Is $limlimits_{xto x_0}f'(x)=f'(x_0)$?












6












$begingroup$


Let $f$ be a function defined in the open interval $(a,b)$ and let $x_0in(a,b)$. Suppose in addition that $f'(x)$ exists for all $x_0neq xin(a,b)$. Is the following statement true:




If $limlimits_{xto x_0}f'(x)$ exists, then $f'(x_0)$ exists and $limlimits_{xto x_0}f'(x)=f'(x_0)$.




Thanks!






calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    The statement in your question is saying that $f'(x)$ is continuous at some $x_0$. Now you must ask yourself if every derivative is continuous.
    $endgroup$
    – Masacroso
    Jun 20 '16 at 11:36






  • 1




    $begingroup$
    No, this is weaker than saying $f'$ is continuous.
    $endgroup$
    – GEdgar
    Jun 20 '16 at 21:07
















6












$begingroup$


Let $f$ be a function defined in the open interval $(a,b)$ and let $x_0in(a,b)$. Suppose in addition that $f'(x)$ exists for all $x_0neq xin(a,b)$. Is the following statement true:




If $limlimits_{xto x_0}f'(x)$ exists, then $f'(x_0)$ exists and $limlimits_{xto x_0}f'(x)=f'(x_0)$.




Thanks!






calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    The statement in your question is saying that $f'(x)$ is continuous at some $x_0$. Now you must ask yourself if every derivative is continuous.
    $endgroup$
    – Masacroso
    Jun 20 '16 at 11:36






  • 1




    $begingroup$
    No, this is weaker than saying $f'$ is continuous.
    $endgroup$
    – GEdgar
    Jun 20 '16 at 21:07














6












6








6


2



$begingroup$


Let $f$ be a function defined in the open interval $(a,b)$ and let $x_0in(a,b)$. Suppose in addition that $f'(x)$ exists for all $x_0neq xin(a,b)$. Is the following statement true:




If $limlimits_{xto x_0}f'(x)$ exists, then $f'(x_0)$ exists and $limlimits_{xto x_0}f'(x)=f'(x_0)$.




Thanks!






calculus







share|cite|improve this question











$endgroup$




Let $f$ be a function defined in the open interval $(a,b)$ and let $x_0in(a,b)$. Suppose in addition that $f'(x)$ exists for all $x_0neq xin(a,b)$. Is the following statement true:




If $limlimits_{xto x_0}f'(x)$ exists, then $f'(x_0)$ exists and $limlimits_{xto x_0}f'(x)=f'(x_0)$.




Thanks!






calculus




calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jun 20 '16 at 20:16









snulty

2,86811338




2,86811338










asked Jun 20 '16 at 11:29









boazboaz

2,400614




2,400614












  • $begingroup$
    The statement in your question is saying that $f'(x)$ is continuous at some $x_0$. Now you must ask yourself if every derivative is continuous.
    $endgroup$
    – Masacroso
    Jun 20 '16 at 11:36






  • 1




    $begingroup$
    No, this is weaker than saying $f'$ is continuous.
    $endgroup$
    – GEdgar
    Jun 20 '16 at 21:07


















  • $begingroup$
    The statement in your question is saying that $f'(x)$ is continuous at some $x_0$. Now you must ask yourself if every derivative is continuous.
    $endgroup$
    – Masacroso
    Jun 20 '16 at 11:36






  • 1




    $begingroup$
    No, this is weaker than saying $f'$ is continuous.
    $endgroup$
    – GEdgar
    Jun 20 '16 at 21:07
















$begingroup$
The statement in your question is saying that $f'(x)$ is continuous at some $x_0$. Now you must ask yourself if every derivative is continuous.
$endgroup$
– Masacroso
Jun 20 '16 at 11:36




$begingroup$
The statement in your question is saying that $f'(x)$ is continuous at some $x_0$. Now you must ask yourself if every derivative is continuous.
$endgroup$
– Masacroso
Jun 20 '16 at 11:36




1




1




$begingroup$
No, this is weaker than saying $f'$ is continuous.
$endgroup$
– GEdgar
Jun 20 '16 at 21:07




$begingroup$
No, this is weaker than saying $f'$ is continuous.
$endgroup$
– GEdgar
Jun 20 '16 at 21:07










3 Answers
3






active

oldest

votes


















8












$begingroup$

A qualified "yes": If $f$ is continuous at $x_{0}$, and if $limlimits_{x to x_{0}}f'(x) = L$ exists, then $f$ is differentiable at $x_{0}$, and $f'(x_{0}) = L$.



Qualitatively, the derivative of a continuous function cannot have a removable discontinuity. (If you don't assume $f$ is continuous, then $f$ itself can have a removable or jump discontinuity.)



The claim follows from the Mean Value Theorem: If $delta > 0$ and $f'(x_{0} + h)$ is defined for $0 < |h| < delta$, then for each such $h$, the Mean Value Theorem (applied to $f$ on the closed interval with endpoints $x_{0}$ and $x_{0} + h$) says there is a $t$ between $x_{0}$ and $x_{0} + h$ such that
$$
frac{f(x_{0} + h) - f(x_{0})}{h} = f'(t).
$$
Since $|t - x_{0}| < |h|$, taking the limit as $h to 0$ forces $t - x_{0} to 0$, as well, so
$$
f'(x_{0}) = lim_{h to 0} frac{f(x_{0} + h) - f(x_{0})}{h}
= lim_{t to x_{0}} f'(t).
$$



(Continuity of $f$ was needed to invoke the Mean Value Theorem.)






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    Thanks Andrew, a great answer.
    $endgroup$
    – boaz
    Jun 20 '16 at 12:04










  • $begingroup$
    You're welcome. :) (And, fixed a couple of typos.)
    $endgroup$
    – Andrew D. Hwang
    Jun 20 '16 at 20:03






  • 1




    $begingroup$
    This is a useful fact to know when you do those popular tricky questions like: is $f(x)=x^2sin(x), x ne 0, f(0)=0$ differentiable everywhere?
    $endgroup$
    – GEdgar
    Jun 20 '16 at 21:10












  • $begingroup$
    Since you assume completeness, then you can also get to the same result by using de l'Hôpital's rule to solve the limit. However the statement works also in a non complete field such as Q, I think. Have a look at my my answer, if you like.
    $endgroup$
    – Matteo
    Dec 18 '18 at 9:42





















0












$begingroup$

No, it is not true. For a simple counterexample take the indicator function $chi_{[0,1)}$ in $xin(-1,1)$ and take $x_0$ to be 0.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Not neccessarily true. We need $f'$ to be continuous at $x_0$. Actually the conditions that $lim_{xto x_0} f'(x)$ and $f'(x_0)$ exist and they are equal to each other are the conditions for the function $f'$ to be continous.



    Here's a counter-example: Consider the function:



    $$f(x) = begin{cases} 0, & mbox{if } mbox{$x le 0$} \ x, & mbox{if } mbox{$x>0$} end{cases}$$



    Obviously $f$ is defined on $mathbb{R}$, but the function isn't differentiable at $0$ and $f'$ is discontinuous at $0$. So by choosing $x_0 = 0$ all the conditions are satisfied, but the result isn't, hence the implication is wrong.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      In the proposed example, $f'$ has one-sided limits, but does not have a two-sided limit...?
      $endgroup$
      – Andrew D. Hwang
      Jun 20 '16 at 11:56










    • $begingroup$
      @AndrewD.Hwang Yeah, both one-sided limits exist, but they are not equal to each other, hence the two-sided limit doesn't exists. At the end by checking the graph we can see that it's continuous, therefore both one-sided limits exist, but it's not "smooth" so we expect them not to be equal to each other.
      $endgroup$
      – Stefan4024
      Jun 20 '16 at 11:59






    • 1




      $begingroup$
      Please, change “$f'$ is continuous” into “$f$ is continuous at $x_0$”.
      $endgroup$
      – egreg
      Jun 20 '16 at 21:25






    • 1




      $begingroup$
      This is an incorrect answer to the question. $lim_{xto 0} f'(x)$ does not exist in your example.
      $endgroup$
      – zhw.
      Jun 20 '16 at 21:39











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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    8












    $begingroup$

    A qualified "yes": If $f$ is continuous at $x_{0}$, and if $limlimits_{x to x_{0}}f'(x) = L$ exists, then $f$ is differentiable at $x_{0}$, and $f'(x_{0}) = L$.



    Qualitatively, the derivative of a continuous function cannot have a removable discontinuity. (If you don't assume $f$ is continuous, then $f$ itself can have a removable or jump discontinuity.)



    The claim follows from the Mean Value Theorem: If $delta > 0$ and $f'(x_{0} + h)$ is defined for $0 < |h| < delta$, then for each such $h$, the Mean Value Theorem (applied to $f$ on the closed interval with endpoints $x_{0}$ and $x_{0} + h$) says there is a $t$ between $x_{0}$ and $x_{0} + h$ such that
    $$
    frac{f(x_{0} + h) - f(x_{0})}{h} = f'(t).
    $$
    Since $|t - x_{0}| < |h|$, taking the limit as $h to 0$ forces $t - x_{0} to 0$, as well, so
    $$
    f'(x_{0}) = lim_{h to 0} frac{f(x_{0} + h) - f(x_{0})}{h}
    = lim_{t to x_{0}} f'(t).
    $$



    (Continuity of $f$ was needed to invoke the Mean Value Theorem.)






    share|cite|improve this answer











    $endgroup$









    • 2




      $begingroup$
      Thanks Andrew, a great answer.
      $endgroup$
      – boaz
      Jun 20 '16 at 12:04










    • $begingroup$
      You're welcome. :) (And, fixed a couple of typos.)
      $endgroup$
      – Andrew D. Hwang
      Jun 20 '16 at 20:03






    • 1




      $begingroup$
      This is a useful fact to know when you do those popular tricky questions like: is $f(x)=x^2sin(x), x ne 0, f(0)=0$ differentiable everywhere?
      $endgroup$
      – GEdgar
      Jun 20 '16 at 21:10












    • $begingroup$
      Since you assume completeness, then you can also get to the same result by using de l'Hôpital's rule to solve the limit. However the statement works also in a non complete field such as Q, I think. Have a look at my my answer, if you like.
      $endgroup$
      – Matteo
      Dec 18 '18 at 9:42


















    8












    $begingroup$

    A qualified "yes": If $f$ is continuous at $x_{0}$, and if $limlimits_{x to x_{0}}f'(x) = L$ exists, then $f$ is differentiable at $x_{0}$, and $f'(x_{0}) = L$.



    Qualitatively, the derivative of a continuous function cannot have a removable discontinuity. (If you don't assume $f$ is continuous, then $f$ itself can have a removable or jump discontinuity.)



    The claim follows from the Mean Value Theorem: If $delta > 0$ and $f'(x_{0} + h)$ is defined for $0 < |h| < delta$, then for each such $h$, the Mean Value Theorem (applied to $f$ on the closed interval with endpoints $x_{0}$ and $x_{0} + h$) says there is a $t$ between $x_{0}$ and $x_{0} + h$ such that
    $$
    frac{f(x_{0} + h) - f(x_{0})}{h} = f'(t).
    $$
    Since $|t - x_{0}| < |h|$, taking the limit as $h to 0$ forces $t - x_{0} to 0$, as well, so
    $$
    f'(x_{0}) = lim_{h to 0} frac{f(x_{0} + h) - f(x_{0})}{h}
    = lim_{t to x_{0}} f'(t).
    $$



    (Continuity of $f$ was needed to invoke the Mean Value Theorem.)






    share|cite|improve this answer











    $endgroup$









    • 2




      $begingroup$
      Thanks Andrew, a great answer.
      $endgroup$
      – boaz
      Jun 20 '16 at 12:04










    • $begingroup$
      You're welcome. :) (And, fixed a couple of typos.)
      $endgroup$
      – Andrew D. Hwang
      Jun 20 '16 at 20:03






    • 1




      $begingroup$
      This is a useful fact to know when you do those popular tricky questions like: is $f(x)=x^2sin(x), x ne 0, f(0)=0$ differentiable everywhere?
      $endgroup$
      – GEdgar
      Jun 20 '16 at 21:10












    • $begingroup$
      Since you assume completeness, then you can also get to the same result by using de l'Hôpital's rule to solve the limit. However the statement works also in a non complete field such as Q, I think. Have a look at my my answer, if you like.
      $endgroup$
      – Matteo
      Dec 18 '18 at 9:42
















    8












    8








    8





    $begingroup$

    A qualified "yes": If $f$ is continuous at $x_{0}$, and if $limlimits_{x to x_{0}}f'(x) = L$ exists, then $f$ is differentiable at $x_{0}$, and $f'(x_{0}) = L$.



    Qualitatively, the derivative of a continuous function cannot have a removable discontinuity. (If you don't assume $f$ is continuous, then $f$ itself can have a removable or jump discontinuity.)



    The claim follows from the Mean Value Theorem: If $delta > 0$ and $f'(x_{0} + h)$ is defined for $0 < |h| < delta$, then for each such $h$, the Mean Value Theorem (applied to $f$ on the closed interval with endpoints $x_{0}$ and $x_{0} + h$) says there is a $t$ between $x_{0}$ and $x_{0} + h$ such that
    $$
    frac{f(x_{0} + h) - f(x_{0})}{h} = f'(t).
    $$
    Since $|t - x_{0}| < |h|$, taking the limit as $h to 0$ forces $t - x_{0} to 0$, as well, so
    $$
    f'(x_{0}) = lim_{h to 0} frac{f(x_{0} + h) - f(x_{0})}{h}
    = lim_{t to x_{0}} f'(t).
    $$



    (Continuity of $f$ was needed to invoke the Mean Value Theorem.)






    share|cite|improve this answer











    $endgroup$



    A qualified "yes": If $f$ is continuous at $x_{0}$, and if $limlimits_{x to x_{0}}f'(x) = L$ exists, then $f$ is differentiable at $x_{0}$, and $f'(x_{0}) = L$.



    Qualitatively, the derivative of a continuous function cannot have a removable discontinuity. (If you don't assume $f$ is continuous, then $f$ itself can have a removable or jump discontinuity.)



    The claim follows from the Mean Value Theorem: If $delta > 0$ and $f'(x_{0} + h)$ is defined for $0 < |h| < delta$, then for each such $h$, the Mean Value Theorem (applied to $f$ on the closed interval with endpoints $x_{0}$ and $x_{0} + h$) says there is a $t$ between $x_{0}$ and $x_{0} + h$ such that
    $$
    frac{f(x_{0} + h) - f(x_{0})}{h} = f'(t).
    $$
    Since $|t - x_{0}| < |h|$, taking the limit as $h to 0$ forces $t - x_{0} to 0$, as well, so
    $$
    f'(x_{0}) = lim_{h to 0} frac{f(x_{0} + h) - f(x_{0})}{h}
    = lim_{t to x_{0}} f'(t).
    $$



    (Continuity of $f$ was needed to invoke the Mean Value Theorem.)







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jun 20 '16 at 20:02

























    answered Jun 20 '16 at 11:53









    Andrew D. HwangAndrew D. Hwang

    53.1k448114




    53.1k448114








    • 2




      $begingroup$
      Thanks Andrew, a great answer.
      $endgroup$
      – boaz
      Jun 20 '16 at 12:04










    • $begingroup$
      You're welcome. :) (And, fixed a couple of typos.)
      $endgroup$
      – Andrew D. Hwang
      Jun 20 '16 at 20:03






    • 1




      $begingroup$
      This is a useful fact to know when you do those popular tricky questions like: is $f(x)=x^2sin(x), x ne 0, f(0)=0$ differentiable everywhere?
      $endgroup$
      – GEdgar
      Jun 20 '16 at 21:10












    • $begingroup$
      Since you assume completeness, then you can also get to the same result by using de l'Hôpital's rule to solve the limit. However the statement works also in a non complete field such as Q, I think. Have a look at my my answer, if you like.
      $endgroup$
      – Matteo
      Dec 18 '18 at 9:42
















    • 2




      $begingroup$
      Thanks Andrew, a great answer.
      $endgroup$
      – boaz
      Jun 20 '16 at 12:04










    • $begingroup$
      You're welcome. :) (And, fixed a couple of typos.)
      $endgroup$
      – Andrew D. Hwang
      Jun 20 '16 at 20:03






    • 1




      $begingroup$
      This is a useful fact to know when you do those popular tricky questions like: is $f(x)=x^2sin(x), x ne 0, f(0)=0$ differentiable everywhere?
      $endgroup$
      – GEdgar
      Jun 20 '16 at 21:10












    • $begingroup$
      Since you assume completeness, then you can also get to the same result by using de l'Hôpital's rule to solve the limit. However the statement works also in a non complete field such as Q, I think. Have a look at my my answer, if you like.
      $endgroup$
      – Matteo
      Dec 18 '18 at 9:42










    2




    2




    $begingroup$
    Thanks Andrew, a great answer.
    $endgroup$
    – boaz
    Jun 20 '16 at 12:04




    $begingroup$
    Thanks Andrew, a great answer.
    $endgroup$
    – boaz
    Jun 20 '16 at 12:04












    $begingroup$
    You're welcome. :) (And, fixed a couple of typos.)
    $endgroup$
    – Andrew D. Hwang
    Jun 20 '16 at 20:03




    $begingroup$
    You're welcome. :) (And, fixed a couple of typos.)
    $endgroup$
    – Andrew D. Hwang
    Jun 20 '16 at 20:03




    1




    1




    $begingroup$
    This is a useful fact to know when you do those popular tricky questions like: is $f(x)=x^2sin(x), x ne 0, f(0)=0$ differentiable everywhere?
    $endgroup$
    – GEdgar
    Jun 20 '16 at 21:10






    $begingroup$
    This is a useful fact to know when you do those popular tricky questions like: is $f(x)=x^2sin(x), x ne 0, f(0)=0$ differentiable everywhere?
    $endgroup$
    – GEdgar
    Jun 20 '16 at 21:10














    $begingroup$
    Since you assume completeness, then you can also get to the same result by using de l'Hôpital's rule to solve the limit. However the statement works also in a non complete field such as Q, I think. Have a look at my my answer, if you like.
    $endgroup$
    – Matteo
    Dec 18 '18 at 9:42






    $begingroup$
    Since you assume completeness, then you can also get to the same result by using de l'Hôpital's rule to solve the limit. However the statement works also in a non complete field such as Q, I think. Have a look at my my answer, if you like.
    $endgroup$
    – Matteo
    Dec 18 '18 at 9:42













    0












    $begingroup$

    No, it is not true. For a simple counterexample take the indicator function $chi_{[0,1)}$ in $xin(-1,1)$ and take $x_0$ to be 0.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      No, it is not true. For a simple counterexample take the indicator function $chi_{[0,1)}$ in $xin(-1,1)$ and take $x_0$ to be 0.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        No, it is not true. For a simple counterexample take the indicator function $chi_{[0,1)}$ in $xin(-1,1)$ and take $x_0$ to be 0.






        share|cite|improve this answer









        $endgroup$



        No, it is not true. For a simple counterexample take the indicator function $chi_{[0,1)}$ in $xin(-1,1)$ and take $x_0$ to be 0.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jun 20 '16 at 11:33







        user335721






























            0












            $begingroup$

            Not neccessarily true. We need $f'$ to be continuous at $x_0$. Actually the conditions that $lim_{xto x_0} f'(x)$ and $f'(x_0)$ exist and they are equal to each other are the conditions for the function $f'$ to be continous.



            Here's a counter-example: Consider the function:



            $$f(x) = begin{cases} 0, & mbox{if } mbox{$x le 0$} \ x, & mbox{if } mbox{$x>0$} end{cases}$$



            Obviously $f$ is defined on $mathbb{R}$, but the function isn't differentiable at $0$ and $f'$ is discontinuous at $0$. So by choosing $x_0 = 0$ all the conditions are satisfied, but the result isn't, hence the implication is wrong.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              In the proposed example, $f'$ has one-sided limits, but does not have a two-sided limit...?
              $endgroup$
              – Andrew D. Hwang
              Jun 20 '16 at 11:56










            • $begingroup$
              @AndrewD.Hwang Yeah, both one-sided limits exist, but they are not equal to each other, hence the two-sided limit doesn't exists. At the end by checking the graph we can see that it's continuous, therefore both one-sided limits exist, but it's not "smooth" so we expect them not to be equal to each other.
              $endgroup$
              – Stefan4024
              Jun 20 '16 at 11:59






            • 1




              $begingroup$
              Please, change “$f'$ is continuous” into “$f$ is continuous at $x_0$”.
              $endgroup$
              – egreg
              Jun 20 '16 at 21:25






            • 1




              $begingroup$
              This is an incorrect answer to the question. $lim_{xto 0} f'(x)$ does not exist in your example.
              $endgroup$
              – zhw.
              Jun 20 '16 at 21:39
















            0












            $begingroup$

            Not neccessarily true. We need $f'$ to be continuous at $x_0$. Actually the conditions that $lim_{xto x_0} f'(x)$ and $f'(x_0)$ exist and they are equal to each other are the conditions for the function $f'$ to be continous.



            Here's a counter-example: Consider the function:



            $$f(x) = begin{cases} 0, & mbox{if } mbox{$x le 0$} \ x, & mbox{if } mbox{$x>0$} end{cases}$$



            Obviously $f$ is defined on $mathbb{R}$, but the function isn't differentiable at $0$ and $f'$ is discontinuous at $0$. So by choosing $x_0 = 0$ all the conditions are satisfied, but the result isn't, hence the implication is wrong.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              In the proposed example, $f'$ has one-sided limits, but does not have a two-sided limit...?
              $endgroup$
              – Andrew D. Hwang
              Jun 20 '16 at 11:56










            • $begingroup$
              @AndrewD.Hwang Yeah, both one-sided limits exist, but they are not equal to each other, hence the two-sided limit doesn't exists. At the end by checking the graph we can see that it's continuous, therefore both one-sided limits exist, but it's not "smooth" so we expect them not to be equal to each other.
              $endgroup$
              – Stefan4024
              Jun 20 '16 at 11:59






            • 1




              $begingroup$
              Please, change “$f'$ is continuous” into “$f$ is continuous at $x_0$”.
              $endgroup$
              – egreg
              Jun 20 '16 at 21:25






            • 1




              $begingroup$
              This is an incorrect answer to the question. $lim_{xto 0} f'(x)$ does not exist in your example.
              $endgroup$
              – zhw.
              Jun 20 '16 at 21:39














            0












            0








            0





            $begingroup$

            Not neccessarily true. We need $f'$ to be continuous at $x_0$. Actually the conditions that $lim_{xto x_0} f'(x)$ and $f'(x_0)$ exist and they are equal to each other are the conditions for the function $f'$ to be continous.



            Here's a counter-example: Consider the function:



            $$f(x) = begin{cases} 0, & mbox{if } mbox{$x le 0$} \ x, & mbox{if } mbox{$x>0$} end{cases}$$



            Obviously $f$ is defined on $mathbb{R}$, but the function isn't differentiable at $0$ and $f'$ is discontinuous at $0$. So by choosing $x_0 = 0$ all the conditions are satisfied, but the result isn't, hence the implication is wrong.






            share|cite|improve this answer











            $endgroup$



            Not neccessarily true. We need $f'$ to be continuous at $x_0$. Actually the conditions that $lim_{xto x_0} f'(x)$ and $f'(x_0)$ exist and they are equal to each other are the conditions for the function $f'$ to be continous.



            Here's a counter-example: Consider the function:



            $$f(x) = begin{cases} 0, & mbox{if } mbox{$x le 0$} \ x, & mbox{if } mbox{$x>0$} end{cases}$$



            Obviously $f$ is defined on $mathbb{R}$, but the function isn't differentiable at $0$ and $f'$ is discontinuous at $0$. So by choosing $x_0 = 0$ all the conditions are satisfied, but the result isn't, hence the implication is wrong.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jun 20 '16 at 21:26

























            answered Jun 20 '16 at 11:35









            Stefan4024Stefan4024

            30.6k63479




            30.6k63479












            • $begingroup$
              In the proposed example, $f'$ has one-sided limits, but does not have a two-sided limit...?
              $endgroup$
              – Andrew D. Hwang
              Jun 20 '16 at 11:56










            • $begingroup$
              @AndrewD.Hwang Yeah, both one-sided limits exist, but they are not equal to each other, hence the two-sided limit doesn't exists. At the end by checking the graph we can see that it's continuous, therefore both one-sided limits exist, but it's not "smooth" so we expect them not to be equal to each other.
              $endgroup$
              – Stefan4024
              Jun 20 '16 at 11:59






            • 1




              $begingroup$
              Please, change “$f'$ is continuous” into “$f$ is continuous at $x_0$”.
              $endgroup$
              – egreg
              Jun 20 '16 at 21:25






            • 1




              $begingroup$
              This is an incorrect answer to the question. $lim_{xto 0} f'(x)$ does not exist in your example.
              $endgroup$
              – zhw.
              Jun 20 '16 at 21:39


















            • $begingroup$
              In the proposed example, $f'$ has one-sided limits, but does not have a two-sided limit...?
              $endgroup$
              – Andrew D. Hwang
              Jun 20 '16 at 11:56










            • $begingroup$
              @AndrewD.Hwang Yeah, both one-sided limits exist, but they are not equal to each other, hence the two-sided limit doesn't exists. At the end by checking the graph we can see that it's continuous, therefore both one-sided limits exist, but it's not "smooth" so we expect them not to be equal to each other.
              $endgroup$
              – Stefan4024
              Jun 20 '16 at 11:59






            • 1




              $begingroup$
              Please, change “$f'$ is continuous” into “$f$ is continuous at $x_0$”.
              $endgroup$
              – egreg
              Jun 20 '16 at 21:25






            • 1




              $begingroup$
              This is an incorrect answer to the question. $lim_{xto 0} f'(x)$ does not exist in your example.
              $endgroup$
              – zhw.
              Jun 20 '16 at 21:39
















            $begingroup$
            In the proposed example, $f'$ has one-sided limits, but does not have a two-sided limit...?
            $endgroup$
            – Andrew D. Hwang
            Jun 20 '16 at 11:56




            $begingroup$
            In the proposed example, $f'$ has one-sided limits, but does not have a two-sided limit...?
            $endgroup$
            – Andrew D. Hwang
            Jun 20 '16 at 11:56












            $begingroup$
            @AndrewD.Hwang Yeah, both one-sided limits exist, but they are not equal to each other, hence the two-sided limit doesn't exists. At the end by checking the graph we can see that it's continuous, therefore both one-sided limits exist, but it's not "smooth" so we expect them not to be equal to each other.
            $endgroup$
            – Stefan4024
            Jun 20 '16 at 11:59




            $begingroup$
            @AndrewD.Hwang Yeah, both one-sided limits exist, but they are not equal to each other, hence the two-sided limit doesn't exists. At the end by checking the graph we can see that it's continuous, therefore both one-sided limits exist, but it's not "smooth" so we expect them not to be equal to each other.
            $endgroup$
            – Stefan4024
            Jun 20 '16 at 11:59




            1




            1




            $begingroup$
            Please, change “$f'$ is continuous” into “$f$ is continuous at $x_0$”.
            $endgroup$
            – egreg
            Jun 20 '16 at 21:25




            $begingroup$
            Please, change “$f'$ is continuous” into “$f$ is continuous at $x_0$”.
            $endgroup$
            – egreg
            Jun 20 '16 at 21:25




            1




            1




            $begingroup$
            This is an incorrect answer to the question. $lim_{xto 0} f'(x)$ does not exist in your example.
            $endgroup$
            – zhw.
            Jun 20 '16 at 21:39




            $begingroup$
            This is an incorrect answer to the question. $lim_{xto 0} f'(x)$ does not exist in your example.
            $endgroup$
            – zhw.
            Jun 20 '16 at 21:39


















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