If derivative goes to $0$, does the function have a limit?












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$begingroup$


Let $f:mathbb{R}tomathbb{R}$ be a differentiable function. Assume that $lim_{xtoinfty}f'(x)=0$, does that mean that $lim_{xtoinfty}f(x)$ exists?



PS. What if we assume that $f$ is also bounded?






real-analysis limits derivatives







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  • 1




    $begingroup$
    By L'Hospital's Rule we get $f(x) /xto 0$ and that's the best we can conclude here.
    $endgroup$
    – Paramanand Singh
    Dec 29 '18 at 1:37
















2












$begingroup$


Let $f:mathbb{R}tomathbb{R}$ be a differentiable function. Assume that $lim_{xtoinfty}f'(x)=0$, does that mean that $lim_{xtoinfty}f(x)$ exists?



PS. What if we assume that $f$ is also bounded?






real-analysis limits derivatives







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    By L'Hospital's Rule we get $f(x) /xto 0$ and that's the best we can conclude here.
    $endgroup$
    – Paramanand Singh
    Dec 29 '18 at 1:37














2












2








2





$begingroup$


Let $f:mathbb{R}tomathbb{R}$ be a differentiable function. Assume that $lim_{xtoinfty}f'(x)=0$, does that mean that $lim_{xtoinfty}f(x)$ exists?



PS. What if we assume that $f$ is also bounded?






real-analysis limits derivatives







share|cite|improve this question











$endgroup$




Let $f:mathbb{R}tomathbb{R}$ be a differentiable function. Assume that $lim_{xtoinfty}f'(x)=0$, does that mean that $lim_{xtoinfty}f(x)$ exists?



PS. What if we assume that $f$ is also bounded?






real-analysis limits derivatives




real-analysis limits derivatives






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 28 '18 at 23:02







David Lingard

















asked Dec 28 '18 at 22:58









David LingardDavid Lingard

676




676








  • 1




    $begingroup$
    By L'Hospital's Rule we get $f(x) /xto 0$ and that's the best we can conclude here.
    $endgroup$
    – Paramanand Singh
    Dec 29 '18 at 1:37














  • 1




    $begingroup$
    By L'Hospital's Rule we get $f(x) /xto 0$ and that's the best we can conclude here.
    $endgroup$
    – Paramanand Singh
    Dec 29 '18 at 1:37








1




1




$begingroup$
By L'Hospital's Rule we get $f(x) /xto 0$ and that's the best we can conclude here.
$endgroup$
– Paramanand Singh
Dec 29 '18 at 1:37




$begingroup$
By L'Hospital's Rule we get $f(x) /xto 0$ and that's the best we can conclude here.
$endgroup$
– Paramanand Singh
Dec 29 '18 at 1:37










4 Answers
4






active

oldest

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11












$begingroup$

Not necessarily. Consider $f(x)=sin(ln(x))$. Then $f'(x)=frac1xcos(ln(x))to0$, but $f(x)$ diverges, as it reaches $-1$ and $1$ infinitely often.






share|cite|improve this answer









$endgroup$





















    7












    $begingroup$

    Not necessarily. See $f(x)=ln x$:



    $$lim_{xto infty}f'(x)=lim_{xtoinfty}frac 1 x=0$$
    $$lim_{xtoinfty}f(x)=+infty$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      ah ok my bad. I missed this example. I wonder what if $f$ is also bounded.
      $endgroup$
      – David Lingard
      Dec 28 '18 at 23:00










    • $begingroup$
      Maybe I should edit the question and add $f$ bounded.
      $endgroup$
      – David Lingard
      Dec 28 '18 at 23:01



















    1












    $begingroup$

    No. Consider the function defined by cos(x) on [0,2pi], cos((x/2)-pi) on [2pi,6pi] and so on.






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      No.



      Consider $lim_{ttoinfty}f'(t)=0$



      and



      $lim_{xtoinfty}f(x)-f(a)=lim_{xtoinfty}int_{a}^{x}f'(t)dt$ and $ageq0$.



      In this case, that integral converges if $f(a)$ exists for all $ageq0$. for example the function $f(x)=ln x$ doesn't satisfy that condition.



      And if that integral converges, the limit exists.






      share|cite|improve this answer









      $endgroup$














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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        11












        $begingroup$

        Not necessarily. Consider $f(x)=sin(ln(x))$. Then $f'(x)=frac1xcos(ln(x))to0$, but $f(x)$ diverges, as it reaches $-1$ and $1$ infinitely often.






        share|cite|improve this answer









        $endgroup$


















          11












          $begingroup$

          Not necessarily. Consider $f(x)=sin(ln(x))$. Then $f'(x)=frac1xcos(ln(x))to0$, but $f(x)$ diverges, as it reaches $-1$ and $1$ infinitely often.






          share|cite|improve this answer









          $endgroup$
















            11












            11








            11





            $begingroup$

            Not necessarily. Consider $f(x)=sin(ln(x))$. Then $f'(x)=frac1xcos(ln(x))to0$, but $f(x)$ diverges, as it reaches $-1$ and $1$ infinitely often.






            share|cite|improve this answer









            $endgroup$



            Not necessarily. Consider $f(x)=sin(ln(x))$. Then $f'(x)=frac1xcos(ln(x))to0$, but $f(x)$ diverges, as it reaches $-1$ and $1$ infinitely often.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 28 '18 at 23:05









            SmileyCraftSmileyCraft

            3,776519




            3,776519























                7












                $begingroup$

                Not necessarily. See $f(x)=ln x$:



                $$lim_{xto infty}f'(x)=lim_{xtoinfty}frac 1 x=0$$
                $$lim_{xtoinfty}f(x)=+infty$$






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  ah ok my bad. I missed this example. I wonder what if $f$ is also bounded.
                  $endgroup$
                  – David Lingard
                  Dec 28 '18 at 23:00










                • $begingroup$
                  Maybe I should edit the question and add $f$ bounded.
                  $endgroup$
                  – David Lingard
                  Dec 28 '18 at 23:01
















                7












                $begingroup$

                Not necessarily. See $f(x)=ln x$:



                $$lim_{xto infty}f'(x)=lim_{xtoinfty}frac 1 x=0$$
                $$lim_{xtoinfty}f(x)=+infty$$






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  ah ok my bad. I missed this example. I wonder what if $f$ is also bounded.
                  $endgroup$
                  – David Lingard
                  Dec 28 '18 at 23:00










                • $begingroup$
                  Maybe I should edit the question and add $f$ bounded.
                  $endgroup$
                  – David Lingard
                  Dec 28 '18 at 23:01














                7












                7








                7





                $begingroup$

                Not necessarily. See $f(x)=ln x$:



                $$lim_{xto infty}f'(x)=lim_{xtoinfty}frac 1 x=0$$
                $$lim_{xtoinfty}f(x)=+infty$$






                share|cite|improve this answer









                $endgroup$



                Not necessarily. See $f(x)=ln x$:



                $$lim_{xto infty}f'(x)=lim_{xtoinfty}frac 1 x=0$$
                $$lim_{xtoinfty}f(x)=+infty$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 28 '18 at 22:59









                Noble MushtakNoble Mushtak

                15.4k1835




                15.4k1835












                • $begingroup$
                  ah ok my bad. I missed this example. I wonder what if $f$ is also bounded.
                  $endgroup$
                  – David Lingard
                  Dec 28 '18 at 23:00










                • $begingroup$
                  Maybe I should edit the question and add $f$ bounded.
                  $endgroup$
                  – David Lingard
                  Dec 28 '18 at 23:01


















                • $begingroup$
                  ah ok my bad. I missed this example. I wonder what if $f$ is also bounded.
                  $endgroup$
                  – David Lingard
                  Dec 28 '18 at 23:00










                • $begingroup$
                  Maybe I should edit the question and add $f$ bounded.
                  $endgroup$
                  – David Lingard
                  Dec 28 '18 at 23:01
















                $begingroup$
                ah ok my bad. I missed this example. I wonder what if $f$ is also bounded.
                $endgroup$
                – David Lingard
                Dec 28 '18 at 23:00




                $begingroup$
                ah ok my bad. I missed this example. I wonder what if $f$ is also bounded.
                $endgroup$
                – David Lingard
                Dec 28 '18 at 23:00












                $begingroup$
                Maybe I should edit the question and add $f$ bounded.
                $endgroup$
                – David Lingard
                Dec 28 '18 at 23:01




                $begingroup$
                Maybe I should edit the question and add $f$ bounded.
                $endgroup$
                – David Lingard
                Dec 28 '18 at 23:01











                1












                $begingroup$

                No. Consider the function defined by cos(x) on [0,2pi], cos((x/2)-pi) on [2pi,6pi] and so on.






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  No. Consider the function defined by cos(x) on [0,2pi], cos((x/2)-pi) on [2pi,6pi] and so on.






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    No. Consider the function defined by cos(x) on [0,2pi], cos((x/2)-pi) on [2pi,6pi] and so on.






                    share|cite|improve this answer









                    $endgroup$



                    No. Consider the function defined by cos(x) on [0,2pi], cos((x/2)-pi) on [2pi,6pi] and so on.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 28 '18 at 23:17









                    Jagol95Jagol95

                    2637




                    2637























                        1












                        $begingroup$

                        No.



                        Consider $lim_{ttoinfty}f'(t)=0$



                        and



                        $lim_{xtoinfty}f(x)-f(a)=lim_{xtoinfty}int_{a}^{x}f'(t)dt$ and $ageq0$.



                        In this case, that integral converges if $f(a)$ exists for all $ageq0$. for example the function $f(x)=ln x$ doesn't satisfy that condition.



                        And if that integral converges, the limit exists.






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          No.



                          Consider $lim_{ttoinfty}f'(t)=0$



                          and



                          $lim_{xtoinfty}f(x)-f(a)=lim_{xtoinfty}int_{a}^{x}f'(t)dt$ and $ageq0$.



                          In this case, that integral converges if $f(a)$ exists for all $ageq0$. for example the function $f(x)=ln x$ doesn't satisfy that condition.



                          And if that integral converges, the limit exists.






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            No.



                            Consider $lim_{ttoinfty}f'(t)=0$



                            and



                            $lim_{xtoinfty}f(x)-f(a)=lim_{xtoinfty}int_{a}^{x}f'(t)dt$ and $ageq0$.



                            In this case, that integral converges if $f(a)$ exists for all $ageq0$. for example the function $f(x)=ln x$ doesn't satisfy that condition.



                            And if that integral converges, the limit exists.






                            share|cite|improve this answer









                            $endgroup$



                            No.



                            Consider $lim_{ttoinfty}f'(t)=0$



                            and



                            $lim_{xtoinfty}f(x)-f(a)=lim_{xtoinfty}int_{a}^{x}f'(t)dt$ and $ageq0$.



                            In this case, that integral converges if $f(a)$ exists for all $ageq0$. for example the function $f(x)=ln x$ doesn't satisfy that condition.



                            And if that integral converges, the limit exists.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 29 '18 at 1:18









                            ZoberZober

                            715




                            715






























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