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If the limit of a function exists as $x$ tends do infinity, must its derivative also exist? [duplicate]
This question already has an answer here:
If a function has a finite limit at infinity, does that imply its derivative goes to zero?
6 answers
I thought of a scenario where: if $ lim_{x-> infty}f(x)$ exists, which I can only imagine is a constant or asymptotic function, then it also seems like the derivative f' must also exist, but I don't know how to prove this either way. I guess if there is just a counterexample that would be easier than proving it. Is there an asymptotic function that somehow has a divergent derivative?
calculus proof-writing
edited Nov 21 '18 at 3:12
Key Flex
7,52441232
asked Nov 21 '18 at 2:56
user608672
64
marked as duplicate by David K, Jean-Claude Arbaut, Tom-Tom, Brahadeesh, José Carlos Santos calculus
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Nov 21 '18 at 10:02
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 |
This question already has an answer here:
If a function has a finite limit at infinity, does that imply its derivative goes to zero?
6 answers
I thought of a scenario where: if $ lim_{x-> infty}f(x)$ exists, which I can only imagine is a constant or asymptotic function, then it also seems like the derivative f' must also exist, but I don't know how to prove this either way. I guess if there is just a counterexample that would be easier than proving it. Is there an asymptotic function that somehow has a divergent derivative?
calculus proof-writing
edited Nov 21 '18 at 3:12
Key Flex
7,52441232
asked Nov 21 '18 at 2:56
user608672
64
marked as duplicate by David K, Jean-Claude Arbaut, Tom-Tom, Brahadeesh, José Carlos Santos calculus
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Nov 21 '18 at 10:02
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.
Are you asking if its derivative exists everywhere?
– Taliant
Nov 21 '18 at 3:00
No, at infinity.
– user608672
Nov 21 '18 at 3:01
Do you mean $lim_{xtoinfty} f'(x)$?
– Taliant
Nov 21 '18 at 3:02
I additionally mean that, yes, but also simultaneously the first case.
– user608672
Nov 21 '18 at 3:03
@JohnNash Not sure about the duplicate. In this question, $f$ is not said to be differentiable. the question is about differentiability itself (whatever the OP means by "differentiable at infinity"), not the value of the derivative.
– Tom-Tom
Nov 21 '18 at 8:49
add a comment |
This question already has an answer here:
If a function has a finite limit at infinity, does that imply its derivative goes to zero?
6 answers
I thought of a scenario where: if $ lim_{x-> infty}f(x)$ exists, which I can only imagine is a constant or asymptotic function, then it also seems like the derivative f' must also exist, but I don't know how to prove this either way. I guess if there is just a counterexample that would be easier than proving it. Is there an asymptotic function that somehow has a divergent derivative?
calculus proof-writing
edited Nov 21 '18 at 3:12
Key Flex
7,52441232
asked Nov 21 '18 at 2:56
user608672
64
This question already has an answer here:
If a function has a finite limit at infinity, does that imply its derivative goes to zero?
6 answers
I thought of a scenario where: if $ lim_{x-> infty}f(x)$ exists, which I can only imagine is a constant or asymptotic function, then it also seems like the derivative f' must also exist, but I don't know how to prove this either way. I guess if there is just a counterexample that would be easier than proving it. Is there an asymptotic function that somehow has a divergent derivative?
This question already has an answer here:
If a function has a finite limit at infinity, does that imply its derivative goes to zero?
6 answers
calculus proof-writing
calculus proof-writing
edited Nov 21 '18 at 3:12
Key Flex
7,52441232
asked Nov 21 '18 at 2:56
user608672
64
edited Nov 21 '18 at 3:12
Key Flex
7,52441232
asked Nov 21 '18 at 2:56
user608672
64
edited Nov 21 '18 at 3:12
Key Flex
7,52441232
edited Nov 21 '18 at 3:12
Key Flex
7,52441232
edited Nov 21 '18 at 3:12
Key Flex
7,52441232
7,52441232
asked Nov 21 '18 at 2:56
user608672
64
asked Nov 21 '18 at 2:56
user608672
64
asked Nov 21 '18 at 2:56
user608672
64
64
marked as duplicate by David K, Jean-Claude Arbaut, Tom-Tom, Brahadeesh, José Carlos Santos calculus
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Nov 21 '18 at 10:02
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 David K, Jean-Claude Arbaut, Tom-Tom, Brahadeesh, José Carlos Santos calculus
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Nov 21 '18 at 10:02
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.
Are you asking if its derivative exists everywhere?
– Taliant
Nov 21 '18 at 3:00
No, at infinity.
– user608672
Nov 21 '18 at 3:01
Do you mean $lim_{xtoinfty} f'(x)$?
– Taliant
Nov 21 '18 at 3:02
I additionally mean that, yes, but also simultaneously the first case.
– user608672
Nov 21 '18 at 3:03
@JohnNash Not sure about the duplicate. In this question, $f$ is not said to be differentiable. the question is about differentiability itself (whatever the OP means by "differentiable at infinity"), not the value of the derivative.
– Tom-Tom
Nov 21 '18 at 8:49
add a comment |
Are you asking if its derivative exists everywhere?
– Taliant
Nov 21 '18 at 3:00
No, at infinity.
– user608672
Nov 21 '18 at 3:01
Do you mean $lim_{xtoinfty} f'(x)$?
– Taliant
Nov 21 '18 at 3:02
I additionally mean that, yes, but also simultaneously the first case.
– user608672
Nov 21 '18 at 3:03
@JohnNash Not sure about the duplicate. In this question, $f$ is not said to be differentiable. the question is about differentiability itself (whatever the OP means by "differentiable at infinity"), not the value of the derivative.
– Tom-Tom
Nov 21 '18 at 8:49
Are you asking if its derivative exists everywhere?
– Taliant
Nov 21 '18 at 3:00
Are you asking if its derivative exists everywhere?
– Taliant
Nov 21 '18 at 3:00
No, at infinity.
– user608672
Nov 21 '18 at 3:01
No, at infinity.
– user608672
Nov 21 '18 at 3:01
Do you mean $lim_{xtoinfty} f'(x)$?
– Taliant
Nov 21 '18 at 3:02
Do you mean $lim_{xtoinfty} f'(x)$?
– Taliant
Nov 21 '18 at 3:02
I additionally mean that, yes, but also simultaneously the first case.
– user608672
Nov 21 '18 at 3:03
I additionally mean that, yes, but also simultaneously the first case.
– user608672
Nov 21 '18 at 3:03
@JohnNash Not sure about the duplicate. In this question, $f$ is not said to be differentiable. the question is about differentiability itself (whatever the OP means by "differentiable at infinity"), not the value of the derivative.
– Tom-Tom
Nov 21 '18 at 8:49
@JohnNash Not sure about the duplicate. In this question, $f$ is not said to be differentiable. the question is about differentiability itself (whatever the OP means by "differentiable at infinity"), not the value of the derivative.
– Tom-Tom
Nov 21 '18 at 8:49
add a comment |
2 Answers
2
active
oldest
votes
No, take for example $f(x)=frac{sin{x^2}}{x}$. Then $lim_{xtoinfty} f(x) = 0$, but $f'(x) = 2 cos{x^2} - frac{sin{x^2}}{x^2}$ and so $lim_{xtoinfty} f'(x)$ does not exist.
answered Nov 21 '18 at 3:06
ignorantFid
21916
add a comment |
No. Consider $f(x)= frac{sin{x^2}}{x}$ and x>0. Then you see it is not differentiable. You can see below If a function has a finite limit at infinity, does that imply its derivative goes to zero?
answered Nov 21 '18 at 3:06
John Nash
7418
The function you chose is a correct example, but I'm puzzled as to why you say it isn't differentiable.
– Matt Samuel
Nov 21 '18 at 4:27
The derivative f'(x) oscillates roughly between 2x and -2x. so limsupf'(x) tends to +∞ as x tends to ∞ and liminff'(x) tends to -∞ as x tends to ∞
– John Nash
Nov 21 '18 at 4:40
Yes, but it's still differentiable.
– Matt Samuel
Nov 21 '18 at 4:41
limsupf'(x)=∞ and liminf'(x)=-∞ as x tens to infinity. both are different one is +ve and second is -ve.
– John Nash
Nov 21 '18 at 4:44
Right, the limit doesn't exist. The limit doesn't exist for $f(x) =x^2$ either, but I'd hope you'd agree that that function is differentiable.
– Matt Samuel
Nov 21 '18 at 4:45
|
show 4 more comments
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
No, take for example $f(x)=frac{sin{x^2}}{x}$. Then $lim_{xtoinfty} f(x) = 0$, but $f'(x) = 2 cos{x^2} - frac{sin{x^2}}{x^2}$ and so $lim_{xtoinfty} f'(x)$ does not exist.
answered Nov 21 '18 at 3:06
ignorantFid
21916
add a comment |
No, take for example $f(x)=frac{sin{x^2}}{x}$. Then $lim_{xtoinfty} f(x) = 0$, but $f'(x) = 2 cos{x^2} - frac{sin{x^2}}{x^2}$ and so $lim_{xtoinfty} f'(x)$ does not exist.
answered Nov 21 '18 at 3:06
ignorantFid
21916
add a comment |
No, take for example $f(x)=frac{sin{x^2}}{x}$. Then $lim_{xtoinfty} f(x) = 0$, but $f'(x) = 2 cos{x^2} - frac{sin{x^2}}{x^2}$ and so $lim_{xtoinfty} f'(x)$ does not exist.
answered Nov 21 '18 at 3:06
ignorantFid
21916
No, take for example $f(x)=frac{sin{x^2}}{x}$. Then $lim_{xtoinfty} f(x) = 0$, but $f'(x) = 2 cos{x^2} - frac{sin{x^2}}{x^2}$ and so $lim_{xtoinfty} f'(x)$ does not exist.
answered Nov 21 '18 at 3:06
ignorantFid
21916
answered Nov 21 '18 at 3:06
ignorantFid
21916
answered Nov 21 '18 at 3:06
ignorantFid
21916
answered Nov 21 '18 at 3:06
ignorantFid
21916
21916
add a comment |
add a comment |
No. Consider $f(x)= frac{sin{x^2}}{x}$ and x>0. Then you see it is not differentiable. You can see below If a function has a finite limit at infinity, does that imply its derivative goes to zero?
answered Nov 21 '18 at 3:06
John Nash
7418
The function you chose is a correct example, but I'm puzzled as to why you say it isn't differentiable.
– Matt Samuel
Nov 21 '18 at 4:27
The derivative f'(x) oscillates roughly between 2x and -2x. so limsupf'(x) tends to +∞ as x tends to ∞ and liminff'(x) tends to -∞ as x tends to ∞
– John Nash
Nov 21 '18 at 4:40
Yes, but it's still differentiable.
– Matt Samuel
Nov 21 '18 at 4:41
limsupf'(x)=∞ and liminf'(x)=-∞ as x tens to infinity. both are different one is +ve and second is -ve.
– John Nash
Nov 21 '18 at 4:44
Right, the limit doesn't exist. The limit doesn't exist for $f(x) =x^2$ either, but I'd hope you'd agree that that function is differentiable.
– Matt Samuel
Nov 21 '18 at 4:45
|
show 4 more comments
No. Consider $f(x)= frac{sin{x^2}}{x}$ and x>0. Then you see it is not differentiable. You can see below If a function has a finite limit at infinity, does that imply its derivative goes to zero?
answered Nov 21 '18 at 3:06
John Nash
7418
The function you chose is a correct example, but I'm puzzled as to why you say it isn't differentiable.
– Matt Samuel
Nov 21 '18 at 4:27
The derivative f'(x) oscillates roughly between 2x and -2x. so limsupf'(x) tends to +∞ as x tends to ∞ and liminff'(x) tends to -∞ as x tends to ∞
– John Nash
Nov 21 '18 at 4:40
Yes, but it's still differentiable.
– Matt Samuel
Nov 21 '18 at 4:41
limsupf'(x)=∞ and liminf'(x)=-∞ as x tens to infinity. both are different one is +ve and second is -ve.
– John Nash
Nov 21 '18 at 4:44
Right, the limit doesn't exist. The limit doesn't exist for $f(x) =x^2$ either, but I'd hope you'd agree that that function is differentiable.
– Matt Samuel
Nov 21 '18 at 4:45
|
show 4 more comments
No. Consider $f(x)= frac{sin{x^2}}{x}$ and x>0. Then you see it is not differentiable. You can see below If a function has a finite limit at infinity, does that imply its derivative goes to zero?
answered Nov 21 '18 at 3:06
John Nash
7418
No. Consider $f(x)= frac{sin{x^2}}{x}$ and x>0. Then you see it is not differentiable. You can see below If a function has a finite limit at infinity, does that imply its derivative goes to zero?
answered Nov 21 '18 at 3:06
John Nash
7418
answered Nov 21 '18 at 3:06
John Nash
7418
answered Nov 21 '18 at 3:06
John Nash
7418
answered Nov 21 '18 at 3:06
John Nash
7418
7418
The function you chose is a correct example, but I'm puzzled as to why you say it isn't differentiable.
– Matt Samuel
Nov 21 '18 at 4:27
The derivative f'(x) oscillates roughly between 2x and -2x. so limsupf'(x) tends to +∞ as x tends to ∞ and liminff'(x) tends to -∞ as x tends to ∞
– John Nash
Nov 21 '18 at 4:40
Yes, but it's still differentiable.
– Matt Samuel
Nov 21 '18 at 4:41
limsupf'(x)=∞ and liminf'(x)=-∞ as x tens to infinity. both are different one is +ve and second is -ve.
– John Nash
Nov 21 '18 at 4:44
Right, the limit doesn't exist. The limit doesn't exist for $f(x) =x^2$ either, but I'd hope you'd agree that that function is differentiable.
– Matt Samuel
Nov 21 '18 at 4:45
|
show 4 more comments
The function you chose is a correct example, but I'm puzzled as to why you say it isn't differentiable.
– Matt Samuel
Nov 21 '18 at 4:27
The derivative f'(x) oscillates roughly between 2x and -2x. so limsupf'(x) tends to +∞ as x tends to ∞ and liminff'(x) tends to -∞ as x tends to ∞
– John Nash
Nov 21 '18 at 4:40
Yes, but it's still differentiable.
– Matt Samuel
Nov 21 '18 at 4:41
limsupf'(x)=∞ and liminf'(x)=-∞ as x tens to infinity. both are different one is +ve and second is -ve.
– John Nash
Nov 21 '18 at 4:44
Right, the limit doesn't exist. The limit doesn't exist for $f(x) =x^2$ either, but I'd hope you'd agree that that function is differentiable.
– Matt Samuel
Nov 21 '18 at 4:45
The function you chose is a correct example, but I'm puzzled as to why you say it isn't differentiable.
– Matt Samuel
Nov 21 '18 at 4:27
The function you chose is a correct example, but I'm puzzled as to why you say it isn't differentiable.
– Matt Samuel
Nov 21 '18 at 4:27
The derivative f'(x) oscillates roughly between 2x and -2x. so limsupf'(x) tends to +∞ as x tends to ∞ and liminff'(x) tends to -∞ as x tends to ∞
– John Nash
Nov 21 '18 at 4:40
The derivative f'(x) oscillates roughly between 2x and -2x. so limsupf'(x) tends to +∞ as x tends to ∞ and liminff'(x) tends to -∞ as x tends to ∞
– John Nash
Nov 21 '18 at 4:40
Yes, but it's still differentiable.
– Matt Samuel
Nov 21 '18 at 4:41
Yes, but it's still differentiable.
– Matt Samuel
Nov 21 '18 at 4:41
limsupf'(x)=∞ and liminf'(x)=-∞ as x tens to infinity. both are different one is +ve and second is -ve.
– John Nash
Nov 21 '18 at 4:44
limsupf'(x)=∞ and liminf'(x)=-∞ as x tens to infinity. both are different one is +ve and second is -ve.
– John Nash
Nov 21 '18 at 4:44
Right, the limit doesn't exist. The limit doesn't exist for $f(x) =x^2$ either, but I'd hope you'd agree that that function is differentiable.
– Matt Samuel
Nov 21 '18 at 4:45
Right, the limit doesn't exist. The limit doesn't exist for $f(x) =x^2$ either, but I'd hope you'd agree that that function is differentiable.
– Matt Samuel
Nov 21 '18 at 4:45
|
show 4 more comments
Are you asking if its derivative exists everywhere?
– Taliant
Nov 21 '18 at 3:00
No, at infinity.
– user608672
Nov 21 '18 at 3:01
Do you mean $lim_{xtoinfty} f'(x)$?
– Taliant
Nov 21 '18 at 3:02
I additionally mean that, yes, but also simultaneously the first case.
– user608672
Nov 21 '18 at 3:03
@JohnNash Not sure about the duplicate. In this question, $f$ is not said to be differentiable. the question is about differentiability itself (whatever the OP means by "differentiable at infinity"), not the value of the derivative.
– Tom-Tom
Nov 21 '18 at 8:49