Proving that $lim_{ntoinfty} left(1+frac{1}{f(n)}right)^{g(n)} = 1$
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9
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I want to prove that $$lim_{ntoinfty} left(1+frac{1}{f(n)}right)^{g(n)} = 1$$ if $f(n)$ grows faster than $g(n)$ for $ntoinfty$ and $lim_{ntoinfty} f(n) = +infty = lim_{ntoinfty}g(n)$.
It is quite easy to see that if $f = g$ the limit is $e$, but I can't find a good strategy to solve this problem.
calculus sequences-and-series limits exponentiation
add a comment |
up vote
9
down vote
favorite
I want to prove that $$lim_{ntoinfty} left(1+frac{1}{f(n)}right)^{g(n)} = 1$$ if $f(n)$ grows faster than $g(n)$ for $ntoinfty$ and $lim_{ntoinfty} f(n) = +infty = lim_{ntoinfty}g(n)$.
It is quite easy to see that if $f = g$ the limit is $e$, but I can't find a good strategy to solve this problem.
calculus sequences-and-series limits exponentiation
What is $lim_{ntoinfty}f(n)$ and $lim_{ntoinfty}g(n)$?
– Thomas Shelby
Dec 1 at 18:24
@ThomasShelby they're both $+infty$; in a special case, I need $n^n$ and $n!$, but a more general case is more interesting.
– Riccardo Cazzin
Dec 1 at 18:26
Yeah, if $f(n)toinfty$ and $g(n)/f(n)to 1,$ your result is true.
– Thomas Andrews
Dec 1 at 18:26
add a comment |
up vote
9
down vote
favorite
up vote
9
down vote
favorite
I want to prove that $$lim_{ntoinfty} left(1+frac{1}{f(n)}right)^{g(n)} = 1$$ if $f(n)$ grows faster than $g(n)$ for $ntoinfty$ and $lim_{ntoinfty} f(n) = +infty = lim_{ntoinfty}g(n)$.
It is quite easy to see that if $f = g$ the limit is $e$, but I can't find a good strategy to solve this problem.
calculus sequences-and-series limits exponentiation
I want to prove that $$lim_{ntoinfty} left(1+frac{1}{f(n)}right)^{g(n)} = 1$$ if $f(n)$ grows faster than $g(n)$ for $ntoinfty$ and $lim_{ntoinfty} f(n) = +infty = lim_{ntoinfty}g(n)$.
It is quite easy to see that if $f = g$ the limit is $e$, but I can't find a good strategy to solve this problem.
calculus sequences-and-series limits exponentiation
calculus sequences-and-series limits exponentiation
edited Dec 2 at 9:28
Asaf Karagila♦
301k32422753
301k32422753
asked Dec 1 at 18:21
Riccardo Cazzin
1905
1905
What is $lim_{ntoinfty}f(n)$ and $lim_{ntoinfty}g(n)$?
– Thomas Shelby
Dec 1 at 18:24
@ThomasShelby they're both $+infty$; in a special case, I need $n^n$ and $n!$, but a more general case is more interesting.
– Riccardo Cazzin
Dec 1 at 18:26
Yeah, if $f(n)toinfty$ and $g(n)/f(n)to 1,$ your result is true.
– Thomas Andrews
Dec 1 at 18:26
add a comment |
What is $lim_{ntoinfty}f(n)$ and $lim_{ntoinfty}g(n)$?
– Thomas Shelby
Dec 1 at 18:24
@ThomasShelby they're both $+infty$; in a special case, I need $n^n$ and $n!$, but a more general case is more interesting.
– Riccardo Cazzin
Dec 1 at 18:26
Yeah, if $f(n)toinfty$ and $g(n)/f(n)to 1,$ your result is true.
– Thomas Andrews
Dec 1 at 18:26
What is $lim_{ntoinfty}f(n)$ and $lim_{ntoinfty}g(n)$?
– Thomas Shelby
Dec 1 at 18:24
What is $lim_{ntoinfty}f(n)$ and $lim_{ntoinfty}g(n)$?
– Thomas Shelby
Dec 1 at 18:24
@ThomasShelby they're both $+infty$; in a special case, I need $n^n$ and $n!$, but a more general case is more interesting.
– Riccardo Cazzin
Dec 1 at 18:26
@ThomasShelby they're both $+infty$; in a special case, I need $n^n$ and $n!$, but a more general case is more interesting.
– Riccardo Cazzin
Dec 1 at 18:26
Yeah, if $f(n)toinfty$ and $g(n)/f(n)to 1,$ your result is true.
– Thomas Andrews
Dec 1 at 18:26
Yeah, if $f(n)toinfty$ and $g(n)/f(n)to 1,$ your result is true.
– Thomas Andrews
Dec 1 at 18:26
add a comment |
5 Answers
5
active
oldest
votes
up vote
14
down vote
accepted
We can use that
$$ left(1+frac{1}{f(n)}right)^{g(n)} =left[left(1+frac{1}{f(n)}right)^{f(n)}right]^{frac{g(n)}{f(n)}}$$
add a comment |
up vote
11
down vote
That depends upon how you defined to grow faster than. But if implies that $lim_{ntoinfty}frac{g(n)}{f(n)}=0$,thenbegin{align}lim_{ntoinfty}left(1+frac1{f(n)}right)^{g(n)}&=lim_{ntoinfty}left(left(1+frac1{f(n)}right)^{f(n)}right)^{frac{g(n)}{f(n)}}\&=e^0\&=1.end{align}
2
The second-to-last step does require a bit of justification though, since $lim_n{a_n}^{b_n} = (lim_n a_n)^{lim_n b_n}$ is not a trivial fact (and is false in general, for instance if limits are in $[0,infty]$).
– Clement C.
Dec 1 at 18:37
@ClementC. Yes, I agree that one should be careful here.
– José Carlos Santos
Dec 1 at 18:48
2
I like how the comment above says that one of the steps need more explanation, but the accepted answer is just the result with zero explanation.
– user1717828
Dec 2 at 1:37
1
@user1717828 Yes.This happens around here. Get used to it.
– José Carlos Santos
Dec 2 at 11:09
@user1717828 what can you do...
– Clement C.
Dec 4 at 22:17
add a comment |
up vote
5
down vote
You have
$$
left(1+frac{1}{f(n)}right)^{g(n)}
= expleft(g(n) ln left(1+frac{1}{f(n)}right) right)
$$
Since $lim_{ntoinfty} f(n) = infty$, we have
$$
g(n) ln left(1+frac{1}{f(n)}right)
= g(n)cdot left(frac{1}{f(n)} + oleft(frac{1}{f(n)}right)right)
= frac{g(n)}{f(n)} + o!left(frac{g(n)}{f(n)}right)
$$
and by your assumption that $f$ "grows faster than $g$", this converges to $ell=0$
(the result holds as long as $lim_{ntoinfty} frac{g(n)}{f(n)}$ exists, not necessarily $0$).
Then,
$$
lim_{ntoinfty }left(1+frac{1}{f(n)}right)^{g(n)}
= e^0 = 1.
$$
2
THIS is the rigorous way (+1)
– Robert Z
Dec 2 at 17:10
add a comment |
up vote
3
down vote
Note that
$$lim_{n to infty}frac{g(n)}{f(n)} = 0 implies bigg(1+frac{1}{f(n)}bigg)^{g(n)} = Biggl[bigg(1+frac{1}{f(n)}bigg)^{f(n)}Biggl]^{frac{g(n)}{f(n)}} = e^frac{g(n)}{f(n)} to e^0 = 1$$
The equality with the exponential is just... false? (and even in terms of limits, there would be assumptions hidden)
– Clement C.
Dec 1 at 18:35
add a comment |
up vote
0
down vote
Taking logarithms it suffices to prove that the limit of the logarithm is zero. To this end note that
$$
g(n)logleft(1+frac{1}{f(n)}right)=frac{g(n)}{f(n)}frac{logleft(1+frac{1}{f(n)}right)}{1/f(n)}to0
$$
since
$$
frac{g(n)}{f(n)}to0
$$
as $f$ grows faster than $g$ and for the second term use the fact $1/f(n)to0$ and
$$
lim_{xto0}frac{log (1+x)-0}{x-0}=1
$$
by definition of the derivative.
So the result is 0*1 ?
– Samy Bencherif
Dec 2 at 6:42
No to the limit is $exp(0times 1)$
– Foobaz John
Dec 2 at 15:57
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
14
down vote
accepted
We can use that
$$ left(1+frac{1}{f(n)}right)^{g(n)} =left[left(1+frac{1}{f(n)}right)^{f(n)}right]^{frac{g(n)}{f(n)}}$$
add a comment |
up vote
14
down vote
accepted
We can use that
$$ left(1+frac{1}{f(n)}right)^{g(n)} =left[left(1+frac{1}{f(n)}right)^{f(n)}right]^{frac{g(n)}{f(n)}}$$
add a comment |
up vote
14
down vote
accepted
up vote
14
down vote
accepted
We can use that
$$ left(1+frac{1}{f(n)}right)^{g(n)} =left[left(1+frac{1}{f(n)}right)^{f(n)}right]^{frac{g(n)}{f(n)}}$$
We can use that
$$ left(1+frac{1}{f(n)}right)^{g(n)} =left[left(1+frac{1}{f(n)}right)^{f(n)}right]^{frac{g(n)}{f(n)}}$$
answered Dec 1 at 18:26
gimusi
92.7k94495
92.7k94495
add a comment |
add a comment |
up vote
11
down vote
That depends upon how you defined to grow faster than. But if implies that $lim_{ntoinfty}frac{g(n)}{f(n)}=0$,thenbegin{align}lim_{ntoinfty}left(1+frac1{f(n)}right)^{g(n)}&=lim_{ntoinfty}left(left(1+frac1{f(n)}right)^{f(n)}right)^{frac{g(n)}{f(n)}}\&=e^0\&=1.end{align}
2
The second-to-last step does require a bit of justification though, since $lim_n{a_n}^{b_n} = (lim_n a_n)^{lim_n b_n}$ is not a trivial fact (and is false in general, for instance if limits are in $[0,infty]$).
– Clement C.
Dec 1 at 18:37
@ClementC. Yes, I agree that one should be careful here.
– José Carlos Santos
Dec 1 at 18:48
2
I like how the comment above says that one of the steps need more explanation, but the accepted answer is just the result with zero explanation.
– user1717828
Dec 2 at 1:37
1
@user1717828 Yes.This happens around here. Get used to it.
– José Carlos Santos
Dec 2 at 11:09
@user1717828 what can you do...
– Clement C.
Dec 4 at 22:17
add a comment |
up vote
11
down vote
That depends upon how you defined to grow faster than. But if implies that $lim_{ntoinfty}frac{g(n)}{f(n)}=0$,thenbegin{align}lim_{ntoinfty}left(1+frac1{f(n)}right)^{g(n)}&=lim_{ntoinfty}left(left(1+frac1{f(n)}right)^{f(n)}right)^{frac{g(n)}{f(n)}}\&=e^0\&=1.end{align}
2
The second-to-last step does require a bit of justification though, since $lim_n{a_n}^{b_n} = (lim_n a_n)^{lim_n b_n}$ is not a trivial fact (and is false in general, for instance if limits are in $[0,infty]$).
– Clement C.
Dec 1 at 18:37
@ClementC. Yes, I agree that one should be careful here.
– José Carlos Santos
Dec 1 at 18:48
2
I like how the comment above says that one of the steps need more explanation, but the accepted answer is just the result with zero explanation.
– user1717828
Dec 2 at 1:37
1
@user1717828 Yes.This happens around here. Get used to it.
– José Carlos Santos
Dec 2 at 11:09
@user1717828 what can you do...
– Clement C.
Dec 4 at 22:17
add a comment |
up vote
11
down vote
up vote
11
down vote
That depends upon how you defined to grow faster than. But if implies that $lim_{ntoinfty}frac{g(n)}{f(n)}=0$,thenbegin{align}lim_{ntoinfty}left(1+frac1{f(n)}right)^{g(n)}&=lim_{ntoinfty}left(left(1+frac1{f(n)}right)^{f(n)}right)^{frac{g(n)}{f(n)}}\&=e^0\&=1.end{align}
That depends upon how you defined to grow faster than. But if implies that $lim_{ntoinfty}frac{g(n)}{f(n)}=0$,thenbegin{align}lim_{ntoinfty}left(1+frac1{f(n)}right)^{g(n)}&=lim_{ntoinfty}left(left(1+frac1{f(n)}right)^{f(n)}right)^{frac{g(n)}{f(n)}}\&=e^0\&=1.end{align}
answered Dec 1 at 18:27
José Carlos Santos
147k22117217
147k22117217
2
The second-to-last step does require a bit of justification though, since $lim_n{a_n}^{b_n} = (lim_n a_n)^{lim_n b_n}$ is not a trivial fact (and is false in general, for instance if limits are in $[0,infty]$).
– Clement C.
Dec 1 at 18:37
@ClementC. Yes, I agree that one should be careful here.
– José Carlos Santos
Dec 1 at 18:48
2
I like how the comment above says that one of the steps need more explanation, but the accepted answer is just the result with zero explanation.
– user1717828
Dec 2 at 1:37
1
@user1717828 Yes.This happens around here. Get used to it.
– José Carlos Santos
Dec 2 at 11:09
@user1717828 what can you do...
– Clement C.
Dec 4 at 22:17
add a comment |
2
The second-to-last step does require a bit of justification though, since $lim_n{a_n}^{b_n} = (lim_n a_n)^{lim_n b_n}$ is not a trivial fact (and is false in general, for instance if limits are in $[0,infty]$).
– Clement C.
Dec 1 at 18:37
@ClementC. Yes, I agree that one should be careful here.
– José Carlos Santos
Dec 1 at 18:48
2
I like how the comment above says that one of the steps need more explanation, but the accepted answer is just the result with zero explanation.
– user1717828
Dec 2 at 1:37
1
@user1717828 Yes.This happens around here. Get used to it.
– José Carlos Santos
Dec 2 at 11:09
@user1717828 what can you do...
– Clement C.
Dec 4 at 22:17
2
2
The second-to-last step does require a bit of justification though, since $lim_n{a_n}^{b_n} = (lim_n a_n)^{lim_n b_n}$ is not a trivial fact (and is false in general, for instance if limits are in $[0,infty]$).
– Clement C.
Dec 1 at 18:37
The second-to-last step does require a bit of justification though, since $lim_n{a_n}^{b_n} = (lim_n a_n)^{lim_n b_n}$ is not a trivial fact (and is false in general, for instance if limits are in $[0,infty]$).
– Clement C.
Dec 1 at 18:37
@ClementC. Yes, I agree that one should be careful here.
– José Carlos Santos
Dec 1 at 18:48
@ClementC. Yes, I agree that one should be careful here.
– José Carlos Santos
Dec 1 at 18:48
2
2
I like how the comment above says that one of the steps need more explanation, but the accepted answer is just the result with zero explanation.
– user1717828
Dec 2 at 1:37
I like how the comment above says that one of the steps need more explanation, but the accepted answer is just the result with zero explanation.
– user1717828
Dec 2 at 1:37
1
1
@user1717828 Yes.This happens around here. Get used to it.
– José Carlos Santos
Dec 2 at 11:09
@user1717828 Yes.This happens around here. Get used to it.
– José Carlos Santos
Dec 2 at 11:09
@user1717828 what can you do...
– Clement C.
Dec 4 at 22:17
@user1717828 what can you do...
– Clement C.
Dec 4 at 22:17
add a comment |
up vote
5
down vote
You have
$$
left(1+frac{1}{f(n)}right)^{g(n)}
= expleft(g(n) ln left(1+frac{1}{f(n)}right) right)
$$
Since $lim_{ntoinfty} f(n) = infty$, we have
$$
g(n) ln left(1+frac{1}{f(n)}right)
= g(n)cdot left(frac{1}{f(n)} + oleft(frac{1}{f(n)}right)right)
= frac{g(n)}{f(n)} + o!left(frac{g(n)}{f(n)}right)
$$
and by your assumption that $f$ "grows faster than $g$", this converges to $ell=0$
(the result holds as long as $lim_{ntoinfty} frac{g(n)}{f(n)}$ exists, not necessarily $0$).
Then,
$$
lim_{ntoinfty }left(1+frac{1}{f(n)}right)^{g(n)}
= e^0 = 1.
$$
2
THIS is the rigorous way (+1)
– Robert Z
Dec 2 at 17:10
add a comment |
up vote
5
down vote
You have
$$
left(1+frac{1}{f(n)}right)^{g(n)}
= expleft(g(n) ln left(1+frac{1}{f(n)}right) right)
$$
Since $lim_{ntoinfty} f(n) = infty$, we have
$$
g(n) ln left(1+frac{1}{f(n)}right)
= g(n)cdot left(frac{1}{f(n)} + oleft(frac{1}{f(n)}right)right)
= frac{g(n)}{f(n)} + o!left(frac{g(n)}{f(n)}right)
$$
and by your assumption that $f$ "grows faster than $g$", this converges to $ell=0$
(the result holds as long as $lim_{ntoinfty} frac{g(n)}{f(n)}$ exists, not necessarily $0$).
Then,
$$
lim_{ntoinfty }left(1+frac{1}{f(n)}right)^{g(n)}
= e^0 = 1.
$$
2
THIS is the rigorous way (+1)
– Robert Z
Dec 2 at 17:10
add a comment |
up vote
5
down vote
up vote
5
down vote
You have
$$
left(1+frac{1}{f(n)}right)^{g(n)}
= expleft(g(n) ln left(1+frac{1}{f(n)}right) right)
$$
Since $lim_{ntoinfty} f(n) = infty$, we have
$$
g(n) ln left(1+frac{1}{f(n)}right)
= g(n)cdot left(frac{1}{f(n)} + oleft(frac{1}{f(n)}right)right)
= frac{g(n)}{f(n)} + o!left(frac{g(n)}{f(n)}right)
$$
and by your assumption that $f$ "grows faster than $g$", this converges to $ell=0$
(the result holds as long as $lim_{ntoinfty} frac{g(n)}{f(n)}$ exists, not necessarily $0$).
Then,
$$
lim_{ntoinfty }left(1+frac{1}{f(n)}right)^{g(n)}
= e^0 = 1.
$$
You have
$$
left(1+frac{1}{f(n)}right)^{g(n)}
= expleft(g(n) ln left(1+frac{1}{f(n)}right) right)
$$
Since $lim_{ntoinfty} f(n) = infty$, we have
$$
g(n) ln left(1+frac{1}{f(n)}right)
= g(n)cdot left(frac{1}{f(n)} + oleft(frac{1}{f(n)}right)right)
= frac{g(n)}{f(n)} + o!left(frac{g(n)}{f(n)}right)
$$
and by your assumption that $f$ "grows faster than $g$", this converges to $ell=0$
(the result holds as long as $lim_{ntoinfty} frac{g(n)}{f(n)}$ exists, not necessarily $0$).
Then,
$$
lim_{ntoinfty }left(1+frac{1}{f(n)}right)^{g(n)}
= e^0 = 1.
$$
answered Dec 1 at 18:32
Clement C.
49.2k33785
49.2k33785
2
THIS is the rigorous way (+1)
– Robert Z
Dec 2 at 17:10
add a comment |
2
THIS is the rigorous way (+1)
– Robert Z
Dec 2 at 17:10
2
2
THIS is the rigorous way (+1)
– Robert Z
Dec 2 at 17:10
THIS is the rigorous way (+1)
– Robert Z
Dec 2 at 17:10
add a comment |
up vote
3
down vote
Note that
$$lim_{n to infty}frac{g(n)}{f(n)} = 0 implies bigg(1+frac{1}{f(n)}bigg)^{g(n)} = Biggl[bigg(1+frac{1}{f(n)}bigg)^{f(n)}Biggl]^{frac{g(n)}{f(n)}} = e^frac{g(n)}{f(n)} to e^0 = 1$$
The equality with the exponential is just... false? (and even in terms of limits, there would be assumptions hidden)
– Clement C.
Dec 1 at 18:35
add a comment |
up vote
3
down vote
Note that
$$lim_{n to infty}frac{g(n)}{f(n)} = 0 implies bigg(1+frac{1}{f(n)}bigg)^{g(n)} = Biggl[bigg(1+frac{1}{f(n)}bigg)^{f(n)}Biggl]^{frac{g(n)}{f(n)}} = e^frac{g(n)}{f(n)} to e^0 = 1$$
The equality with the exponential is just... false? (and even in terms of limits, there would be assumptions hidden)
– Clement C.
Dec 1 at 18:35
add a comment |
up vote
3
down vote
up vote
3
down vote
Note that
$$lim_{n to infty}frac{g(n)}{f(n)} = 0 implies bigg(1+frac{1}{f(n)}bigg)^{g(n)} = Biggl[bigg(1+frac{1}{f(n)}bigg)^{f(n)}Biggl]^{frac{g(n)}{f(n)}} = e^frac{g(n)}{f(n)} to e^0 = 1$$
Note that
$$lim_{n to infty}frac{g(n)}{f(n)} = 0 implies bigg(1+frac{1}{f(n)}bigg)^{g(n)} = Biggl[bigg(1+frac{1}{f(n)}bigg)^{f(n)}Biggl]^{frac{g(n)}{f(n)}} = e^frac{g(n)}{f(n)} to e^0 = 1$$
answered Dec 1 at 18:32
KM101
3,882417
3,882417
The equality with the exponential is just... false? (and even in terms of limits, there would be assumptions hidden)
– Clement C.
Dec 1 at 18:35
add a comment |
The equality with the exponential is just... false? (and even in terms of limits, there would be assumptions hidden)
– Clement C.
Dec 1 at 18:35
The equality with the exponential is just... false? (and even in terms of limits, there would be assumptions hidden)
– Clement C.
Dec 1 at 18:35
The equality with the exponential is just... false? (and even in terms of limits, there would be assumptions hidden)
– Clement C.
Dec 1 at 18:35
add a comment |
up vote
0
down vote
Taking logarithms it suffices to prove that the limit of the logarithm is zero. To this end note that
$$
g(n)logleft(1+frac{1}{f(n)}right)=frac{g(n)}{f(n)}frac{logleft(1+frac{1}{f(n)}right)}{1/f(n)}to0
$$
since
$$
frac{g(n)}{f(n)}to0
$$
as $f$ grows faster than $g$ and for the second term use the fact $1/f(n)to0$ and
$$
lim_{xto0}frac{log (1+x)-0}{x-0}=1
$$
by definition of the derivative.
So the result is 0*1 ?
– Samy Bencherif
Dec 2 at 6:42
No to the limit is $exp(0times 1)$
– Foobaz John
Dec 2 at 15:57
add a comment |
up vote
0
down vote
Taking logarithms it suffices to prove that the limit of the logarithm is zero. To this end note that
$$
g(n)logleft(1+frac{1}{f(n)}right)=frac{g(n)}{f(n)}frac{logleft(1+frac{1}{f(n)}right)}{1/f(n)}to0
$$
since
$$
frac{g(n)}{f(n)}to0
$$
as $f$ grows faster than $g$ and for the second term use the fact $1/f(n)to0$ and
$$
lim_{xto0}frac{log (1+x)-0}{x-0}=1
$$
by definition of the derivative.
So the result is 0*1 ?
– Samy Bencherif
Dec 2 at 6:42
No to the limit is $exp(0times 1)$
– Foobaz John
Dec 2 at 15:57
add a comment |
up vote
0
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Taking logarithms it suffices to prove that the limit of the logarithm is zero. To this end note that
$$
g(n)logleft(1+frac{1}{f(n)}right)=frac{g(n)}{f(n)}frac{logleft(1+frac{1}{f(n)}right)}{1/f(n)}to0
$$
since
$$
frac{g(n)}{f(n)}to0
$$
as $f$ grows faster than $g$ and for the second term use the fact $1/f(n)to0$ and
$$
lim_{xto0}frac{log (1+x)-0}{x-0}=1
$$
by definition of the derivative.
Taking logarithms it suffices to prove that the limit of the logarithm is zero. To this end note that
$$
g(n)logleft(1+frac{1}{f(n)}right)=frac{g(n)}{f(n)}frac{logleft(1+frac{1}{f(n)}right)}{1/f(n)}to0
$$
since
$$
frac{g(n)}{f(n)}to0
$$
as $f$ grows faster than $g$ and for the second term use the fact $1/f(n)to0$ and
$$
lim_{xto0}frac{log (1+x)-0}{x-0}=1
$$
by definition of the derivative.
answered Dec 1 at 18:38
Foobaz John
20.4k41250
20.4k41250
So the result is 0*1 ?
– Samy Bencherif
Dec 2 at 6:42
No to the limit is $exp(0times 1)$
– Foobaz John
Dec 2 at 15:57
add a comment |
So the result is 0*1 ?
– Samy Bencherif
Dec 2 at 6:42
No to the limit is $exp(0times 1)$
– Foobaz John
Dec 2 at 15:57
So the result is 0*1 ?
– Samy Bencherif
Dec 2 at 6:42
So the result is 0*1 ?
– Samy Bencherif
Dec 2 at 6:42
No to the limit is $exp(0times 1)$
– Foobaz John
Dec 2 at 15:57
No to the limit is $exp(0times 1)$
– Foobaz John
Dec 2 at 15:57
add a comment |
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What is $lim_{ntoinfty}f(n)$ and $lim_{ntoinfty}g(n)$?
– Thomas Shelby
Dec 1 at 18:24
@ThomasShelby they're both $+infty$; in a special case, I need $n^n$ and $n!$, but a more general case is more interesting.
– Riccardo Cazzin
Dec 1 at 18:26
Yeah, if $f(n)toinfty$ and $g(n)/f(n)to 1,$ your result is true.
– Thomas Andrews
Dec 1 at 18:26