$underset{xrightarrowinfty}{lim}frac{f(x)}{x}=0$ Implies $underset{xrightarrowinfty}{f'(x)}=0$ [duplicate]












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  • Proving that $limlimits_{xtoinfty}f'(x) = 0$ when $limlimits_{xtoinfty}f(x)$ and $limlimits_{xtoinfty}f'(x)$ exist

    6 answers




Let $f:mathbb{R}rightarrowmathbb{R}$ be a continuously differentiable function such that $underset{xrightarrowinfty}{lim}frac{f(x)}{x}=0$ and suppose $underset{xrightarrowinfty}{f'(x)}$ exists. Then Prove that $underset{xrightarrowinfty}{f'(x)}=0$



I can see that if we apply L'hoptal's theorem directly to $frac{f(x)}{x}$ then we can get the answer. But is it possible to do so without knowing the value of $underset{xrightarrowinfty}{f(x)}$



On the similar problem: found here, they have given the existence of $lim_{xrightarrowinfty} f(x)$. But in this particular problem they haven't










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marked as duplicate by Math Lover, Cesareo, José Carlos Santos calculus
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Jan 3 at 1:17


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.















  • $begingroup$
    It is true if you assume that $lim_{x to infty} f'(x)$ exists, see e.g. math.stackexchange.com/questions/42277/…. Also related: math.stackexchange.com/q/313676/42969.
    $endgroup$
    – Martin R
    Jan 2 at 21:35












  • $begingroup$
    Thanks and I edited
    $endgroup$
    – DD90
    Jan 2 at 21:37
















4












$begingroup$



This question already has an answer here:




  • Proving that $limlimits_{xtoinfty}f'(x) = 0$ when $limlimits_{xtoinfty}f(x)$ and $limlimits_{xtoinfty}f'(x)$ exist

    6 answers




Let $f:mathbb{R}rightarrowmathbb{R}$ be a continuously differentiable function such that $underset{xrightarrowinfty}{lim}frac{f(x)}{x}=0$ and suppose $underset{xrightarrowinfty}{f'(x)}$ exists. Then Prove that $underset{xrightarrowinfty}{f'(x)}=0$



I can see that if we apply L'hoptal's theorem directly to $frac{f(x)}{x}$ then we can get the answer. But is it possible to do so without knowing the value of $underset{xrightarrowinfty}{f(x)}$



On the similar problem: found here, they have given the existence of $lim_{xrightarrowinfty} f(x)$. But in this particular problem they haven't










share|cite|improve this question











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marked as duplicate by Math Lover, Cesareo, José Carlos Santos calculus
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Jan 3 at 1:17


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.















  • $begingroup$
    It is true if you assume that $lim_{x to infty} f'(x)$ exists, see e.g. math.stackexchange.com/questions/42277/…. Also related: math.stackexchange.com/q/313676/42969.
    $endgroup$
    – Martin R
    Jan 2 at 21:35












  • $begingroup$
    Thanks and I edited
    $endgroup$
    – DD90
    Jan 2 at 21:37














4












4








4





$begingroup$



This question already has an answer here:




  • Proving that $limlimits_{xtoinfty}f'(x) = 0$ when $limlimits_{xtoinfty}f(x)$ and $limlimits_{xtoinfty}f'(x)$ exist

    6 answers




Let $f:mathbb{R}rightarrowmathbb{R}$ be a continuously differentiable function such that $underset{xrightarrowinfty}{lim}frac{f(x)}{x}=0$ and suppose $underset{xrightarrowinfty}{f'(x)}$ exists. Then Prove that $underset{xrightarrowinfty}{f'(x)}=0$



I can see that if we apply L'hoptal's theorem directly to $frac{f(x)}{x}$ then we can get the answer. But is it possible to do so without knowing the value of $underset{xrightarrowinfty}{f(x)}$



On the similar problem: found here, they have given the existence of $lim_{xrightarrowinfty} f(x)$. But in this particular problem they haven't










share|cite|improve this question











$endgroup$





This question already has an answer here:




  • Proving that $limlimits_{xtoinfty}f'(x) = 0$ when $limlimits_{xtoinfty}f(x)$ and $limlimits_{xtoinfty}f'(x)$ exist

    6 answers




Let $f:mathbb{R}rightarrowmathbb{R}$ be a continuously differentiable function such that $underset{xrightarrowinfty}{lim}frac{f(x)}{x}=0$ and suppose $underset{xrightarrowinfty}{f'(x)}$ exists. Then Prove that $underset{xrightarrowinfty}{f'(x)}=0$



I can see that if we apply L'hoptal's theorem directly to $frac{f(x)}{x}$ then we can get the answer. But is it possible to do so without knowing the value of $underset{xrightarrowinfty}{f(x)}$



On the similar problem: found here, they have given the existence of $lim_{xrightarrowinfty} f(x)$. But in this particular problem they haven't





This question already has an answer here:




  • Proving that $limlimits_{xtoinfty}f'(x) = 0$ when $limlimits_{xtoinfty}f(x)$ and $limlimits_{xtoinfty}f'(x)$ exist

    6 answers








real-analysis calculus analysis






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edited Jan 2 at 21:52







DD90

















asked Jan 2 at 21:30









DD90DD90

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marked as duplicate by Math Lover, Cesareo, José Carlos Santos calculus
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Jan 3 at 1:17


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 Math Lover, Cesareo, José Carlos Santos calculus
Users with the  calculus badge can single-handedly close calculus questions as duplicates and reopen them as needed.

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Jan 3 at 1:17


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.














  • $begingroup$
    It is true if you assume that $lim_{x to infty} f'(x)$ exists, see e.g. math.stackexchange.com/questions/42277/…. Also related: math.stackexchange.com/q/313676/42969.
    $endgroup$
    – Martin R
    Jan 2 at 21:35












  • $begingroup$
    Thanks and I edited
    $endgroup$
    – DD90
    Jan 2 at 21:37


















  • $begingroup$
    It is true if you assume that $lim_{x to infty} f'(x)$ exists, see e.g. math.stackexchange.com/questions/42277/…. Also related: math.stackexchange.com/q/313676/42969.
    $endgroup$
    – Martin R
    Jan 2 at 21:35












  • $begingroup$
    Thanks and I edited
    $endgroup$
    – DD90
    Jan 2 at 21:37
















$begingroup$
It is true if you assume that $lim_{x to infty} f'(x)$ exists, see e.g. math.stackexchange.com/questions/42277/…. Also related: math.stackexchange.com/q/313676/42969.
$endgroup$
– Martin R
Jan 2 at 21:35






$begingroup$
It is true if you assume that $lim_{x to infty} f'(x)$ exists, see e.g. math.stackexchange.com/questions/42277/…. Also related: math.stackexchange.com/q/313676/42969.
$endgroup$
– Martin R
Jan 2 at 21:35














$begingroup$
Thanks and I edited
$endgroup$
– DD90
Jan 2 at 21:37




$begingroup$
Thanks and I edited
$endgroup$
– DD90
Jan 2 at 21:37










2 Answers
2






active

oldest

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$begingroup$

As stated the result is wrong. Take $f(x)=sin x^2$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Sorry I edited.
    $endgroup$
    – DD90
    Jan 2 at 21:52



















2












$begingroup$

Suppose $lim_{x to infty} f'(x) >a >0$. Then $f(n)-f(n-1) > a $ for $n$ suffciently large, say for $n geq n_0$ [This is by MVT]. Then $f(n) geq f(n_0)+ a (n-n_0)$ for $n geq n_0$ which contradicts the hypothesis that $frac {f(x)} xto 0$. For the case $lim_{x to infty} f'(x) <0$ simply replace $f$ by $-f$.






share|cite|improve this answer









$endgroup$




















    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    As stated the result is wrong. Take $f(x)=sin x^2$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Sorry I edited.
      $endgroup$
      – DD90
      Jan 2 at 21:52
















    4












    $begingroup$

    As stated the result is wrong. Take $f(x)=sin x^2$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Sorry I edited.
      $endgroup$
      – DD90
      Jan 2 at 21:52














    4












    4








    4





    $begingroup$

    As stated the result is wrong. Take $f(x)=sin x^2$.






    share|cite|improve this answer









    $endgroup$



    As stated the result is wrong. Take $f(x)=sin x^2$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 2 at 21:34









    mathcounterexamples.netmathcounterexamples.net

    25.6k21953




    25.6k21953












    • $begingroup$
      Sorry I edited.
      $endgroup$
      – DD90
      Jan 2 at 21:52


















    • $begingroup$
      Sorry I edited.
      $endgroup$
      – DD90
      Jan 2 at 21:52
















    $begingroup$
    Sorry I edited.
    $endgroup$
    – DD90
    Jan 2 at 21:52




    $begingroup$
    Sorry I edited.
    $endgroup$
    – DD90
    Jan 2 at 21:52











    2












    $begingroup$

    Suppose $lim_{x to infty} f'(x) >a >0$. Then $f(n)-f(n-1) > a $ for $n$ suffciently large, say for $n geq n_0$ [This is by MVT]. Then $f(n) geq f(n_0)+ a (n-n_0)$ for $n geq n_0$ which contradicts the hypothesis that $frac {f(x)} xto 0$. For the case $lim_{x to infty} f'(x) <0$ simply replace $f$ by $-f$.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Suppose $lim_{x to infty} f'(x) >a >0$. Then $f(n)-f(n-1) > a $ for $n$ suffciently large, say for $n geq n_0$ [This is by MVT]. Then $f(n) geq f(n_0)+ a (n-n_0)$ for $n geq n_0$ which contradicts the hypothesis that $frac {f(x)} xto 0$. For the case $lim_{x to infty} f'(x) <0$ simply replace $f$ by $-f$.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Suppose $lim_{x to infty} f'(x) >a >0$. Then $f(n)-f(n-1) > a $ for $n$ suffciently large, say for $n geq n_0$ [This is by MVT]. Then $f(n) geq f(n_0)+ a (n-n_0)$ for $n geq n_0$ which contradicts the hypothesis that $frac {f(x)} xto 0$. For the case $lim_{x to infty} f'(x) <0$ simply replace $f$ by $-f$.






        share|cite|improve this answer









        $endgroup$



        Suppose $lim_{x to infty} f'(x) >a >0$. Then $f(n)-f(n-1) > a $ for $n$ suffciently large, say for $n geq n_0$ [This is by MVT]. Then $f(n) geq f(n_0)+ a (n-n_0)$ for $n geq n_0$ which contradicts the hypothesis that $frac {f(x)} xto 0$. For the case $lim_{x to infty} f'(x) <0$ simply replace $f$ by $-f$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 2 at 23:51









        Kavi Rama MurthyKavi Rama Murthy

        53.5k32055




        53.5k32055















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