0:57
HenryHenry
99.6k479165
99.6k479165
1
$begingroup$
though this does not use $e^{log_eleft(left(frac{x}{x-1}right)^x right)} = e^{xlog_eleft(frac{x}{x-1}right)}$
$endgroup$
– Henry
Dec 7 '18 at 1:01
1
$begingroup$
+1 You beat me by a minute while I was writing the same answer.
$endgroup$
– Ethan Bolker
Dec 7 '18 at 1:01
add a comment |
1
$begingroup$
though this does not use $e^{log_eleft(left(frac{x}{x-1}right)^x right)} = e^{xlog_eleft(frac{x}{x-1}right)}$
$endgroup$
– Henry
Dec 7 '18 at 1:01
1
$begingroup$
+1 You beat me by a minute while I was writing the same answer.
$endgroup$
– Ethan Bolker
Dec 7 '18 at 1:01
1
1
$begingroup$
though this does not use $e^{log_eleft(left(frac{x}{x-1}right)^x right)} = e^{xlog_eleft(frac{x}{x-1}right)}$
$endgroup$
– Henry
Dec 7 '18 at 1:01
$begingroup$
though this does not use $e^{log_eleft(left(frac{x}{x-1}right)^x right)} = e^{xlog_eleft(frac{x}{x-1}right)}$
$endgroup$
– Henry
Dec 7 '18 at 1:01
1
1
$begingroup$
+1 You beat me by a minute while I was writing the same answer.
$endgroup$
– Ethan Bolker
Dec 7 '18 at 1:01
$begingroup$
+1 You beat me by a minute while I was writing the same answer.
$endgroup$
– Ethan Bolker
Dec 7 '18 at 1:01
add a comment |
$begingroup$
Remember, for continuous function $f$, we can say, when we compose the function with some other function $g$,
$$lim_{x to c} f(g(x)) = f left( lim_{x to c} g(x) right)$$
In your case,
$$lim_{x to infty} ln left( frac{x}{x-1} right) = ln left( lim_{x to infty} left( frac{x}{x-1} right) right) $$
Alternatively (or in conjunction with this), in line with what other people have suggested, you could also observe
$$frac{x}{x-1} = frac{x-1+1}{x-1} = 1 + frac{1}{x-1}$$
answered Dec 7 '18 at 0:49
Eevee TrainerEevee Trainer
5,7961936
$endgroup$
add a comment |
$begingroup$
Remember, for continuous function $f$, we can say, when we compose the function with some other function $g$,
$$lim_{x to c} f(g(x)) = f left( lim_{x to c} g(x) right)$$
In your case,
$$lim_{x to infty} ln left( frac{x}{x-1} right) = ln left( lim_{x to infty} left( frac{x}{x-1} right) right) $$
Alternatively (or in conjunction with this), in line with what other people have suggested, you could also observe
$$frac{x}{x-1} = frac{x-1+1}{x-1} = 1 + frac{1}{x-1}$$
answered Dec 7 '18 at 0:49
Eevee TrainerEevee Trainer
5,7961936
$endgroup$
add a comment |
$begingroup$
Remember, for continuous function $f$, we can say, when we compose the function with some other function $g$,
$$lim_{x to c} f(g(x)) = f left( lim_{x to c} g(x) right)$$
In your case,
$$lim_{x to infty} ln left( frac{x}{x-1} right) = ln left( lim_{x to infty} left( frac{x}{x-1} right) right) $$
Alternatively (or in conjunction with this), in line with what other people have suggested, you could also observe
$$frac{x}{x-1} = frac{x-1+1}{x-1} = 1 + frac{1}{x-1}$$
answered Dec 7 '18 at 0:49
Eevee TrainerEevee Trainer
5,7961936
$endgroup$
Remember, for continuous function $f$, we can say, when we compose the function with some other function $g$,
$$lim_{x to c} f(g(x)) = f left( lim_{x to c} g(x) right)$$
In your case,
$$lim_{x to infty} ln left( frac{x}{x-1} right) = ln left( lim_{x to infty} left( frac{x}{x-1} right) right) $$
Alternatively (or in conjunction with this), in line with what other people have suggested, you could also observe
$$frac{x}{x-1} = frac{x-1+1}{x-1} = 1 + frac{1}{x-1}$$
answered Dec 7 '18 at 0:49
Eevee TrainerEevee Trainer
5,7961936
answered Dec 7 '18 at 0:49
Eevee TrainerEevee Trainer
5,7961936
answered Dec 7 '18 at 0:49
Eevee TrainerEevee Trainer
5,7961936
answered Dec 7 '18 at 0:49
Eevee TrainerEevee Trainer
5,7961936
5,7961936
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$ln left(frac{x-1}{x}right)=ln left(1-frac{1}{x}right)$. As $x to infty$, the expression approaches $ln 1=0$.
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– Anurag A
Dec 7 '18 at 0:47
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Do the answerers seriously think that this limit would not have been handled multiple times here already? Shame on you!
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– Jyrki Lahtonen
Dec 7 '18 at 10:27