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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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2

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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

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asked Jul 19 at 15:56

Lucas PereiroLucas Pereiro

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  The function $x mapsto 0$ is differentiable everywhere.
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  – copper.hat
  Jul 19 at 16:15
  

  
  
  
  


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  @copper.hat: The limits at $pminfty$ of $xmapsto0$ (both exist and) are both finite; they are both $0.$
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  – Will R
  Jul 20 at 0:42
  
  

  
  
  
  


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  Yes, I do know what I was thinking.
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  – copper.hat
  Jul 20 at 1:05
  

  
  
  
  


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  Do not I mean...
  $endgroup$
  – copper.hat
  Jul 20 at 1:06
  

  
  
  
  

add a comment | 

2

$begingroup$

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

share|cite|improve this question

asked Jul 19 at 15:56

Lucas PereiroLucas Pereiro

4955 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The function $x mapsto 0$ is differentiable everywhere.
  $endgroup$
  – copper.hat
  Jul 19 at 16:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @copper.hat: The limits at $pminfty$ of $xmapsto0$ (both exist and) are both finite; they are both $0.$
  $endgroup$
  – Will R
  Jul 20 at 0:42
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, I do know what I was thinking.
  $endgroup$
  – copper.hat
  Jul 20 at 1:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do not I mean...
  $endgroup$
  – copper.hat
  Jul 20 at 1:06
  

  
  
  
  

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

share|cite|improve this question

asked Jul 19 at 15:56

Lucas PereiroLucas Pereiro

4955 bronze badges

$endgroup$

share|cite|improve this question

asked Jul 19 at 15:56

Lucas PereiroLucas Pereiro

4955 bronze badges

share|cite|improve this question

asked Jul 19 at 15:56

Lucas PereiroLucas Pereiro

4955 bronze badges

asked Jul 19 at 15:56

Lucas PereiroLucas Pereiro

4955 bronze badges

asked Jul 19 at 15:56

Lucas PereiroLucas Pereiro

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3 Answers
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4

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No, as the sine function shows. It has no limit at $pminfty$.

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answered Jul 19 at 15:57

José Carlos SantosJosé Carlos Santos

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4

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No, it doesn't. For example, $sin$ and $cos$ are infinitely differentiable functions, but they have no limit at $pm infty$.

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answered Jul 19 at 15:58

peek-a-boopeek-a-boo

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Neither. Counter-examples: $f(x)=e^-x^2,$ and $f(x)=x^2.$

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answered Jul 19 at 15:58

Adrian KeisterAdrian Keister

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  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
  Jul 19 at 16:19
  

  
  
  
  


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  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
  Jul 19 at 16:25
  

  
  
  
  


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  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
  Jul 19 at 16:35
  

  
  
  
  

add a comment | 

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4

$begingroup$

No, as the sine function shows. It has no limit at $pminfty$.

share|cite|improve this answer

answered Jul 19 at 15:57

José Carlos SantosJosé Carlos Santos

204k2525 gold badges159159 silver badges280280 bronze badges

$endgroup$

add a comment | 

4

$begingroup$

No, as the sine function shows. It has no limit at $pminfty$.

share|cite|improve this answer

answered Jul 19 at 15:57

José Carlos SantosJosé Carlos Santos

204k2525 gold badges159159 silver badges280280 bronze badges

$endgroup$

add a comment | 

$begingroup$

No, as the sine function shows. It has no limit at $pminfty$.

share|cite|improve this answer

answered Jul 19 at 15:57

José Carlos SantosJosé Carlos Santos

204k2525 gold badges159159 silver badges280280 bronze badges

$endgroup$

share|cite|improve this answer

answered Jul 19 at 15:57

José Carlos SantosJosé Carlos Santos

204k2525 gold badges159159 silver badges280280 bronze badges

answered Jul 19 at 15:57

José Carlos SantosJosé Carlos Santos

204k2525 gold badges159159 silver badges280280 bronze badges

answered Jul 19 at 15:57

José Carlos SantosJosé Carlos Santos

204k2525 gold badges159159 silver badges280280 bronze badges

4

$begingroup$

No, it doesn't. For example, $sin$ and $cos$ are infinitely differentiable functions, but they have no limit at $pm infty$.

share|cite|improve this answer

answered Jul 19 at 15:58

peek-a-boopeek-a-boo

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add a comment | 

4

$begingroup$

No, it doesn't. For example, $sin$ and $cos$ are infinitely differentiable functions, but they have no limit at $pm infty$.

share|cite|improve this answer

answered Jul 19 at 15:58

peek-a-boopeek-a-boo

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$endgroup$

add a comment | 

$begingroup$

No, it doesn't. For example, $sin$ and $cos$ are infinitely differentiable functions, but they have no limit at $pm infty$.

share|cite|improve this answer

answered Jul 19 at 15:58

peek-a-boopeek-a-boo

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$endgroup$

share|cite|improve this answer

answered Jul 19 at 15:58

peek-a-boopeek-a-boo

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answered Jul 19 at 15:58

peek-a-boopeek-a-boo

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answered Jul 19 at 15:58

peek-a-boopeek-a-boo

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1

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Neither. Counter-examples: $f(x)=e^-x^2,$ and $f(x)=x^2.$

share|cite|improve this answer

answered Jul 19 at 15:58

Adrian KeisterAdrian Keister

6,36277 gold badges2222 silver badges3333 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
  Jul 19 at 16:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Jul 19 at 16:25
  

  
  
  
  


• 
  
  
  

  
  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
  Jul 19 at 16:35
  

  
  
  
  

add a comment | 

1

$begingroup$

Neither. Counter-examples: $f(x)=e^-x^2,$ and $f(x)=x^2.$

share|cite|improve this answer

answered Jul 19 at 15:58

Adrian KeisterAdrian Keister

6,36277 gold badges2222 silver badges3333 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
  Jul 19 at 16:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Jul 19 at 16:25
  

  
  
  
  


• 
  
  
  

  
  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
  Jul 19 at 16:35
  

  
  
  
  

add a comment | 

$begingroup$

Neither. Counter-examples: $f(x)=e^-x^2,$ and $f(x)=x^2.$

share|cite|improve this answer

answered Jul 19 at 15:58

Adrian KeisterAdrian Keister

6,36277 gold badges2222 silver badges3333 bronze badges

$endgroup$

share|cite|improve this answer

answered Jul 19 at 15:58

Adrian KeisterAdrian Keister

6,36277 gold badges2222 silver badges3333 bronze badges

answered Jul 19 at 15:58

Adrian KeisterAdrian Keister

6,36277 gold badges2222 silver badges3333 bronze badges

answered Jul 19 at 15:58

Adrian KeisterAdrian Keister

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