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Does any $f(x)$ which for large $x$ grows slower than any power of $x$ necessarily satisfy $lim_xtoinftyf'(x)=0$?

Prove: If $lim_ntoinftya_n = L$ and $a_n > a$ for all $n$ then $L geq a$Show that two bounded sequences have convergent subsequences with the same index sequenceWhat happen to composite of infinite number of continuous functions?Limit of the ratio of the logarithms of two functions versus the limit of the ratio of the functionsFor continuous $f:[0, infty) to mathbbR$ with finite $a = lim_x to infty f(x)$ the IVT appliesIs there any function which grows 'slower' than its derivative?Do square integrable function and its derivative imply the function tends to $0$ for $|x|$ large?Function grows slower than $ln(x)$is it for granted that $log x$ grows slower than $x^a$?Sequence with two different norm limits

0

$begingroup$

The title is essentially the question. Intuitively I would await that any function $f(x)$, which for large $x$ grows slower than any power of $x$, must necessarily satisfy

$lim_xtoinftyf'(x)=0$.

This holds at least for prominent examples like $ln(x)$ and $sqrtx$. But I don't know of a statement that it holds in general. Does anyone know or have a proof or counter examples?

Thanks for help!

EDIT: I forgot to mention that $f$ should also be strictly monotonically increasing, for large $x$, and also that it should be smooth ($C^infty(mathbb R)$).

limits

share|cite|improve this question

edited Mar 28 at 22:09

Britzel

asked Mar 28 at 21:23

BritzelBritzel

375

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What do you mean explicitly?
  $endgroup$
  – Jorge Fernández Hidalgo
  Mar 28 at 21:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is $f$ increasing?
  $endgroup$
  – Jorge Fernández Hidalgo
  Mar 28 at 21:27
  

  
  
  
  

add a comment | 

0

$begingroup$

The title is essentially the question. Intuitively I would await that any function $f(x)$, which for large $x$ grows slower than any power of $x$, must necessarily satisfy

$lim_xtoinftyf'(x)=0$.

This holds at least for prominent examples like $ln(x)$ and $sqrtx$. But I don't know of a statement that it holds in general. Does anyone know or have a proof or counter examples?

Thanks for help!

EDIT: I forgot to mention that $f$ should also be strictly monotonically increasing, for large $x$, and also that it should be smooth ($C^infty(mathbb R)$).

limits

share|cite|improve this question

edited Mar 28 at 22:09

Britzel

asked Mar 28 at 21:23

BritzelBritzel

375

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What do you mean explicitly?
  $endgroup$
  – Jorge Fernández Hidalgo
  Mar 28 at 21:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is $f$ increasing?
  $endgroup$
  – Jorge Fernández Hidalgo
  Mar 28 at 21:27
  

  
  
  
  

add a comment | 

$begingroup$

The title is essentially the question. Intuitively I would await that any function $f(x)$, which for large $x$ grows slower than any power of $x$, must necessarily satisfy

$lim_xtoinftyf'(x)=0$.

This holds at least for prominent examples like $ln(x)$ and $sqrtx$. But I don't know of a statement that it holds in general. Does anyone know or have a proof or counter examples?

Thanks for help!

EDIT: I forgot to mention that $f$ should also be strictly monotonically increasing, for large $x$, and also that it should be smooth ($C^infty(mathbb R)$).

limits

share|cite|improve this question

edited Mar 28 at 22:09

Britzel

asked Mar 28 at 21:23

BritzelBritzel

375

$endgroup$

share|cite|improve this question

edited Mar 28 at 22:09

Britzel

asked Mar 28 at 21:23

BritzelBritzel

375

share|cite|improve this question

edited Mar 28 at 22:09

Britzel

asked Mar 28 at 21:23

BritzelBritzel

375

asked Mar 28 at 21:23

BritzelBritzel

375

asked Mar 28 at 21:23

BritzelBritzel

375

2 Answers
2

active

oldest

votes

3

$begingroup$

I assume that $f$ is differentiable (otherwise $f'(x)$ is meaningless).

Assuming the limit exists, it must be $0$.

Otherwise, if $$lim_x to +infty f'(x)=L >0$$
then there exists $x_0 >0$ such that for $x>x_0$ necessarily
$$f'(x) > fracL2$$This implies that $$f(x)=f(x_0)+ int_x_0^x f'(t) mathrm dt>f(x_0)+fracL2(x-x_0)$$
which contradicts our hypothesis that $f$ grows slower than $x$.

However, nothing ensures that the limit exists. As a counterexample, you could consider
$$f(x) = int_0^x |sin t|^t mathrm dt$$

share|cite|improve this answer

answered Mar 28 at 21:34

CrostulCrostul

28.2k22352

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hi there, thanks for the detailed answer! The counter example you show is monotonically increasing, but not strictly monotonically. But I suppose the situation wouldn't change, if I would only consider strictly monotonically increasing functions, right? (They should also be $C^infty(mathbb R)$ actually.)
  $endgroup$
  – Britzel
  Mar 28 at 22:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Actually my last counterexample is strictly monotonic. Anyway, one can construct a $C^infty$ function which is positive, has no limit as $x to infty$ but whose improper integral is convergent. Then, any of its primitives is a counterexample to your problem.
  $endgroup$
  – Crostul
  Mar 28 at 22:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Very interesting. Thank you so much for your help again!
  $endgroup$
  – Britzel
  Mar 28 at 23:51
  

  
  
  
  

add a comment | 

1

$begingroup$

take function which has unit steps with unit derivative at increasingly large intervals. Limit doesn't exist, function grows as slow as you like.

share|cite|improve this answer

answered Mar 28 at 21:26

RadostRadost

65512

$endgroup$

add a comment | 

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3

$begingroup$

I assume that $f$ is differentiable (otherwise $f'(x)$ is meaningless).

Assuming the limit exists, it must be $0$.

Otherwise, if $$lim_x to +infty f'(x)=L >0$$
then there exists $x_0 >0$ such that for $x>x_0$ necessarily
$$f'(x) > fracL2$$This implies that $$f(x)=f(x_0)+ int_x_0^x f'(t) mathrm dt>f(x_0)+fracL2(x-x_0)$$
which contradicts our hypothesis that $f$ grows slower than $x$.

However, nothing ensures that the limit exists. As a counterexample, you could consider
$$f(x) = int_0^x |sin t|^t mathrm dt$$

share|cite|improve this answer

answered Mar 28 at 21:34

CrostulCrostul

28.2k22352

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hi there, thanks for the detailed answer! The counter example you show is monotonically increasing, but not strictly monotonically. But I suppose the situation wouldn't change, if I would only consider strictly monotonically increasing functions, right? (They should also be $C^infty(mathbb R)$ actually.)
  $endgroup$
  – Britzel
  Mar 28 at 22:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Actually my last counterexample is strictly monotonic. Anyway, one can construct a $C^infty$ function which is positive, has no limit as $x to infty$ but whose improper integral is convergent. Then, any of its primitives is a counterexample to your problem.
  $endgroup$
  – Crostul
  Mar 28 at 22:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Very interesting. Thank you so much for your help again!
  $endgroup$
  – Britzel
  Mar 28 at 23:51
  

  
  
  
  

add a comment | 

3

$begingroup$

I assume that $f$ is differentiable (otherwise $f'(x)$ is meaningless).

Assuming the limit exists, it must be $0$.

Otherwise, if $$lim_x to +infty f'(x)=L >0$$
then there exists $x_0 >0$ such that for $x>x_0$ necessarily
$$f'(x) > fracL2$$This implies that $$f(x)=f(x_0)+ int_x_0^x f'(t) mathrm dt>f(x_0)+fracL2(x-x_0)$$
which contradicts our hypothesis that $f$ grows slower than $x$.

However, nothing ensures that the limit exists. As a counterexample, you could consider
$$f(x) = int_0^x |sin t|^t mathrm dt$$

share|cite|improve this answer

answered Mar 28 at 21:34

CrostulCrostul

28.2k22352

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hi there, thanks for the detailed answer! The counter example you show is monotonically increasing, but not strictly monotonically. But I suppose the situation wouldn't change, if I would only consider strictly monotonically increasing functions, right? (They should also be $C^infty(mathbb R)$ actually.)
  $endgroup$
  – Britzel
  Mar 28 at 22:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Actually my last counterexample is strictly monotonic. Anyway, one can construct a $C^infty$ function which is positive, has no limit as $x to infty$ but whose improper integral is convergent. Then, any of its primitives is a counterexample to your problem.
  $endgroup$
  – Crostul
  Mar 28 at 22:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Very interesting. Thank you so much for your help again!
  $endgroup$
  – Britzel
  Mar 28 at 23:51
  

  
  
  
  

add a comment | 

$begingroup$

I assume that $f$ is differentiable (otherwise $f'(x)$ is meaningless).

Assuming the limit exists, it must be $0$.

Otherwise, if $$lim_x to +infty f'(x)=L >0$$
then there exists $x_0 >0$ such that for $x>x_0$ necessarily
$$f'(x) > fracL2$$This implies that $$f(x)=f(x_0)+ int_x_0^x f'(t) mathrm dt>f(x_0)+fracL2(x-x_0)$$
which contradicts our hypothesis that $f$ grows slower than $x$.

However, nothing ensures that the limit exists. As a counterexample, you could consider
$$f(x) = int_0^x |sin t|^t mathrm dt$$

share|cite|improve this answer

answered Mar 28 at 21:34

CrostulCrostul

28.2k22352

$endgroup$

share|cite|improve this answer

answered Mar 28 at 21:34

CrostulCrostul

28.2k22352

answered Mar 28 at 21:34

CrostulCrostul

28.2k22352

answered Mar 28 at 21:34

CrostulCrostul

28.2k22352

1

$begingroup$

take function which has unit steps with unit derivative at increasingly large intervals. Limit doesn't exist, function grows as slow as you like.

share|cite|improve this answer

answered Mar 28 at 21:26

RadostRadost

65512

$endgroup$

add a comment | 

1

$begingroup$

take function which has unit steps with unit derivative at increasingly large intervals. Limit doesn't exist, function grows as slow as you like.

share|cite|improve this answer

answered Mar 28 at 21:26

RadostRadost

65512

$endgroup$

add a comment | 

$begingroup$

take function which has unit steps with unit derivative at increasingly large intervals. Limit doesn't exist, function grows as slow as you like.

share|cite|improve this answer

answered Mar 28 at 21:26

RadostRadost

65512

$endgroup$

share|cite|improve this answer

answered Mar 28 at 21:26

RadostRadost

65512

answered Mar 28 at 21:26

RadostRadost

65512

answered Mar 28 at 21:26

RadostRadost

65512