Differentiable functions and existence of limitsReal Analysis - limits and differentiationModifications of Weierstrass's continuous, nowhere differentiable functionsExamples of differentiable functions that are not of bounded variationFor a function which is everywhere right-differentiable, what can be said about the existence of points where it is differentiable?Limits, derivatives and oscillationsPointwise limits of differentiable functions under constraintFunctions Which are non differentiable on a Given Set.Nowhere differentiable continuous functions and local extrema$f(x,y) = frac(xy^3)(x^2 + y^4)$ except at $(0,0)$ where it is equal to 0, show it is continuous, is it differentiable at origin?bounded differentiable functions
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Differentiable functions and existence of limits
Real Analysis - limits and differentiationModifications of Weierstrass's continuous, nowhere differentiable functionsExamples of differentiable functions that are not of bounded variationFor a function which is everywhere right-differentiable, what can be said about the existence of points where it is differentiable?Limits, derivatives and oscillationsPointwise limits of differentiable functions under constraintFunctions Which are non differentiable on a Given Set.Nowhere differentiable continuous functions and local extrema$f(x,y) = frac(xy^3)(x^2 + y^4)$ except at $(0,0)$ where it is equal to 0, show it is continuous, is it differentiable at origin?bounded differentiable functions
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$begingroup$
If a function is differentiable everywhere, does it imply that the limit at $pm infty$ is either finite or it diverges to $pm infty$?
real-analysis limits derivatives
asked 9 hours ago
Lucas PereiroLucas Pereiro
475 bronze badges
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add a comment |
$begingroup$
If a function is differentiable everywhere, does it imply that the limit at $pm infty$ is either finite or it diverges to $pm infty$?
real-analysis limits derivatives
asked 9 hours ago
Lucas PereiroLucas Pereiro
475 bronze badges
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$begingroup$
The function $x mapsto 0$ is differentiable everywhere.
$endgroup$
– copper.hat
8 hours ago
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@copper.hat: The limits at $pminfty$ of $xmapsto0$ (both exist and) are both finite; they are both $0.$
$endgroup$
– Will R
23 mins ago
$begingroup$
Yes, I do know what I was thinking.
$endgroup$
– copper.hat
45 secs ago
$begingroup$
Do not I mean...
$endgroup$
– copper.hat
18 secs ago
add a comment |
$begingroup$
If a function is differentiable everywhere, does it imply that the limit at $pm infty$ is either finite or it diverges to $pm infty$?
real-analysis limits derivatives
asked 9 hours ago
Lucas PereiroLucas Pereiro
475 bronze badges
$endgroup$
If a function is differentiable everywhere, does it imply that the limit at $pm infty$ is either finite or it diverges to $pm infty$?
real-analysis limits derivatives
real-analysis limits derivatives
asked 9 hours ago
Lucas PereiroLucas Pereiro
475 bronze badges
asked 9 hours ago
Lucas PereiroLucas Pereiro
475 bronze badges
asked 9 hours ago
Lucas PereiroLucas Pereiro
475 bronze badges
asked 9 hours ago
Lucas PereiroLucas Pereiro
475 bronze badges
asked 9 hours ago
Lucas PereiroLucas Pereiro
475 bronze badges
475 bronze badges
$begingroup$
The function $x mapsto 0$ is differentiable everywhere.
$endgroup$
– copper.hat
8 hours ago
$begingroup$
@copper.hat: The limits at $pminfty$ of $xmapsto0$ (both exist and) are both finite; they are both $0.$
$endgroup$
– Will R
23 mins ago
$begingroup$
Yes, I do know what I was thinking.
$endgroup$
– copper.hat
45 secs ago
$begingroup$
Do not I mean...
$endgroup$
– copper.hat
18 secs ago
add a comment |
$begingroup$
The function $x mapsto 0$ is differentiable everywhere.
$endgroup$
– copper.hat
8 hours ago
$begingroup$
@copper.hat: The limits at $pminfty$ of $xmapsto0$ (both exist and) are both finite; they are both $0.$
$endgroup$
– Will R
23 mins ago
$begingroup$
Yes, I do know what I was thinking.
$endgroup$
– copper.hat
45 secs ago
$begingroup$
Do not I mean...
$endgroup$
– copper.hat
18 secs ago
$begingroup$
The function $x mapsto 0$ is differentiable everywhere.
$endgroup$
– copper.hat
8 hours ago
$begingroup$
The function $x mapsto 0$ is differentiable everywhere.
$endgroup$
– copper.hat
8 hours ago
$begingroup$
@copper.hat: The limits at $pminfty$ of $xmapsto0$ (both exist and) are both finite; they are both $0.$
$endgroup$
– Will R
23 mins ago
$begingroup$
@copper.hat: The limits at $pminfty$ of $xmapsto0$ (both exist and) are both finite; they are both $0.$
$endgroup$
– Will R
23 mins ago
$begingroup$
Yes, I do know what I was thinking.
$endgroup$
– copper.hat
45 secs ago
$begingroup$
Yes, I do know what I was thinking.
$endgroup$
– copper.hat
45 secs ago
$begingroup$
Do not I mean...
$endgroup$
– copper.hat
18 secs ago
$begingroup$
Do not I mean...
$endgroup$
– copper.hat
18 secs ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
No, as the sine function shows. It has no limit at $pminfty$.
answered 9 hours ago
José Carlos SantosJosé Carlos Santos
201k25 gold badges159 silver badges278 bronze badges
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add a comment |
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No, it doesn't. For example, $sin$ and $cos$ are infinitely differentiable functions, but they have no limit at $pm infty$.
answered 9 hours ago
peek-a-boopeek-a-boo
4,2463 silver badges17 bronze badges
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add a comment |
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Neither. Counter-examples: $f(x)=e^-x^2,$ and $f(x)=x^2.$
answered 9 hours ago
Adrian KeisterAdrian Keister
6,3287 gold badges22 silver badges33 bronze badges
$endgroup$
1
$begingroup$
But $e^-x^2$ does have a finite limit at $pminfty$ and $x^2$ diverges to $infty$ at $pminfty$. Is this not what the OP was suggesting?
$endgroup$
– Jam
8 hours ago
$begingroup$
What I'm saying is that one of the functions is a counterexample to one side, and the other function is a counterexample to the other. $e^-x^2$ shows that a differentiable function need not have infinite limits at infinity. $x^2$ shows that a differential function need not have finite limits at infinity.
$endgroup$
– Adrian Keister
8 hours ago
1
$begingroup$
Oh, I see what you mean. I was under the impression that the OP was separately considering each limit at $pminfty$, as opposed to considering them together. I think it's not entirely clear from their question.
$endgroup$
– Jam
8 hours ago
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No, as the sine function shows. It has no limit at $pminfty$.
answered 9 hours ago
José Carlos SantosJosé Carlos Santos
201k25 gold badges159 silver badges278 bronze badges
$endgroup$
add a comment |
$begingroup$
No, as the sine function shows. It has no limit at $pminfty$.
answered 9 hours ago
José Carlos SantosJosé Carlos Santos
201k25 gold badges159 silver badges278 bronze badges
$endgroup$
add a comment |
$begingroup$
No, as the sine function shows. It has no limit at $pminfty$.
answered 9 hours ago
José Carlos SantosJosé Carlos Santos
201k25 gold badges159 silver badges278 bronze badges
$endgroup$
No, as the sine function shows. It has no limit at $pminfty$.
answered 9 hours ago
José Carlos SantosJosé Carlos Santos
201k25 gold badges159 silver badges278 bronze badges
answered 9 hours ago
José Carlos SantosJosé Carlos Santos
201k25 gold badges159 silver badges278 bronze badges
answered 9 hours ago
José Carlos SantosJosé Carlos Santos
201k25 gold badges159 silver badges278 bronze badges
answered 9 hours ago
José Carlos SantosJosé Carlos Santos
201k25 gold badges159 silver badges278 bronze badges
201k25 gold badges159 silver badges278 bronze badges
add a comment |
add a comment |
$begingroup$
No, it doesn't. For example, $sin$ and $cos$ are infinitely differentiable functions, but they have no limit at $pm infty$.
answered 9 hours ago
peek-a-boopeek-a-boo
4,2463 silver badges17 bronze badges
$endgroup$
add a comment |
$begingroup$
No, it doesn't. For example, $sin$ and $cos$ are infinitely differentiable functions, but they have no limit at $pm infty$.
answered 9 hours ago
peek-a-boopeek-a-boo
4,2463 silver badges17 bronze badges
$endgroup$
add a comment |
$begingroup$
No, it doesn't. For example, $sin$ and $cos$ are infinitely differentiable functions, but they have no limit at $pm infty$.
answered 9 hours ago
peek-a-boopeek-a-boo
4,2463 silver badges17 bronze badges
$endgroup$
No, it doesn't. For example, $sin$ and $cos$ are infinitely differentiable functions, but they have no limit at $pm infty$.
answered 9 hours ago
peek-a-boopeek-a-boo
4,2463 silver badges17 bronze badges
answered 9 hours ago
peek-a-boopeek-a-boo
4,2463 silver badges17 bronze badges
answered 9 hours ago
peek-a-boopeek-a-boo
4,2463 silver badges17 bronze badges
answered 9 hours ago
peek-a-boopeek-a-boo
4,2463 silver badges17 bronze badges
4,2463 silver badges17 bronze badges
add a comment |
add a comment |
$begingroup$
Neither. Counter-examples: $f(x)=e^-x^2,$ and $f(x)=x^2.$
answered 9 hours ago
Adrian KeisterAdrian Keister
6,3287 gold badges22 silver badges33 bronze badges
$endgroup$
1
$begingroup$
But $e^-x^2$ does have a finite limit at $pminfty$ and $x^2$ diverges to $infty$ at $pminfty$. Is this not what the OP was suggesting?
$endgroup$
– Jam
8 hours ago
$begingroup$
What I'm saying is that one of the functions is a counterexample to one side, and the other function is a counterexample to the other. $e^-x^2$ shows that a differentiable function need not have infinite limits at infinity. $x^2$ shows that a differential function need not have finite limits at infinity.
$endgroup$
– Adrian Keister
8 hours ago
1
$begingroup$
Oh, I see what you mean. I was under the impression that the OP was separately considering each limit at $pminfty$, as opposed to considering them together. I think it's not entirely clear from their question.
$endgroup$
– Jam
8 hours ago
add a comment |
$begingroup$
Neither. Counter-examples: $f(x)=e^-x^2,$ and $f(x)=x^2.$
answered 9 hours ago
Adrian KeisterAdrian Keister
6,3287 gold badges22 silver badges33 bronze badges
$endgroup$
1
$begingroup$
But $e^-x^2$ does have a finite limit at $pminfty$ and $x^2$ diverges to $infty$ at $pminfty$. Is this not what the OP was suggesting?
$endgroup$
– Jam
8 hours ago
$begingroup$
What I'm saying is that one of the functions is a counterexample to one side, and the other function is a counterexample to the other. $e^-x^2$ shows that a differentiable function need not have infinite limits at infinity. $x^2$ shows that a differential function need not have finite limits at infinity.
$endgroup$
– Adrian Keister
8 hours ago
1
$begingroup$
Oh, I see what you mean. I was under the impression that the OP was separately considering each limit at $pminfty$, as opposed to considering them together. I think it's not entirely clear from their question.
$endgroup$
– Jam
8 hours ago
add a comment |
$begingroup$
Neither. Counter-examples: $f(x)=e^-x^2,$ and $f(x)=x^2.$
answered 9 hours ago
Adrian KeisterAdrian Keister
6,3287 gold badges22 silver badges33 bronze badges
$endgroup$
Neither. Counter-examples: $f(x)=e^-x^2,$ and $f(x)=x^2.$
answered 9 hours ago
Adrian KeisterAdrian Keister
6,3287 gold badges22 silver badges33 bronze badges
answered 9 hours ago
Adrian KeisterAdrian Keister
6,3287 gold badges22 silver badges33 bronze badges
answered 9 hours ago
Adrian KeisterAdrian Keister
6,3287 gold badges22 silver badges33 bronze badges
answered 9 hours ago
Adrian KeisterAdrian Keister
6,3287 gold badges22 silver badges33 bronze badges
6,3287 gold badges22 silver badges33 bronze badges
1
$begingroup$
But $e^-x^2$ does have a finite limit at $pminfty$ and $x^2$ diverges to $infty$ at $pminfty$. Is this not what the OP was suggesting?
$endgroup$
– Jam
8 hours ago
$begingroup$
What I'm saying is that one of the functions is a counterexample to one side, and the other function is a counterexample to the other. $e^-x^2$ shows that a differentiable function need not have infinite limits at infinity. $x^2$ shows that a differential function need not have finite limits at infinity.
$endgroup$
– Adrian Keister
8 hours ago
1
$begingroup$
Oh, I see what you mean. I was under the impression that the OP was separately considering each limit at $pminfty$, as opposed to considering them together. I think it's not entirely clear from their question.
$endgroup$
– Jam
8 hours ago
add a comment |
1
$begingroup$
But $e^-x^2$ does have a finite limit at $pminfty$ and $x^2$ diverges to $infty$ at $pminfty$. Is this not what the OP was suggesting?
$endgroup$
– Jam
8 hours ago
$begingroup$
What I'm saying is that one of the functions is a counterexample to one side, and the other function is a counterexample to the other. $e^-x^2$ shows that a differentiable function need not have infinite limits at infinity. $x^2$ shows that a differential function need not have finite limits at infinity.
$endgroup$
– Adrian Keister
8 hours ago
1
$begingroup$
Oh, I see what you mean. I was under the impression that the OP was separately considering each limit at $pminfty$, as opposed to considering them together. I think it's not entirely clear from their question.
$endgroup$
– Jam
8 hours ago
1
1
$begingroup$
But $e^-x^2$ does have a finite limit at $pminfty$ and $x^2$ diverges to $infty$ at $pminfty$. Is this not what the OP was suggesting?
$endgroup$
– Jam
8 hours ago
$begingroup$
But $e^-x^2$ does have a finite limit at $pminfty$ and $x^2$ diverges to $infty$ at $pminfty$. Is this not what the OP was suggesting?
$endgroup$
– Jam
8 hours ago
$begingroup$
What I'm saying is that one of the functions is a counterexample to one side, and the other function is a counterexample to the other. $e^-x^2$ shows that a differentiable function need not have infinite limits at infinity. $x^2$ shows that a differential function need not have finite limits at infinity.
$endgroup$
– Adrian Keister
8 hours ago
$begingroup$
What I'm saying is that one of the functions is a counterexample to one side, and the other function is a counterexample to the other. $e^-x^2$ shows that a differentiable function need not have infinite limits at infinity. $x^2$ shows that a differential function need not have finite limits at infinity.
$endgroup$
– Adrian Keister
8 hours ago
1
1
$begingroup$
Oh, I see what you mean. I was under the impression that the OP was separately considering each limit at $pminfty$, as opposed to considering them together. I think it's not entirely clear from their question.
$endgroup$
– Jam
8 hours ago
$begingroup$
Oh, I see what you mean. I was under the impression that the OP was separately considering each limit at $pminfty$, as opposed to considering them together. I think it's not entirely clear from their question.
$endgroup$
– Jam
8 hours ago
add a comment |
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$begingroup$
The function $x mapsto 0$ is differentiable everywhere.
$endgroup$
– copper.hat
8 hours ago
$begingroup$
@copper.hat: The limits at $pminfty$ of $xmapsto0$ (both exist and) are both finite; they are both $0.$
$endgroup$
– Will R
23 mins ago
$begingroup$
Yes, I do know what I was thinking.
$endgroup$
– copper.hat
45 secs ago
$begingroup$
Do not I mean...
$endgroup$
– copper.hat
18 secs ago