$f(x)geq e^x$ implies $xf'(x)/f(x)to infty$ [closed]

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Is it true that $f(x)geq e^x$ for large $x$ implies $xf'(x)/f(x)to infty$? We assume that $f$ is smooth.

calculus differential-equations

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asked Jul 15 at 3:45

Dong Li

702413

closed as off-topic by Shailesh, heropup, John Ma, RRL, Parcly Taxel Jul 15 at 7:16

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• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shailesh, heropup, John Ma, RRL, Parcly Taxel

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  Certainly not. It's easy to imagine a function $f(x)$ dominating $e^x$ but having periodic "humps" where $f'=0$. A better question is if $limsup fracxf'(x)f(x) = infty$.
  – MathematicsStudent1122
  Jul 15 at 3:48
  

  
  
  
  

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up vote
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Is it true that $f(x)geq e^x$ for large $x$ implies $xf'(x)/f(x)to infty$? We assume that $f$ is smooth.

calculus differential-equations

share|cite|improve this question

asked Jul 15 at 3:45

Dong Li

702413

closed as off-topic by Shailesh, heropup, John Ma, RRL, Parcly Taxel Jul 15 at 7:16

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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shailesh, heropup, John Ma, RRL, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Certainly not. It's easy to imagine a function $f(x)$ dominating $e^x$ but having periodic "humps" where $f'=0$. A better question is if $limsup fracxf'(x)f(x) = infty$.
  – MathematicsStudent1122
  Jul 15 at 3:48
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

1

up vote
0
down vote

favorite

1

1

Is it true that $f(x)geq e^x$ for large $x$ implies $xf'(x)/f(x)to infty$? We assume that $f$ is smooth.

calculus differential-equations

share|cite|improve this question

asked Jul 15 at 3:45

Dong Li

702413

Is it true that $f(x)geq e^x$ for large $x$ implies $xf'(x)/f(x)to infty$? We assume that $f$ is smooth.

calculus differential-equations

share|cite|improve this question

asked Jul 15 at 3:45

Dong Li

702413

share|cite|improve this question

share|cite|improve this question

asked Jul 15 at 3:45

Dong Li

702413

asked Jul 15 at 3:45

Dong Li

702413

asked Jul 15 at 3:45

Dong Li

702413

702413

closed as off-topic by Shailesh, heropup, John Ma, RRL, Parcly Taxel Jul 15 at 7:16

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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shailesh, heropup, John Ma, RRL, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by Shailesh, heropup, John Ma, RRL, Parcly Taxel Jul 15 at 7:16

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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shailesh, heropup, John Ma, RRL, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Certainly not. It's easy to imagine a function $f(x)$ dominating $e^x$ but having periodic "humps" where $f'=0$. A better question is if $limsup fracxf'(x)f(x) = infty$.
  – MathematicsStudent1122
  Jul 15 at 3:48
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Certainly not. It's easy to imagine a function $f(x)$ dominating $e^x$ but having periodic "humps" where $f'=0$. A better question is if $limsup fracxf'(x)f(x) = infty$.
  – MathematicsStudent1122
  Jul 15 at 3:48
  

  
  
  
  

2

2

Certainly not. It's easy to imagine a function $f(x)$ dominating $e^x$ but having periodic "humps" where $f'=0$. A better question is if $limsup fracxf'(x)f(x) = infty$.
– MathematicsStudent1122
Jul 15 at 3:48

Certainly not. It's easy to imagine a function $f(x)$ dominating $e^x$ but having periodic "humps" where $f'=0$. A better question is if $limsup fracxf'(x)f(x) = infty$.
– MathematicsStudent1122
Jul 15 at 3:48

add a comment | 

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I'll show $limsup_x to infty fracxf'(x)f(x) = +infty$. If we let $g(x) = fracf(x)e^x$, then $g(x) ge 1$ and $xfracf'(x)f(x) = x(1+fracg'(x)g(x))$ for each $x$. If the limsup of this were not $+infty$, then $g'(x)$ must be less than $-g(x) le -1$ for all large $x$. But this contradicts $g(x) ge 1$ for all large $x$.

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answered Jul 15 at 4:03

mathworker21

6,4601727

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1 Answer
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1 Answer
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active

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active

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up vote
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I'll show $limsup_x to infty fracxf'(x)f(x) = +infty$. If we let $g(x) = fracf(x)e^x$, then $g(x) ge 1$ and $xfracf'(x)f(x) = x(1+fracg'(x)g(x))$ for each $x$. If the limsup of this were not $+infty$, then $g'(x)$ must be less than $-g(x) le -1$ for all large $x$. But this contradicts $g(x) ge 1$ for all large $x$.

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answered Jul 15 at 4:03

mathworker21

6,4601727

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I'll show $limsup_x to infty fracxf'(x)f(x) = +infty$. If we let $g(x) = fracf(x)e^x$, then $g(x) ge 1$ and $xfracf'(x)f(x) = x(1+fracg'(x)g(x))$ for each $x$. If the limsup of this were not $+infty$, then $g'(x)$ must be less than $-g(x) le -1$ for all large $x$. But this contradicts $g(x) ge 1$ for all large $x$.

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answered Jul 15 at 4:03

mathworker21

6,4601727

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up vote
1
down vote

up vote
1
down vote

I'll show $limsup_x to infty fracxf'(x)f(x) = +infty$. If we let $g(x) = fracf(x)e^x$, then $g(x) ge 1$ and $xfracf'(x)f(x) = x(1+fracg'(x)g(x))$ for each $x$. If the limsup of this were not $+infty$, then $g'(x)$ must be less than $-g(x) le -1$ for all large $x$. But this contradicts $g(x) ge 1$ for all large $x$.

share|cite|improve this answer

answered Jul 15 at 4:03

mathworker21

6,4601727

I'll show $limsup_x to infty fracxf'(x)f(x) = +infty$. If we let $g(x) = fracf(x)e^x$, then $g(x) ge 1$ and $xfracf'(x)f(x) = x(1+fracg'(x)g(x))$ for each $x$. If the limsup of this were not $+infty$, then $g'(x)$ must be less than $-g(x) le -1$ for all large $x$. But this contradicts $g(x) ge 1$ for all large $x$.

share|cite|improve this answer

answered Jul 15 at 4:03

mathworker21

6,4601727

share|cite|improve this answer

share|cite|improve this answer

answered Jul 15 at 4:03

mathworker21

6,4601727

answered Jul 15 at 4:03

mathworker21

6,4601727

answered Jul 15 at 4:03

mathworker21

6,4601727

6,4601727

add a comment | 

add a comment |