Why is this not correct？












5












$begingroup$


We are assigned to deal with the following task.




Assume that $f(x)$ is derivable for any $x in mathbb{R}$. We want to
research $limlimits_{x to x_0}f'(x)$ where $x_0 in mathbb{R}$.




Notice that
begin{align*}
f'(x_0)=lim_{x to x_0}frac{f(x)-f(x_0)}{x-x_0}=lim_{x to x_0}frac{f'(xi)(x-x_0)}{x-x_0}=lim_{x to x_0}f'(xi),
end{align*}
where we applied Lagrange's Mean Value Theorem, and $ x_0 lessgtr xi lessgtr x.$ Since $xi$ is squeezed by $x_0$ and $x$, then $x to x_0$ implies $xi to x_0$. Thus
$$f'(x_0)=lim_{x to x_0}f'(xi)=lim_{xi to x_0}f'(xi).$$



What does this say? It shows that $f'(x)$ is always continuous at any point $x=x_0$, which is an absurd conclusion, because we know safely $f'(x)$ may probably has the discontinuity point (of the second kind). But where dose the mistake occur during the reasoning above?






calculus







share|cite|improve this question









$endgroup$












  • $begingroup$
    Recall that $xi$ is actually a function of $x$. Thus, $lim_{xrightarrow x_0} f'(xi(x))$ does not imply $lim_{xi rightarrow x_0} f'(xi)$
    $endgroup$
    – Severin Schraven
    Dec 15 '18 at 9:08












  • $begingroup$
    it is just a $ksi$ not every $x$...
    $endgroup$
    – dmtri
    Dec 15 '18 at 9:09










  • $begingroup$
    @SeverinSchraven Do you know the squeeze theorem? $x_0 lessgtr xi(x) lessgtr x$. Let $x to x_0$，then $xi(x) to x_0$. It has no fault here.
    $endgroup$
    – mengdie1982
    Dec 15 '18 at 9:25






  • 2




    $begingroup$
    Reread my comment, I did NOT claim that $xi(x) rightarrow x_0$ is wrong. Your line of reasoning is the following: Let $$ f(x) = begin{cases} 0;& xin mathbb{Q} \ 1,& xin mathbb{R}setminus mathbb{Q} end{cases} $$ now take $x_0=0$ and $xi(x)$ be a rational number between $0$ and $x$. Clearly we have $lim_{xrightarrow 0} f(xi(x))=0$ and thus $lim_{xi rightarrow 0} f(xi)= lim_{x rightarrow 0} f(xi(x))$. But $lim_{xi rightarrow 0} f(xi)$ does not even exist. That is, because you do not hit every number with $xi(x)$.
    $endgroup$
    – Severin Schraven
    Dec 15 '18 at 9:32












  • $begingroup$
    @SeverinSchraven You have pointed out some essential thing！
    $endgroup$
    – mengdie1982
    Dec 15 '18 at 9:36
















5












$begingroup$


We are assigned to deal with the following task.




Assume that $f(x)$ is derivable for any $x in mathbb{R}$. We want to
research $limlimits_{x to x_0}f'(x)$ where $x_0 in mathbb{R}$.




Notice that
begin{align*}
f'(x_0)=lim_{x to x_0}frac{f(x)-f(x_0)}{x-x_0}=lim_{x to x_0}frac{f'(xi)(x-x_0)}{x-x_0}=lim_{x to x_0}f'(xi),
end{align*}
where we applied Lagrange's Mean Value Theorem, and $ x_0 lessgtr xi lessgtr x.$ Since $xi$ is squeezed by $x_0$ and $x$, then $x to x_0$ implies $xi to x_0$. Thus
$$f'(x_0)=lim_{x to x_0}f'(xi)=lim_{xi to x_0}f'(xi).$$



What does this say? It shows that $f'(x)$ is always continuous at any point $x=x_0$, which is an absurd conclusion, because we know safely $f'(x)$ may probably has the discontinuity point (of the second kind). But where dose the mistake occur during the reasoning above?






calculus







share|cite|improve this question









$endgroup$












  • $begingroup$
    Recall that $xi$ is actually a function of $x$. Thus, $lim_{xrightarrow x_0} f'(xi(x))$ does not imply $lim_{xi rightarrow x_0} f'(xi)$
    $endgroup$
    – Severin Schraven
    Dec 15 '18 at 9:08












  • $begingroup$
    it is just a $ksi$ not every $x$...
    $endgroup$
    – dmtri
    Dec 15 '18 at 9:09










  • $begingroup$
    @SeverinSchraven Do you know the squeeze theorem? $x_0 lessgtr xi(x) lessgtr x$. Let $x to x_0$，then $xi(x) to x_0$. It has no fault here.
    $endgroup$
    – mengdie1982
    Dec 15 '18 at 9:25






  • 2




    $begingroup$
    Reread my comment, I did NOT claim that $xi(x) rightarrow x_0$ is wrong. Your line of reasoning is the following: Let $$ f(x) = begin{cases} 0;& xin mathbb{Q} \ 1,& xin mathbb{R}setminus mathbb{Q} end{cases} $$ now take $x_0=0$ and $xi(x)$ be a rational number between $0$ and $x$. Clearly we have $lim_{xrightarrow 0} f(xi(x))=0$ and thus $lim_{xi rightarrow 0} f(xi)= lim_{x rightarrow 0} f(xi(x))$. But $lim_{xi rightarrow 0} f(xi)$ does not even exist. That is, because you do not hit every number with $xi(x)$.
    $endgroup$
    – Severin Schraven
    Dec 15 '18 at 9:32












  • $begingroup$
    @SeverinSchraven You have pointed out some essential thing！
    $endgroup$
    – mengdie1982
    Dec 15 '18 at 9:36














5












5








5





$begingroup$


We are assigned to deal with the following task.




Assume that $f(x)$ is derivable for any $x in mathbb{R}$. We want to
research $limlimits_{x to x_0}f'(x)$ where $x_0 in mathbb{R}$.




Notice that
begin{align*}
f'(x_0)=lim_{x to x_0}frac{f(x)-f(x_0)}{x-x_0}=lim_{x to x_0}frac{f'(xi)(x-x_0)}{x-x_0}=lim_{x to x_0}f'(xi),
end{align*}
where we applied Lagrange's Mean Value Theorem, and $ x_0 lessgtr xi lessgtr x.$ Since $xi$ is squeezed by $x_0$ and $x$, then $x to x_0$ implies $xi to x_0$. Thus
$$f'(x_0)=lim_{x to x_0}f'(xi)=lim_{xi to x_0}f'(xi).$$



What does this say? It shows that $f'(x)$ is always continuous at any point $x=x_0$, which is an absurd conclusion, because we know safely $f'(x)$ may probably has the discontinuity point (of the second kind). But where dose the mistake occur during the reasoning above?






calculus







share|cite|improve this question









$endgroup$




We are assigned to deal with the following task.




Assume that $f(x)$ is derivable for any $x in mathbb{R}$. We want to
research $limlimits_{x to x_0}f'(x)$ where $x_0 in mathbb{R}$.




Notice that
begin{align*}
f'(x_0)=lim_{x to x_0}frac{f(x)-f(x_0)}{x-x_0}=lim_{x to x_0}frac{f'(xi)(x-x_0)}{x-x_0}=lim_{x to x_0}f'(xi),
end{align*}
where we applied Lagrange's Mean Value Theorem, and $ x_0 lessgtr xi lessgtr x.$ Since $xi$ is squeezed by $x_0$ and $x$, then $x to x_0$ implies $xi to x_0$. Thus
$$f'(x_0)=lim_{x to x_0}f'(xi)=lim_{xi to x_0}f'(xi).$$



What does this say? It shows that $f'(x)$ is always continuous at any point $x=x_0$, which is an absurd conclusion, because we know safely $f'(x)$ may probably has the discontinuity point (of the second kind). But where dose the mistake occur during the reasoning above?






calculus




calculus






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 15 '18 at 9:03









mengdie1982mengdie1982

4,932618




4,932618












  • $begingroup$
    Recall that $xi$ is actually a function of $x$. Thus, $lim_{xrightarrow x_0} f'(xi(x))$ does not imply $lim_{xi rightarrow x_0} f'(xi)$
    $endgroup$
    – Severin Schraven
    Dec 15 '18 at 9:08












  • $begingroup$
    it is just a $ksi$ not every $x$...
    $endgroup$
    – dmtri
    Dec 15 '18 at 9:09










  • $begingroup$
    @SeverinSchraven Do you know the squeeze theorem? $x_0 lessgtr xi(x) lessgtr x$. Let $x to x_0$，then $xi(x) to x_0$. It has no fault here.
    $endgroup$
    – mengdie1982
    Dec 15 '18 at 9:25






  • 2




    $begingroup$
    Reread my comment, I did NOT claim that $xi(x) rightarrow x_0$ is wrong. Your line of reasoning is the following: Let $$ f(x) = begin{cases} 0;& xin mathbb{Q} \ 1,& xin mathbb{R}setminus mathbb{Q} end{cases} $$ now take $x_0=0$ and $xi(x)$ be a rational number between $0$ and $x$. Clearly we have $lim_{xrightarrow 0} f(xi(x))=0$ and thus $lim_{xi rightarrow 0} f(xi)= lim_{x rightarrow 0} f(xi(x))$. But $lim_{xi rightarrow 0} f(xi)$ does not even exist. That is, because you do not hit every number with $xi(x)$.
    $endgroup$
    – Severin Schraven
    Dec 15 '18 at 9:32












  • $begingroup$
    @SeverinSchraven You have pointed out some essential thing！
    $endgroup$
    – mengdie1982
    Dec 15 '18 at 9:36


















  • $begingroup$
    Recall that $xi$ is actually a function of $x$. Thus, $lim_{xrightarrow x_0} f'(xi(x))$ does not imply $lim_{xi rightarrow x_0} f'(xi)$
    $endgroup$
    – Severin Schraven
    Dec 15 '18 at 9:08












  • $begingroup$
    it is just a $ksi$ not every $x$...
    $endgroup$
    – dmtri
    Dec 15 '18 at 9:09










  • $begingroup$
    @SeverinSchraven Do you know the squeeze theorem? $x_0 lessgtr xi(x) lessgtr x$. Let $x to x_0$，then $xi(x) to x_0$. It has no fault here.
    $endgroup$
    – mengdie1982
    Dec 15 '18 at 9:25






  • 2




    $begingroup$
    Reread my comment, I did NOT claim that $xi(x) rightarrow x_0$ is wrong. Your line of reasoning is the following: Let $$ f(x) = begin{cases} 0;& xin mathbb{Q} \ 1,& xin mathbb{R}setminus mathbb{Q} end{cases} $$ now take $x_0=0$ and $xi(x)$ be a rational number between $0$ and $x$. Clearly we have $lim_{xrightarrow 0} f(xi(x))=0$ and thus $lim_{xi rightarrow 0} f(xi)= lim_{x rightarrow 0} f(xi(x))$. But $lim_{xi rightarrow 0} f(xi)$ does not even exist. That is, because you do not hit every number with $xi(x)$.
    $endgroup$
    – Severin Schraven
    Dec 15 '18 at 9:32












  • $begingroup$
    @SeverinSchraven You have pointed out some essential thing！
    $endgroup$
    – mengdie1982
    Dec 15 '18 at 9:36
















$begingroup$
Recall that $xi$ is actually a function of $x$. Thus, $lim_{xrightarrow x_0} f'(xi(x))$ does not imply $lim_{xi rightarrow x_0} f'(xi)$
$endgroup$
– Severin Schraven
Dec 15 '18 at 9:08






$begingroup$
Recall that $xi$ is actually a function of $x$. Thus, $lim_{xrightarrow x_0} f'(xi(x))$ does not imply $lim_{xi rightarrow x_0} f'(xi)$
$endgroup$
– Severin Schraven
Dec 15 '18 at 9:08














$begingroup$
it is just a $ksi$ not every $x$...
$endgroup$
– dmtri
Dec 15 '18 at 9:09




$begingroup$
it is just a $ksi$ not every $x$...
$endgroup$
– dmtri
Dec 15 '18 at 9:09












$begingroup$
@SeverinSchraven Do you know the squeeze theorem? $x_0 lessgtr xi(x) lessgtr x$. Let $x to x_0$，then $xi(x) to x_0$. It has no fault here.
$endgroup$
– mengdie1982
Dec 15 '18 at 9:25




$begingroup$
@SeverinSchraven Do you know the squeeze theorem? $x_0 lessgtr xi(x) lessgtr x$. Let $x to x_0$，then $xi(x) to x_0$. It has no fault here.
$endgroup$
– mengdie1982
Dec 15 '18 at 9:25




2




2




$begingroup$
Reread my comment, I did NOT claim that $xi(x) rightarrow x_0$ is wrong. Your line of reasoning is the following: Let $$ f(x) = begin{cases} 0;& xin mathbb{Q} \ 1,& xin mathbb{R}setminus mathbb{Q} end{cases} $$ now take $x_0=0$ and $xi(x)$ be a rational number between $0$ and $x$. Clearly we have $lim_{xrightarrow 0} f(xi(x))=0$ and thus $lim_{xi rightarrow 0} f(xi)= lim_{x rightarrow 0} f(xi(x))$. But $lim_{xi rightarrow 0} f(xi)$ does not even exist. That is, because you do not hit every number with $xi(x)$.
$endgroup$
– Severin Schraven
Dec 15 '18 at 9:32






$begingroup$
Reread my comment, I did NOT claim that $xi(x) rightarrow x_0$ is wrong. Your line of reasoning is the following: Let $$ f(x) = begin{cases} 0;& xin mathbb{Q} \ 1,& xin mathbb{R}setminus mathbb{Q} end{cases} $$ now take $x_0=0$ and $xi(x)$ be a rational number between $0$ and $x$. Clearly we have $lim_{xrightarrow 0} f(xi(x))=0$ and thus $lim_{xi rightarrow 0} f(xi)= lim_{x rightarrow 0} f(xi(x))$. But $lim_{xi rightarrow 0} f(xi)$ does not even exist. That is, because you do not hit every number with $xi(x)$.
$endgroup$
– Severin Schraven
Dec 15 '18 at 9:32














$begingroup$
@SeverinSchraven You have pointed out some essential thing！
$endgroup$
– mengdie1982
Dec 15 '18 at 9:36




$begingroup$
@SeverinSchraven You have pointed out some essential thing！
$endgroup$
– mengdie1982
Dec 15 '18 at 9:36










3 Answers
3






active

oldest

votes


















2












$begingroup$

Your reasoning only shows that if $lim_{x to x_0} f'(x)$ exists, then it's equal to $f'(x_0)$. But, as you say, it may not exist.



This is a rather classical exercise, which has been treated many times on this site, for example in this question.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Actually，we can make a comment for the reasoning like this:



    The fact that $limlimits_{xi to x_0}f'(xi)$ exists dose not imply $limlimits_{x to x_0}f'(x)$ also exsits, because, according to Heine's Theorem, the latter one necessitates that $f'(x_n)$ converges for any sequence $x_n to x_0$. As we can see, $xi_n$ is only a specific sequence. Even though $f'(xi_n)$ is convergent, this is not enough to guarantee $f'(x_n)$ converges as well.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      You mean $lim_{xto x_0}f'(xi)$ instead of $lim_{xito x_0}f'(xi)$.
      $endgroup$
      – Paramanand Singh
      Dec 20 '18 at 9:42



















    0












    $begingroup$

    Actually, you are right, this also called 'theorem of derivative limitation' in some textbooks.
    If you already assumed that $f(x)$ is derivable for any $x in mathbb{R}$.



    With an additional condition the limitation of derivative $lim_{x to x_{0}}f'(x)$ exist.



    You can write your conclusion:



    $$lim_{x to x_{0}}f'(x)=f'(x_{0})$$



    The condition $f(x)$ is derivable for any $x in mathbb{R}$ can also be weaken to $f(x)$ is continuous in $U(x_{0})$ which is the neighborhood of $x_{0}$, and derivable in $U°(x_{0})$ which is the punctured neighborhood of $x_{0}$.



    The only problem is that you have not assumed the limitation of your derivative should always exist, which means the derivative can not divergence to infinity, like you say sometimes it would be the discontinuity point of the second kind.



    And this theorem can be simply remembered as in other 'not so strict (please notice the comment below)' ways like:



    If a continuous function has (finite) derivative, its derivative function is also continuous. (but maybe not derivable)






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      It would be fair to say that you're wrong. The fact $f(x)$ is derivable does not imply $f'(x)$ is continuous. Here is a counterexample: $$f(x)=begin{cases}x^2sindfrac{1}{x},&x neq 0,\0,&x=0.end{cases}.$$ You may verify that $f(x)$ is derivable over $(-infty,+infty)$，but $f'(x)$ is not continuous at $x=0$.
      $endgroup$
      – mengdie1982
      Dec 15 '18 at 16:16












    • $begingroup$
      But of course, if we add a condition that $limlimits_{x to x_0}f'(x)$ exists, then $limlimits_{x to x_0}f'(x)=f'(x_0)$. In another word, $f'(x)$ is continuous at $x=x_0$ now！
      $endgroup$
      – mengdie1982
      Dec 15 '18 at 16:24












    • $begingroup$
      @mengdie1982 You still remind a good example for me. Yes, I should explain that the 'not so strict' place this the limitation of derivative must be exist not only the derivative exist. Your comment point out it.
      $endgroup$
      – Nanayajitzuki
      Dec 15 '18 at 16:30











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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Your reasoning only shows that if $lim_{x to x_0} f'(x)$ exists, then it's equal to $f'(x_0)$. But, as you say, it may not exist.



    This is a rather classical exercise, which has been treated many times on this site, for example in this question.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Your reasoning only shows that if $lim_{x to x_0} f'(x)$ exists, then it's equal to $f'(x_0)$. But, as you say, it may not exist.



      This is a rather classical exercise, which has been treated many times on this site, for example in this question.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Your reasoning only shows that if $lim_{x to x_0} f'(x)$ exists, then it's equal to $f'(x_0)$. But, as you say, it may not exist.



        This is a rather classical exercise, which has been treated many times on this site, for example in this question.






        share|cite|improve this answer









        $endgroup$



        Your reasoning only shows that if $lim_{x to x_0} f'(x)$ exists, then it's equal to $f'(x_0)$. But, as you say, it may not exist.



        This is a rather classical exercise, which has been treated many times on this site, for example in this question.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 15 '18 at 9:48









        Hans LundmarkHans Lundmark

        35.8k564115




        35.8k564115























            2












            $begingroup$

            Actually，we can make a comment for the reasoning like this:



            The fact that $limlimits_{xi to x_0}f'(xi)$ exists dose not imply $limlimits_{x to x_0}f'(x)$ also exsits, because, according to Heine's Theorem, the latter one necessitates that $f'(x_n)$ converges for any sequence $x_n to x_0$. As we can see, $xi_n$ is only a specific sequence. Even though $f'(xi_n)$ is convergent, this is not enough to guarantee $f'(x_n)$ converges as well.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              You mean $lim_{xto x_0}f'(xi)$ instead of $lim_{xito x_0}f'(xi)$.
              $endgroup$
              – Paramanand Singh
              Dec 20 '18 at 9:42
















            2












            $begingroup$

            Actually，we can make a comment for the reasoning like this:



            The fact that $limlimits_{xi to x_0}f'(xi)$ exists dose not imply $limlimits_{x to x_0}f'(x)$ also exsits, because, according to Heine's Theorem, the latter one necessitates that $f'(x_n)$ converges for any sequence $x_n to x_0$. As we can see, $xi_n$ is only a specific sequence. Even though $f'(xi_n)$ is convergent, this is not enough to guarantee $f'(x_n)$ converges as well.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              You mean $lim_{xto x_0}f'(xi)$ instead of $lim_{xito x_0}f'(xi)$.
              $endgroup$
              – Paramanand Singh
              Dec 20 '18 at 9:42














            2












            2








            2





            $begingroup$

            Actually，we can make a comment for the reasoning like this:



            The fact that $limlimits_{xi to x_0}f'(xi)$ exists dose not imply $limlimits_{x to x_0}f'(x)$ also exsits, because, according to Heine's Theorem, the latter one necessitates that $f'(x_n)$ converges for any sequence $x_n to x_0$. As we can see, $xi_n$ is only a specific sequence. Even though $f'(xi_n)$ is convergent, this is not enough to guarantee $f'(x_n)$ converges as well.






            share|cite|improve this answer









            $endgroup$



            Actually，we can make a comment for the reasoning like this:



            The fact that $limlimits_{xi to x_0}f'(xi)$ exists dose not imply $limlimits_{x to x_0}f'(x)$ also exsits, because, according to Heine's Theorem, the latter one necessitates that $f'(x_n)$ converges for any sequence $x_n to x_0$. As we can see, $xi_n$ is only a specific sequence. Even though $f'(xi_n)$ is convergent, this is not enough to guarantee $f'(x_n)$ converges as well.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 15 '18 at 11:36









            mengdie1982mengdie1982

            4,932618




            4,932618












            • $begingroup$
              You mean $lim_{xto x_0}f'(xi)$ instead of $lim_{xito x_0}f'(xi)$.
              $endgroup$
              – Paramanand Singh
              Dec 20 '18 at 9:42


















            • $begingroup$
              You mean $lim_{xto x_0}f'(xi)$ instead of $lim_{xito x_0}f'(xi)$.
              $endgroup$
              – Paramanand Singh
              Dec 20 '18 at 9:42
















            $begingroup$
            You mean $lim_{xto x_0}f'(xi)$ instead of $lim_{xito x_0}f'(xi)$.
            $endgroup$
            – Paramanand Singh
            Dec 20 '18 at 9:42




            $begingroup$
            You mean $lim_{xto x_0}f'(xi)$ instead of $lim_{xito x_0}f'(xi)$.
            $endgroup$
            – Paramanand Singh
            Dec 20 '18 at 9:42











            0












            $begingroup$

            Actually, you are right, this also called 'theorem of derivative limitation' in some textbooks.
            If you already assumed that $f(x)$ is derivable for any $x in mathbb{R}$.



            With an additional condition the limitation of derivative $lim_{x to x_{0}}f'(x)$ exist.



            You can write your conclusion:



            $$lim_{x to x_{0}}f'(x)=f'(x_{0})$$



            The condition $f(x)$ is derivable for any $x in mathbb{R}$ can also be weaken to $f(x)$ is continuous in $U(x_{0})$ which is the neighborhood of $x_{0}$, and derivable in $U°(x_{0})$ which is the punctured neighborhood of $x_{0}$.



            The only problem is that you have not assumed the limitation of your derivative should always exist, which means the derivative can not divergence to infinity, like you say sometimes it would be the discontinuity point of the second kind.



            And this theorem can be simply remembered as in other 'not so strict (please notice the comment below)' ways like:



            If a continuous function has (finite) derivative, its derivative function is also continuous. (but maybe not derivable)






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              It would be fair to say that you're wrong. The fact $f(x)$ is derivable does not imply $f'(x)$ is continuous. Here is a counterexample: $$f(x)=begin{cases}x^2sindfrac{1}{x},&x neq 0,\0,&x=0.end{cases}.$$ You may verify that $f(x)$ is derivable over $(-infty,+infty)$，but $f'(x)$ is not continuous at $x=0$.
              $endgroup$
              – mengdie1982
              Dec 15 '18 at 16:16












            • $begingroup$
              But of course, if we add a condition that $limlimits_{x to x_0}f'(x)$ exists, then $limlimits_{x to x_0}f'(x)=f'(x_0)$. In another word, $f'(x)$ is continuous at $x=x_0$ now！
              $endgroup$
              – mengdie1982
              Dec 15 '18 at 16:24












            • $begingroup$
              @mengdie1982 You still remind a good example for me. Yes, I should explain that the 'not so strict' place this the limitation of derivative must be exist not only the derivative exist. Your comment point out it.
              $endgroup$
              – Nanayajitzuki
              Dec 15 '18 at 16:30
















            0












            $begingroup$

            Actually, you are right, this also called 'theorem of derivative limitation' in some textbooks.
            If you already assumed that $f(x)$ is derivable for any $x in mathbb{R}$.



            With an additional condition the limitation of derivative $lim_{x to x_{0}}f'(x)$ exist.



            You can write your conclusion:



            $$lim_{x to x_{0}}f'(x)=f'(x_{0})$$



            The condition $f(x)$ is derivable for any $x in mathbb{R}$ can also be weaken to $f(x)$ is continuous in $U(x_{0})$ which is the neighborhood of $x_{0}$, and derivable in $U°(x_{0})$ which is the punctured neighborhood of $x_{0}$.



            The only problem is that you have not assumed the limitation of your derivative should always exist, which means the derivative can not divergence to infinity, like you say sometimes it would be the discontinuity point of the second kind.



            And this theorem can be simply remembered as in other 'not so strict (please notice the comment below)' ways like:



            If a continuous function has (finite) derivative, its derivative function is also continuous. (but maybe not derivable)






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              It would be fair to say that you're wrong. The fact $f(x)$ is derivable does not imply $f'(x)$ is continuous. Here is a counterexample: $$f(x)=begin{cases}x^2sindfrac{1}{x},&x neq 0,\0,&x=0.end{cases}.$$ You may verify that $f(x)$ is derivable over $(-infty,+infty)$，but $f'(x)$ is not continuous at $x=0$.
              $endgroup$
              – mengdie1982
              Dec 15 '18 at 16:16












            • $begingroup$
              But of course, if we add a condition that $limlimits_{x to x_0}f'(x)$ exists, then $limlimits_{x to x_0}f'(x)=f'(x_0)$. In another word, $f'(x)$ is continuous at $x=x_0$ now！
              $endgroup$
              – mengdie1982
              Dec 15 '18 at 16:24












            • $begingroup$
              @mengdie1982 You still remind a good example for me. Yes, I should explain that the 'not so strict' place this the limitation of derivative must be exist not only the derivative exist. Your comment point out it.
              $endgroup$
              – Nanayajitzuki
              Dec 15 '18 at 16:30














            0












            0








            0





            $begingroup$

            Actually, you are right, this also called 'theorem of derivative limitation' in some textbooks.
            If you already assumed that $f(x)$ is derivable for any $x in mathbb{R}$.



            With an additional condition the limitation of derivative $lim_{x to x_{0}}f'(x)$ exist.



            You can write your conclusion:



            $$lim_{x to x_{0}}f'(x)=f'(x_{0})$$



            The condition $f(x)$ is derivable for any $x in mathbb{R}$ can also be weaken to $f(x)$ is continuous in $U(x_{0})$ which is the neighborhood of $x_{0}$, and derivable in $U°(x_{0})$ which is the punctured neighborhood of $x_{0}$.



            The only problem is that you have not assumed the limitation of your derivative should always exist, which means the derivative can not divergence to infinity, like you say sometimes it would be the discontinuity point of the second kind.



            And this theorem can be simply remembered as in other 'not so strict (please notice the comment below)' ways like:



            If a continuous function has (finite) derivative, its derivative function is also continuous. (but maybe not derivable)






            share|cite|improve this answer











            $endgroup$



            Actually, you are right, this also called 'theorem of derivative limitation' in some textbooks.
            If you already assumed that $f(x)$ is derivable for any $x in mathbb{R}$.



            With an additional condition the limitation of derivative $lim_{x to x_{0}}f'(x)$ exist.



            You can write your conclusion:



            $$lim_{x to x_{0}}f'(x)=f'(x_{0})$$



            The condition $f(x)$ is derivable for any $x in mathbb{R}$ can also be weaken to $f(x)$ is continuous in $U(x_{0})$ which is the neighborhood of $x_{0}$, and derivable in $U°(x_{0})$ which is the punctured neighborhood of $x_{0}$.



            The only problem is that you have not assumed the limitation of your derivative should always exist, which means the derivative can not divergence to infinity, like you say sometimes it would be the discontinuity point of the second kind.



            And this theorem can be simply remembered as in other 'not so strict (please notice the comment below)' ways like:



            If a continuous function has (finite) derivative, its derivative function is also continuous. (but maybe not derivable)







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 15 '18 at 16:54

























            answered Dec 15 '18 at 15:20









            NanayajitzukiNanayajitzuki

            3185




            3185












            • $begingroup$
              It would be fair to say that you're wrong. The fact $f(x)$ is derivable does not imply $f'(x)$ is continuous. Here is a counterexample: $$f(x)=begin{cases}x^2sindfrac{1}{x},&x neq 0,\0,&x=0.end{cases}.$$ You may verify that $f(x)$ is derivable over $(-infty,+infty)$，but $f'(x)$ is not continuous at $x=0$.
              $endgroup$
              – mengdie1982
              Dec 15 '18 at 16:16












            • $begingroup$
              But of course, if we add a condition that $limlimits_{x to x_0}f'(x)$ exists, then $limlimits_{x to x_0}f'(x)=f'(x_0)$. In another word, $f'(x)$ is continuous at $x=x_0$ now！
              $endgroup$
              – mengdie1982
              Dec 15 '18 at 16:24












            • $begingroup$
              @mengdie1982 You still remind a good example for me. Yes, I should explain that the 'not so strict' place this the limitation of derivative must be exist not only the derivative exist. Your comment point out it.
              $endgroup$
              – Nanayajitzuki
              Dec 15 '18 at 16:30


















            • $begingroup$
              It would be fair to say that you're wrong. The fact $f(x)$ is derivable does not imply $f'(x)$ is continuous. Here is a counterexample: $$f(x)=begin{cases}x^2sindfrac{1}{x},&x neq 0,\0,&x=0.end{cases}.$$ You may verify that $f(x)$ is derivable over $(-infty,+infty)$，but $f'(x)$ is not continuous at $x=0$.
              $endgroup$
              – mengdie1982
              Dec 15 '18 at 16:16












            • $begingroup$
              But of course, if we add a condition that $limlimits_{x to x_0}f'(x)$ exists, then $limlimits_{x to x_0}f'(x)=f'(x_0)$. In another word, $f'(x)$ is continuous at $x=x_0$ now！
              $endgroup$
              – mengdie1982
              Dec 15 '18 at 16:24












            • $begingroup$
              @mengdie1982 You still remind a good example for me. Yes, I should explain that the 'not so strict' place this the limitation of derivative must be exist not only the derivative exist. Your comment point out it.
              $endgroup$
              – Nanayajitzuki
              Dec 15 '18 at 16:30
















            $begingroup$
            It would be fair to say that you're wrong. The fact $f(x)$ is derivable does not imply $f'(x)$ is continuous. Here is a counterexample: $$f(x)=begin{cases}x^2sindfrac{1}{x},&x neq 0,\0,&x=0.end{cases}.$$ You may verify that $f(x)$ is derivable over $(-infty,+infty)$，but $f'(x)$ is not continuous at $x=0$.
            $endgroup$
            – mengdie1982
            Dec 15 '18 at 16:16






            $begingroup$
            It would be fair to say that you're wrong. The fact $f(x)$ is derivable does not imply $f'(x)$ is continuous. Here is a counterexample: $$f(x)=begin{cases}x^2sindfrac{1}{x},&x neq 0,\0,&x=0.end{cases}.$$ You may verify that $f(x)$ is derivable over $(-infty,+infty)$，but $f'(x)$ is not continuous at $x=0$.
            $endgroup$
            – mengdie1982
            Dec 15 '18 at 16:16














            $begingroup$
            But of course, if we add a condition that $limlimits_{x to x_0}f'(x)$ exists, then $limlimits_{x to x_0}f'(x)=f'(x_0)$. In another word, $f'(x)$ is continuous at $x=x_0$ now！
            $endgroup$
            – mengdie1982
            Dec 15 '18 at 16:24






            $begingroup$
            But of course, if we add a condition that $limlimits_{x to x_0}f'(x)$ exists, then $limlimits_{x to x_0}f'(x)=f'(x_0)$. In another word, $f'(x)$ is continuous at $x=x_0$ now！
            $endgroup$
            – mengdie1982
            Dec 15 '18 at 16:24














            $begingroup$
            @mengdie1982 You still remind a good example for me. Yes, I should explain that the 'not so strict' place this the limitation of derivative must be exist not only the derivative exist. Your comment point out it.
            $endgroup$
            – Nanayajitzuki
            Dec 15 '18 at 16:30




            $begingroup$
            @mengdie1982 You still remind a good example for me. Yes, I should explain that the 'not so strict' place this the limitation of derivative must be exist not only the derivative exist. Your comment point out it.
            $endgroup$
            – Nanayajitzuki
            Dec 15 '18 at 16:30


















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