Does there exist the limit at infinity of a function $f:[1,+infty)to[0,+infty)$ such that...
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Let $f:[1,+infty)to[0,+infty)$ be a function such that $int_{1}^{+infty}f(x)dx$ is convergent. I have 3 questions as follows:
- Does $f(x)$ have limit at infinity?
- If the answer of question (1) is yes, then is $lim_{xtoinfty}f(x)=0$ true?
- If the answers of questions (1) and (2) are NEGATIVE, What can we say about their answers, if $f$ is a continuous function?
I found that the most function under this condition have limit at infinity, also their limit equal to zero. Moreover, I tried to apply the definition of improper integral
$$int_{1}^{infty}f(x)dx=lim_{btoinfty}int_{1}^{b}f(x)dx$$
to find Contradictionو but I could not arrive at the aim.
calculus improper-integrals
asked Jan 18 at 16:37
soodehMehboodisoodehMehboodi
65848
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closed as off-topic by RRL, A. Pongrácz, Cesareo, Adrian Keister, metamorphy Jan 19 at 15:17
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, A. Pongrácz, Cesareo, Adrian Keister, metamorphy
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Let $f:[1,+infty)to[0,+infty)$ be a function such that $int_{1}^{+infty}f(x)dx$ is convergent. I have 3 questions as follows:
- Does $f(x)$ have limit at infinity?
- If the answer of question (1) is yes, then is $lim_{xtoinfty}f(x)=0$ true?
- If the answers of questions (1) and (2) are NEGATIVE, What can we say about their answers, if $f$ is a continuous function?
I found that the most function under this condition have limit at infinity, also their limit equal to zero. Moreover, I tried to apply the definition of improper integral
$$int_{1}^{infty}f(x)dx=lim_{btoinfty}int_{1}^{b}f(x)dx$$
to find Contradictionو but I could not arrive at the aim.
calculus improper-integrals
asked Jan 18 at 16:37
soodehMehboodisoodehMehboodi
65848
$endgroup$
closed as off-topic by RRL, A. Pongrácz, Cesareo, Adrian Keister, metamorphy Jan 19 at 15:17
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, A. Pongrácz, Cesareo, Adrian Keister, metamorphy
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Let $f:[1,+infty)to[0,+infty)$ be a function such that $int_{1}^{+infty}f(x)dx$ is convergent. I have 3 questions as follows:
- Does $f(x)$ have limit at infinity?
- If the answer of question (1) is yes, then is $lim_{xtoinfty}f(x)=0$ true?
- If the answers of questions (1) and (2) are NEGATIVE, What can we say about their answers, if $f$ is a continuous function?
I found that the most function under this condition have limit at infinity, also their limit equal to zero. Moreover, I tried to apply the definition of improper integral
$$int_{1}^{infty}f(x)dx=lim_{btoinfty}int_{1}^{b}f(x)dx$$
to find Contradictionو but I could not arrive at the aim.
calculus improper-integrals
asked Jan 18 at 16:37
soodehMehboodisoodehMehboodi
65848
$endgroup$
Let $f:[1,+infty)to[0,+infty)$ be a function such that $int_{1}^{+infty}f(x)dx$ is convergent. I have 3 questions as follows:
- Does $f(x)$ have limit at infinity?
- If the answer of question (1) is yes, then is $lim_{xtoinfty}f(x)=0$ true?
- If the answers of questions (1) and (2) are NEGATIVE, What can we say about their answers, if $f$ is a continuous function?
I found that the most function under this condition have limit at infinity, also their limit equal to zero. Moreover, I tried to apply the definition of improper integral
$$int_{1}^{infty}f(x)dx=lim_{btoinfty}int_{1}^{b}f(x)dx$$
to find Contradictionو but I could not arrive at the aim.
calculus improper-integrals
calculus improper-integrals
asked Jan 18 at 16:37
soodehMehboodisoodehMehboodi
65848
asked Jan 18 at 16:37
soodehMehboodisoodehMehboodi
65848
edited Jan 19 at 17:31
soodehMehboodi
asked Jan 18 at 16:37
soodehMehboodisoodehMehboodi
65848
asked Jan 18 at 16:37
soodehMehboodisoodehMehboodi
65848
asked Jan 18 at 16:37
soodehMehboodisoodehMehboodi
65848
65848
closed as off-topic by RRL, A. Pongrácz, Cesareo, Adrian Keister, metamorphy Jan 19 at 15:17
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, A. Pongrácz, Cesareo, Adrian Keister, metamorphy
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by RRL, A. Pongrácz, Cesareo, Adrian Keister, metamorphy Jan 19 at 15:17
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, A. Pongrácz, Cesareo, Adrian Keister, metamorphy
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
2 Answers
2
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oldest
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Excellent question! Turns out the answer to (1) is negative. See this post. However, if $f$ does have a limit at $infty$, it turns out it is actually $0$, as you suspected. This is an interesting exercise, try solving it :)
answered Jan 18 at 16:40
Yuval GatYuval Gat
1,0441213
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Are you sure about your sentence "However, if $f$ does have a limit at $infty$, it turns out it is actually $0$". If yes, would you prove that please.
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– soodehMehboodi
Jan 18 at 18:33
1
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Yes. Here's a sketch of the proof: if the limit is $+infty$, $f$ is greater than $1$ from some point $M$. But the integral of $1$ from $M$ to $infty$ diverges, so by comparison test so does the integral of $f$, contradiction. If the limit is $-infty$, examine $-f$ and you'll get a contradiction from what we proved earlier. I'll leave the finite case to you, as it is very similar.
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– Yuval Gat
Jan 18 at 18:42
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@YuvalGat: The limit being $pm infty$ is far from the only option. Did you mean to say that if the limit is $A>0$, then $f(x)>A/2$ from some point $M$?
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– Hans Lundmark
Jan 18 at 18:51
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@HansLundmark Of course the infinities aren’t the only options! I left the finite case as an exercise, as I stated in my original comment. Perhaps you missed that last sentence there :)
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– Yuval Gat
Jan 18 at 18:54
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@YuvalGat: Indeed I did! Sorry if I spoiled the exercise... ;-)
$endgroup$
– Hans Lundmark
Jan 19 at 9:19
add a comment |
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There are simple counterexamples to your first question. For instance, the indicator function $f = 1_{mathbb Z}$ does not have a limit at infinity, but is integrable (Riemann and Lesbegue) with $int_1^infty f, dx = 0$.
The case when $f$ is continuous does not hold either, but here the counterexamples are technical. See the answers to this question.
answered Jan 18 at 17:39
o.h.o.h.
6917
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Excellent question! Turns out the answer to (1) is negative. See this post. However, if $f$ does have a limit at $infty$, it turns out it is actually $0$, as you suspected. This is an interesting exercise, try solving it :)
answered Jan 18 at 16:40
Yuval GatYuval Gat
1,0441213
$endgroup$
$begingroup$
Are you sure about your sentence "However, if $f$ does have a limit at $infty$, it turns out it is actually $0$". If yes, would you prove that please.
$endgroup$
– soodehMehboodi
Jan 18 at 18:33
1
$begingroup$
Yes. Here's a sketch of the proof: if the limit is $+infty$, $f$ is greater than $1$ from some point $M$. But the integral of $1$ from $M$ to $infty$ diverges, so by comparison test so does the integral of $f$, contradiction. If the limit is $-infty$, examine $-f$ and you'll get a contradiction from what we proved earlier. I'll leave the finite case to you, as it is very similar.
$endgroup$
– Yuval Gat
Jan 18 at 18:42
$begingroup$
@YuvalGat: The limit being $pm infty$ is far from the only option. Did you mean to say that if the limit is $A>0$, then $f(x)>A/2$ from some point $M$?
$endgroup$
– Hans Lundmark
Jan 18 at 18:51
$begingroup$
@HansLundmark Of course the infinities aren’t the only options! I left the finite case as an exercise, as I stated in my original comment. Perhaps you missed that last sentence there :)
$endgroup$
– Yuval Gat
Jan 18 at 18:54
$begingroup$
@YuvalGat: Indeed I did! Sorry if I spoiled the exercise... ;-)
$endgroup$
– Hans Lundmark
Jan 19 at 9:19
add a comment |
$begingroup$
Excellent question! Turns out the answer to (1) is negative. See this post. However, if $f$ does have a limit at $infty$, it turns out it is actually $0$, as you suspected. This is an interesting exercise, try solving it :)
answered Jan 18 at 16:40
Yuval GatYuval Gat
1,0441213
$endgroup$
$begingroup$
Are you sure about your sentence "However, if $f$ does have a limit at $infty$, it turns out it is actually $0$". If yes, would you prove that please.
$endgroup$
– soodehMehboodi
Jan 18 at 18:33
1
$begingroup$
Yes. Here's a sketch of the proof: if the limit is $+infty$, $f$ is greater than $1$ from some point $M$. But the integral of $1$ from $M$ to $infty$ diverges, so by comparison test so does the integral of $f$, contradiction. If the limit is $-infty$, examine $-f$ and you'll get a contradiction from what we proved earlier. I'll leave the finite case to you, as it is very similar.
$endgroup$
– Yuval Gat
Jan 18 at 18:42
$begingroup$
@YuvalGat: The limit being $pm infty$ is far from the only option. Did you mean to say that if the limit is $A>0$, then $f(x)>A/2$ from some point $M$?
$endgroup$
– Hans Lundmark
Jan 18 at 18:51
$begingroup$
@HansLundmark Of course the infinities aren’t the only options! I left the finite case as an exercise, as I stated in my original comment. Perhaps you missed that last sentence there :)
$endgroup$
– Yuval Gat
Jan 18 at 18:54
$begingroup$
@YuvalGat: Indeed I did! Sorry if I spoiled the exercise... ;-)
$endgroup$
– Hans Lundmark
Jan 19 at 9:19
add a comment |
$begingroup$
Excellent question! Turns out the answer to (1) is negative. See this post. However, if $f$ does have a limit at $infty$, it turns out it is actually $0$, as you suspected. This is an interesting exercise, try solving it :)
answered Jan 18 at 16:40
Yuval GatYuval Gat
1,0441213
$endgroup$
Excellent question! Turns out the answer to (1) is negative. See this post. However, if $f$ does have a limit at $infty$, it turns out it is actually $0$, as you suspected. This is an interesting exercise, try solving it :)
answered Jan 18 at 16:40
Yuval GatYuval Gat
1,0441213
answered Jan 18 at 16:40
Yuval GatYuval Gat
1,0441213
answered Jan 18 at 16:40
Yuval GatYuval Gat
1,0441213
answered Jan 18 at 16:40
Yuval GatYuval Gat
1,0441213
1,0441213
$begingroup$
Are you sure about your sentence "However, if $f$ does have a limit at $infty$, it turns out it is actually $0$". If yes, would you prove that please.
$endgroup$
– soodehMehboodi
Jan 18 at 18:33
1
$begingroup$
Yes. Here's a sketch of the proof: if the limit is $+infty$, $f$ is greater than $1$ from some point $M$. But the integral of $1$ from $M$ to $infty$ diverges, so by comparison test so does the integral of $f$, contradiction. If the limit is $-infty$, examine $-f$ and you'll get a contradiction from what we proved earlier. I'll leave the finite case to you, as it is very similar.
$endgroup$
– Yuval Gat
Jan 18 at 18:42
$begingroup$
@YuvalGat: The limit being $pm infty$ is far from the only option. Did you mean to say that if the limit is $A>0$, then $f(x)>A/2$ from some point $M$?
$endgroup$
– Hans Lundmark
Jan 18 at 18:51
$begingroup$
@HansLundmark Of course the infinities aren’t the only options! I left the finite case as an exercise, as I stated in my original comment. Perhaps you missed that last sentence there :)
$endgroup$
– Yuval Gat
Jan 18 at 18:54
$begingroup$
@YuvalGat: Indeed I did! Sorry if I spoiled the exercise... ;-)
$endgroup$
– Hans Lundmark
Jan 19 at 9:19
add a comment |
$begingroup$
Are you sure about your sentence "However, if $f$ does have a limit at $infty$, it turns out it is actually $0$". If yes, would you prove that please.
$endgroup$
– soodehMehboodi
Jan 18 at 18:33
1
$begingroup$
Yes. Here's a sketch of the proof: if the limit is $+infty$, $f$ is greater than $1$ from some point $M$. But the integral of $1$ from $M$ to $infty$ diverges, so by comparison test so does the integral of $f$, contradiction. If the limit is $-infty$, examine $-f$ and you'll get a contradiction from what we proved earlier. I'll leave the finite case to you, as it is very similar.
$endgroup$
– Yuval Gat
Jan 18 at 18:42
$begingroup$
@YuvalGat: The limit being $pm infty$ is far from the only option. Did you mean to say that if the limit is $A>0$, then $f(x)>A/2$ from some point $M$?
$endgroup$
– Hans Lundmark
Jan 18 at 18:51
$begingroup$
@HansLundmark Of course the infinities aren’t the only options! I left the finite case as an exercise, as I stated in my original comment. Perhaps you missed that last sentence there :)
$endgroup$
– Yuval Gat
Jan 18 at 18:54
$begingroup$
@YuvalGat: Indeed I did! Sorry if I spoiled the exercise... ;-)
$endgroup$
– Hans Lundmark
Jan 19 at 9:19
$begingroup$
Are you sure about your sentence "However, if $f$ does have a limit at $infty$, it turns out it is actually $0$". If yes, would you prove that please.
$endgroup$
– soodehMehboodi
Jan 18 at 18:33
$begingroup$
Are you sure about your sentence "However, if $f$ does have a limit at $infty$, it turns out it is actually $0$". If yes, would you prove that please.
$endgroup$
– soodehMehboodi
Jan 18 at 18:33
1
1
$begingroup$
Yes. Here's a sketch of the proof: if the limit is $+infty$, $f$ is greater than $1$ from some point $M$. But the integral of $1$ from $M$ to $infty$ diverges, so by comparison test so does the integral of $f$, contradiction. If the limit is $-infty$, examine $-f$ and you'll get a contradiction from what we proved earlier. I'll leave the finite case to you, as it is very similar.
$endgroup$
– Yuval Gat
Jan 18 at 18:42
$begingroup$
Yes. Here's a sketch of the proof: if the limit is $+infty$, $f$ is greater than $1$ from some point $M$. But the integral of $1$ from $M$ to $infty$ diverges, so by comparison test so does the integral of $f$, contradiction. If the limit is $-infty$, examine $-f$ and you'll get a contradiction from what we proved earlier. I'll leave the finite case to you, as it is very similar.
$endgroup$
– Yuval Gat
Jan 18 at 18:42
$begingroup$
@YuvalGat: The limit being $pm infty$ is far from the only option. Did you mean to say that if the limit is $A>0$, then $f(x)>A/2$ from some point $M$?
$endgroup$
– Hans Lundmark
Jan 18 at 18:51
$begingroup$
@YuvalGat: The limit being $pm infty$ is far from the only option. Did you mean to say that if the limit is $A>0$, then $f(x)>A/2$ from some point $M$?
$endgroup$
– Hans Lundmark
Jan 18 at 18:51
$begingroup$
@HansLundmark Of course the infinities aren’t the only options! I left the finite case as an exercise, as I stated in my original comment. Perhaps you missed that last sentence there :)
$endgroup$
– Yuval Gat
Jan 18 at 18:54
$begingroup$
@HansLundmark Of course the infinities aren’t the only options! I left the finite case as an exercise, as I stated in my original comment. Perhaps you missed that last sentence there :)
$endgroup$
– Yuval Gat
Jan 18 at 18:54
$begingroup$
@YuvalGat: Indeed I did! Sorry if I spoiled the exercise... ;-)
$endgroup$
– Hans Lundmark
Jan 19 at 9:19
$begingroup$
@YuvalGat: Indeed I did! Sorry if I spoiled the exercise... ;-)
$endgroup$
– Hans Lundmark
Jan 19 at 9:19
add a comment |
$begingroup$
There are simple counterexamples to your first question. For instance, the indicator function $f = 1_{mathbb Z}$ does not have a limit at infinity, but is integrable (Riemann and Lesbegue) with $int_1^infty f, dx = 0$.
The case when $f$ is continuous does not hold either, but here the counterexamples are technical. See the answers to this question.
answered Jan 18 at 17:39
o.h.o.h.
6917
$endgroup$
add a comment |
$begingroup$
There are simple counterexamples to your first question. For instance, the indicator function $f = 1_{mathbb Z}$ does not have a limit at infinity, but is integrable (Riemann and Lesbegue) with $int_1^infty f, dx = 0$.
The case when $f$ is continuous does not hold either, but here the counterexamples are technical. See the answers to this question.
answered Jan 18 at 17:39
o.h.o.h.
6917
$endgroup$
add a comment |
$begingroup$
There are simple counterexamples to your first question. For instance, the indicator function $f = 1_{mathbb Z}$ does not have a limit at infinity, but is integrable (Riemann and Lesbegue) with $int_1^infty f, dx = 0$.
The case when $f$ is continuous does not hold either, but here the counterexamples are technical. See the answers to this question.
answered Jan 18 at 17:39
o.h.o.h.
6917
$endgroup$
There are simple counterexamples to your first question. For instance, the indicator function $f = 1_{mathbb Z}$ does not have a limit at infinity, but is integrable (Riemann and Lesbegue) with $int_1^infty f, dx = 0$.
The case when $f$ is continuous does not hold either, but here the counterexamples are technical. See the answers to this question.
answered Jan 18 at 17:39
o.h.o.h.
6917
answered Jan 18 at 17:39
o.h.o.h.
6917
answered Jan 18 at 17:39
o.h.o.h.
6917
answered Jan 18 at 17:39
o.h.o.h.
6917
6917
add a comment |
add a comment |
