How to give an example of a $f$ differentiable in a deleted neighborhood of $x_0$ such that $lim_{xto...











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How would I give a simple example of a function $f$ differentiable in a deleted neighborhood of $x_0$ such that $lim_{xto x_0}f^prime(x)$ does not exist? I cannot seem to think of an example.



A delete neighborhood is an open interval about $x_0$ which does not contain $x_0$. So, $(x_0-delta,x_0+delta)-{x_0}$ for some $delta>0$.



How would something be differentiable in a deleted neighborhood if at the point of the derivative, the limit does not exist. Presumably, the derivative ends up looking something like $lim_{xto x_0} dfrac{1}{x}$, if it does not exist.






calculus derivatives examples-counterexamples







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  • Must be your function continuous?
    – Dog_69
    yesterday












  • @Dog_69 No, it can be any function we can dream up
    – kaisa
    yesterday










  • I was thinking about the Heaviside's function but I will say the absolute value $|x|$ around $x=0$.
    – Dog_69
    yesterday

















up vote
1
down vote

favorite












How would I give a simple example of a function $f$ differentiable in a deleted neighborhood of $x_0$ such that $lim_{xto x_0}f^prime(x)$ does not exist? I cannot seem to think of an example.



A delete neighborhood is an open interval about $x_0$ which does not contain $x_0$. So, $(x_0-delta,x_0+delta)-{x_0}$ for some $delta>0$.



How would something be differentiable in a deleted neighborhood if at the point of the derivative, the limit does not exist. Presumably, the derivative ends up looking something like $lim_{xto x_0} dfrac{1}{x}$, if it does not exist.






calculus derivatives examples-counterexamples







share|cite|improve this question
























  • Must be your function continuous?
    – Dog_69
    yesterday












  • @Dog_69 No, it can be any function we can dream up
    – kaisa
    yesterday










  • I was thinking about the Heaviside's function but I will say the absolute value $|x|$ around $x=0$.
    – Dog_69
    yesterday















up vote
1
down vote

favorite









up vote
1
down vote

favorite











How would I give a simple example of a function $f$ differentiable in a deleted neighborhood of $x_0$ such that $lim_{xto x_0}f^prime(x)$ does not exist? I cannot seem to think of an example.



A delete neighborhood is an open interval about $x_0$ which does not contain $x_0$. So, $(x_0-delta,x_0+delta)-{x_0}$ for some $delta>0$.



How would something be differentiable in a deleted neighborhood if at the point of the derivative, the limit does not exist. Presumably, the derivative ends up looking something like $lim_{xto x_0} dfrac{1}{x}$, if it does not exist.






calculus derivatives examples-counterexamples







share|cite|improve this question















How would I give a simple example of a function $f$ differentiable in a deleted neighborhood of $x_0$ such that $lim_{xto x_0}f^prime(x)$ does not exist? I cannot seem to think of an example.



A delete neighborhood is an open interval about $x_0$ which does not contain $x_0$. So, $(x_0-delta,x_0+delta)-{x_0}$ for some $delta>0$.



How would something be differentiable in a deleted neighborhood if at the point of the derivative, the limit does not exist. Presumably, the derivative ends up looking something like $lim_{xto x_0} dfrac{1}{x}$, if it does not exist.






calculus derivatives examples-counterexamples




calculus derivatives examples-counterexamples






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share|cite|improve this question













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edited yesterday









GNUSupporter 8964民主女神 地下教會

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asked yesterday









kaisa

425




425












  • Must be your function continuous?
    – Dog_69
    yesterday












  • @Dog_69 No, it can be any function we can dream up
    – kaisa
    yesterday










  • I was thinking about the Heaviside's function but I will say the absolute value $|x|$ around $x=0$.
    – Dog_69
    yesterday




















  • Must be your function continuous?
    – Dog_69
    yesterday












  • @Dog_69 No, it can be any function we can dream up
    – kaisa
    yesterday










  • I was thinking about the Heaviside's function but I will say the absolute value $|x|$ around $x=0$.
    – Dog_69
    yesterday


















Must be your function continuous?
– Dog_69
yesterday






Must be your function continuous?
– Dog_69
yesterday














@Dog_69 No, it can be any function we can dream up
– kaisa
yesterday




@Dog_69 No, it can be any function we can dream up
– kaisa
yesterday












I was thinking about the Heaviside's function but I will say the absolute value $|x|$ around $x=0$.
– Dog_69
yesterday






I was thinking about the Heaviside's function but I will say the absolute value $|x|$ around $x=0$.
– Dog_69
yesterday












5 Answers
5






active

oldest

votes

















up vote
4
down vote



accepted










Classic example:
$$sqrt[3]{(x-x_0)^2}$$






share|cite|improve this answer




























    up vote
    6
    down vote













    Take $f(x) = x sin (1/x)$ near $0$






    share|cite|improve this answer




























      up vote
      3
      down vote













      You may try $f(x)=x^2cos(1/x)$, so that $f'(x)=2xcos(1/x)-sin(1/x)$ has a point of discontinuity at $x=0$.






      share|cite|improve this answer

















      • 2




        This is a slightly better example than mine, in fact, because $f$ is differentiable at zero as well.
        – Richard Martin
        yesterday


















      up vote
      2
      down vote













      Does $f(x)=x^frac 12 $ count?
      $f'(x)=frac 1{2x^frac 12}$ which is discontinuous at $x=0$






      share|cite|improve this answer

















      • 1




        Yes, it does! But see my remark on $x^2 cos 1/x$
        – Richard Martin
        yesterday


















      up vote
      0
      down vote













      $$ln'(x) = dfrac{1}{x}$$



      If you are looking for that exact derivative.






      share|cite|improve this answer





















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        5 Answers
        5






        active

        oldest

        votes








        5 Answers
        5






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        4
        down vote



        accepted










        Classic example:
        $$sqrt[3]{(x-x_0)^2}$$






        share|cite|improve this answer

























          up vote
          4
          down vote



          accepted










          Classic example:
          $$sqrt[3]{(x-x_0)^2}$$






          share|cite|improve this answer























            up vote
            4
            down vote



            accepted







            up vote
            4
            down vote



            accepted






            Classic example:
            $$sqrt[3]{(x-x_0)^2}$$






            share|cite|improve this answer












            Classic example:
            $$sqrt[3]{(x-x_0)^2}$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered yesterday









            trancelocation

            8,0561519




            8,0561519






















                up vote
                6
                down vote













                Take $f(x) = x sin (1/x)$ near $0$






                share|cite|improve this answer

























                  up vote
                  6
                  down vote













                  Take $f(x) = x sin (1/x)$ near $0$






                  share|cite|improve this answer























                    up vote
                    6
                    down vote










                    up vote
                    6
                    down vote









                    Take $f(x) = x sin (1/x)$ near $0$






                    share|cite|improve this answer












                    Take $f(x) = x sin (1/x)$ near $0$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered yesterday









                    Richard Martin

                    1,3238




                    1,3238






















                        up vote
                        3
                        down vote













                        You may try $f(x)=x^2cos(1/x)$, so that $f'(x)=2xcos(1/x)-sin(1/x)$ has a point of discontinuity at $x=0$.






                        share|cite|improve this answer

















                        • 2




                          This is a slightly better example than mine, in fact, because $f$ is differentiable at zero as well.
                          – Richard Martin
                          yesterday















                        up vote
                        3
                        down vote













                        You may try $f(x)=x^2cos(1/x)$, so that $f'(x)=2xcos(1/x)-sin(1/x)$ has a point of discontinuity at $x=0$.






                        share|cite|improve this answer

















                        • 2




                          This is a slightly better example than mine, in fact, because $f$ is differentiable at zero as well.
                          – Richard Martin
                          yesterday













                        up vote
                        3
                        down vote










                        up vote
                        3
                        down vote









                        You may try $f(x)=x^2cos(1/x)$, so that $f'(x)=2xcos(1/x)-sin(1/x)$ has a point of discontinuity at $x=0$.






                        share|cite|improve this answer












                        You may try $f(x)=x^2cos(1/x)$, so that $f'(x)=2xcos(1/x)-sin(1/x)$ has a point of discontinuity at $x=0$.







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered yesterday









                        GNUSupporter 8964民主女神 地下教會

                        12.3k72343




                        12.3k72343








                        • 2




                          This is a slightly better example than mine, in fact, because $f$ is differentiable at zero as well.
                          – Richard Martin
                          yesterday














                        • 2




                          This is a slightly better example than mine, in fact, because $f$ is differentiable at zero as well.
                          – Richard Martin
                          yesterday








                        2




                        2




                        This is a slightly better example than mine, in fact, because $f$ is differentiable at zero as well.
                        – Richard Martin
                        yesterday




                        This is a slightly better example than mine, in fact, because $f$ is differentiable at zero as well.
                        – Richard Martin
                        yesterday










                        up vote
                        2
                        down vote













                        Does $f(x)=x^frac 12 $ count?
                        $f'(x)=frac 1{2x^frac 12}$ which is discontinuous at $x=0$






                        share|cite|improve this answer

















                        • 1




                          Yes, it does! But see my remark on $x^2 cos 1/x$
                          – Richard Martin
                          yesterday















                        up vote
                        2
                        down vote













                        Does $f(x)=x^frac 12 $ count?
                        $f'(x)=frac 1{2x^frac 12}$ which is discontinuous at $x=0$






                        share|cite|improve this answer

















                        • 1




                          Yes, it does! But see my remark on $x^2 cos 1/x$
                          – Richard Martin
                          yesterday













                        up vote
                        2
                        down vote










                        up vote
                        2
                        down vote









                        Does $f(x)=x^frac 12 $ count?
                        $f'(x)=frac 1{2x^frac 12}$ which is discontinuous at $x=0$






                        share|cite|improve this answer












                        Does $f(x)=x^frac 12 $ count?
                        $f'(x)=frac 1{2x^frac 12}$ which is discontinuous at $x=0$







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered yesterday









                        SmarthBansal

                        36412




                        36412








                        • 1




                          Yes, it does! But see my remark on $x^2 cos 1/x$
                          – Richard Martin
                          yesterday














                        • 1




                          Yes, it does! But see my remark on $x^2 cos 1/x$
                          – Richard Martin
                          yesterday








                        1




                        1




                        Yes, it does! But see my remark on $x^2 cos 1/x$
                        – Richard Martin
                        yesterday




                        Yes, it does! But see my remark on $x^2 cos 1/x$
                        – Richard Martin
                        yesterday










                        up vote
                        0
                        down vote













                        $$ln'(x) = dfrac{1}{x}$$



                        If you are looking for that exact derivative.






                        share|cite|improve this answer

























                          up vote
                          0
                          down vote













                          $$ln'(x) = dfrac{1}{x}$$



                          If you are looking for that exact derivative.






                          share|cite|improve this answer























                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            $$ln'(x) = dfrac{1}{x}$$



                            If you are looking for that exact derivative.






                            share|cite|improve this answer












                            $$ln'(x) = dfrac{1}{x}$$



                            If you are looking for that exact derivative.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered yesterday









                            rustypaper

                            84




                            84






























                                 

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