Error in the derivative proof












1












$begingroup$


Assume $h$ is a function on an open interval $K$, and differentiable on $K$. Therefore $h'$ is cont on $K$.



The faulty proof goes as follows:



Let $ain K$



By the definition of the derivative:



$$h'(a)=lim_{xto a}frac{h(x)-h(a)}{x-a}$$



Since $h$ is cont, limit of numerator is $0$, same goes for denominator. Since $h$ is diff, we can apply LH rule.



$$h'(a)=lim_{xto a}frac{h(x)-h(a)}{x-a}to(LH)to lim_{xto a}frac{h'(x)-0}{1-0}=lim_{xto a}h'(x)$$



We have proven that $h'(a)=lim_{xto a}h'(x)$. By defn, $h'$ is cont.




There is an error in the proof, and I also need to fix the proof.



I don't see something right away.



My guess is that $x-aneq 0$, but I am not sure how to prove this.






limits derivatives proof-verification fake-proofs







share|cite|improve this question











$endgroup$












  • $begingroup$
    Comments are not for extended discussion; this conversation has been moved to chat.
    $endgroup$
    – Aloizio Macedo♦
    Jan 10 at 5:58
















1












$begingroup$


Assume $h$ is a function on an open interval $K$, and differentiable on $K$. Therefore $h'$ is cont on $K$.



The faulty proof goes as follows:



Let $ain K$



By the definition of the derivative:



$$h'(a)=lim_{xto a}frac{h(x)-h(a)}{x-a}$$



Since $h$ is cont, limit of numerator is $0$, same goes for denominator. Since $h$ is diff, we can apply LH rule.



$$h'(a)=lim_{xto a}frac{h(x)-h(a)}{x-a}to(LH)to lim_{xto a}frac{h'(x)-0}{1-0}=lim_{xto a}h'(x)$$



We have proven that $h'(a)=lim_{xto a}h'(x)$. By defn, $h'$ is cont.




There is an error in the proof, and I also need to fix the proof.



I don't see something right away.



My guess is that $x-aneq 0$, but I am not sure how to prove this.






limits derivatives proof-verification fake-proofs







share|cite|improve this question











$endgroup$












  • $begingroup$
    Comments are not for extended discussion; this conversation has been moved to chat.
    $endgroup$
    – Aloizio Macedo♦
    Jan 10 at 5:58














1












1








1





$begingroup$


Assume $h$ is a function on an open interval $K$, and differentiable on $K$. Therefore $h'$ is cont on $K$.



The faulty proof goes as follows:



Let $ain K$



By the definition of the derivative:



$$h'(a)=lim_{xto a}frac{h(x)-h(a)}{x-a}$$



Since $h$ is cont, limit of numerator is $0$, same goes for denominator. Since $h$ is diff, we can apply LH rule.



$$h'(a)=lim_{xto a}frac{h(x)-h(a)}{x-a}to(LH)to lim_{xto a}frac{h'(x)-0}{1-0}=lim_{xto a}h'(x)$$



We have proven that $h'(a)=lim_{xto a}h'(x)$. By defn, $h'$ is cont.




There is an error in the proof, and I also need to fix the proof.



I don't see something right away.



My guess is that $x-aneq 0$, but I am not sure how to prove this.






limits derivatives proof-verification fake-proofs







share|cite|improve this question











$endgroup$




Assume $h$ is a function on an open interval $K$, and differentiable on $K$. Therefore $h'$ is cont on $K$.



The faulty proof goes as follows:



Let $ain K$



By the definition of the derivative:



$$h'(a)=lim_{xto a}frac{h(x)-h(a)}{x-a}$$



Since $h$ is cont, limit of numerator is $0$, same goes for denominator. Since $h$ is diff, we can apply LH rule.



$$h'(a)=lim_{xto a}frac{h(x)-h(a)}{x-a}to(LH)to lim_{xto a}frac{h'(x)-0}{1-0}=lim_{xto a}h'(x)$$



We have proven that $h'(a)=lim_{xto a}h'(x)$. By defn, $h'$ is cont.




There is an error in the proof, and I also need to fix the proof.



I don't see something right away.



My guess is that $x-aneq 0$, but I am not sure how to prove this.






limits derivatives proof-verification fake-proofs




limits derivatives proof-verification fake-proofs






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 8 at 20:34







K Split X

















asked Jan 8 at 20:21









K Split XK Split X

4,19611131




4,19611131












  • $begingroup$
    Comments are not for extended discussion; this conversation has been moved to chat.
    $endgroup$
    – Aloizio Macedo♦
    Jan 10 at 5:58


















  • $begingroup$
    Comments are not for extended discussion; this conversation has been moved to chat.
    $endgroup$
    – Aloizio Macedo♦
    Jan 10 at 5:58
















$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Jan 10 at 5:58




$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Jan 10 at 5:58










2 Answers
2






active

oldest

votes


















3












$begingroup$

L'Hospital's rule says that if $f'/g'$ converges to a limit $A$ (and some other assumptions are fulfilled), then also $f/g to A$.



In your “proof”, $displaystylelim_{x to a} h'(x)$ need not exist.



(But if it does, then the claim is true: Prove that $f'(a)=lim_{xrightarrow a}f'(x)$.)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    By need not exist, do you mean infinity?
    $endgroup$
    – K Split X
    Jan 8 at 20:44










  • $begingroup$
    No, I mean “not exist”. Like for the standard example $h(x) = x^2 sin(x)$ for $x neq 0$, $h(0)=0$, where $h'(x)$ will oscillate as $x to 0$, so that there is no limit (neither finite nor infinite).
    $endgroup$
    – Hans Lundmark
    Jan 8 at 20:47






  • 3




    $begingroup$
    Oops, that should have been $h(x)=x^2 sin(1/x)$, of course.
    $endgroup$
    – Hans Lundmark
    Jan 8 at 20:53






  • 2




    $begingroup$
    +1 for pointing out the exact flaw. Also the argument used by asker proves that if the limit of derivative exists then derivative is continuous. This is very special of derivatives that they can't have what is usually called simple / jump discontinuity.
    $endgroup$
    – Paramanand Singh
    Jan 9 at 19:13



















0












$begingroup$

Let $a=0$, $h(x) = begin{cases} x^2 sin {1 over x} , & x neq 0 \
0, & x = 0end{cases}$ on $K=(-1,1)$. Then $h$ is differentiable on $K$ but ${h(x) over x}$ has no limit as $x to 0$.






share|cite|improve this answer











$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    L'Hospital's rule says that if $f'/g'$ converges to a limit $A$ (and some other assumptions are fulfilled), then also $f/g to A$.



    In your “proof”, $displaystylelim_{x to a} h'(x)$ need not exist.



    (But if it does, then the claim is true: Prove that $f'(a)=lim_{xrightarrow a}f'(x)$.)






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      By need not exist, do you mean infinity?
      $endgroup$
      – K Split X
      Jan 8 at 20:44










    • $begingroup$
      No, I mean “not exist”. Like for the standard example $h(x) = x^2 sin(x)$ for $x neq 0$, $h(0)=0$, where $h'(x)$ will oscillate as $x to 0$, so that there is no limit (neither finite nor infinite).
      $endgroup$
      – Hans Lundmark
      Jan 8 at 20:47






    • 3




      $begingroup$
      Oops, that should have been $h(x)=x^2 sin(1/x)$, of course.
      $endgroup$
      – Hans Lundmark
      Jan 8 at 20:53






    • 2




      $begingroup$
      +1 for pointing out the exact flaw. Also the argument used by asker proves that if the limit of derivative exists then derivative is continuous. This is very special of derivatives that they can't have what is usually called simple / jump discontinuity.
      $endgroup$
      – Paramanand Singh
      Jan 9 at 19:13
















    3












    $begingroup$

    L'Hospital's rule says that if $f'/g'$ converges to a limit $A$ (and some other assumptions are fulfilled), then also $f/g to A$.



    In your “proof”, $displaystylelim_{x to a} h'(x)$ need not exist.



    (But if it does, then the claim is true: Prove that $f'(a)=lim_{xrightarrow a}f'(x)$.)






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      By need not exist, do you mean infinity?
      $endgroup$
      – K Split X
      Jan 8 at 20:44










    • $begingroup$
      No, I mean “not exist”. Like for the standard example $h(x) = x^2 sin(x)$ for $x neq 0$, $h(0)=0$, where $h'(x)$ will oscillate as $x to 0$, so that there is no limit (neither finite nor infinite).
      $endgroup$
      – Hans Lundmark
      Jan 8 at 20:47






    • 3




      $begingroup$
      Oops, that should have been $h(x)=x^2 sin(1/x)$, of course.
      $endgroup$
      – Hans Lundmark
      Jan 8 at 20:53






    • 2




      $begingroup$
      +1 for pointing out the exact flaw. Also the argument used by asker proves that if the limit of derivative exists then derivative is continuous. This is very special of derivatives that they can't have what is usually called simple / jump discontinuity.
      $endgroup$
      – Paramanand Singh
      Jan 9 at 19:13














    3












    3








    3





    $begingroup$

    L'Hospital's rule says that if $f'/g'$ converges to a limit $A$ (and some other assumptions are fulfilled), then also $f/g to A$.



    In your “proof”, $displaystylelim_{x to a} h'(x)$ need not exist.



    (But if it does, then the claim is true: Prove that $f'(a)=lim_{xrightarrow a}f'(x)$.)






    share|cite|improve this answer









    $endgroup$



    L'Hospital's rule says that if $f'/g'$ converges to a limit $A$ (and some other assumptions are fulfilled), then also $f/g to A$.



    In your “proof”, $displaystylelim_{x to a} h'(x)$ need not exist.



    (But if it does, then the claim is true: Prove that $f'(a)=lim_{xrightarrow a}f'(x)$.)







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 8 at 20:42









    Hans LundmarkHans Lundmark

    35.3k564114




    35.3k564114












    • $begingroup$
      By need not exist, do you mean infinity?
      $endgroup$
      – K Split X
      Jan 8 at 20:44










    • $begingroup$
      No, I mean “not exist”. Like for the standard example $h(x) = x^2 sin(x)$ for $x neq 0$, $h(0)=0$, where $h'(x)$ will oscillate as $x to 0$, so that there is no limit (neither finite nor infinite).
      $endgroup$
      – Hans Lundmark
      Jan 8 at 20:47






    • 3




      $begingroup$
      Oops, that should have been $h(x)=x^2 sin(1/x)$, of course.
      $endgroup$
      – Hans Lundmark
      Jan 8 at 20:53






    • 2




      $begingroup$
      +1 for pointing out the exact flaw. Also the argument used by asker proves that if the limit of derivative exists then derivative is continuous. This is very special of derivatives that they can't have what is usually called simple / jump discontinuity.
      $endgroup$
      – Paramanand Singh
      Jan 9 at 19:13


















    • $begingroup$
      By need not exist, do you mean infinity?
      $endgroup$
      – K Split X
      Jan 8 at 20:44










    • $begingroup$
      No, I mean “not exist”. Like for the standard example $h(x) = x^2 sin(x)$ for $x neq 0$, $h(0)=0$, where $h'(x)$ will oscillate as $x to 0$, so that there is no limit (neither finite nor infinite).
      $endgroup$
      – Hans Lundmark
      Jan 8 at 20:47






    • 3




      $begingroup$
      Oops, that should have been $h(x)=x^2 sin(1/x)$, of course.
      $endgroup$
      – Hans Lundmark
      Jan 8 at 20:53






    • 2




      $begingroup$
      +1 for pointing out the exact flaw. Also the argument used by asker proves that if the limit of derivative exists then derivative is continuous. This is very special of derivatives that they can't have what is usually called simple / jump discontinuity.
      $endgroup$
      – Paramanand Singh
      Jan 9 at 19:13
















    $begingroup$
    By need not exist, do you mean infinity?
    $endgroup$
    – K Split X
    Jan 8 at 20:44




    $begingroup$
    By need not exist, do you mean infinity?
    $endgroup$
    – K Split X
    Jan 8 at 20:44












    $begingroup$
    No, I mean “not exist”. Like for the standard example $h(x) = x^2 sin(x)$ for $x neq 0$, $h(0)=0$, where $h'(x)$ will oscillate as $x to 0$, so that there is no limit (neither finite nor infinite).
    $endgroup$
    – Hans Lundmark
    Jan 8 at 20:47




    $begingroup$
    No, I mean “not exist”. Like for the standard example $h(x) = x^2 sin(x)$ for $x neq 0$, $h(0)=0$, where $h'(x)$ will oscillate as $x to 0$, so that there is no limit (neither finite nor infinite).
    $endgroup$
    – Hans Lundmark
    Jan 8 at 20:47




    3




    3




    $begingroup$
    Oops, that should have been $h(x)=x^2 sin(1/x)$, of course.
    $endgroup$
    – Hans Lundmark
    Jan 8 at 20:53




    $begingroup$
    Oops, that should have been $h(x)=x^2 sin(1/x)$, of course.
    $endgroup$
    – Hans Lundmark
    Jan 8 at 20:53




    2




    2




    $begingroup$
    +1 for pointing out the exact flaw. Also the argument used by asker proves that if the limit of derivative exists then derivative is continuous. This is very special of derivatives that they can't have what is usually called simple / jump discontinuity.
    $endgroup$
    – Paramanand Singh
    Jan 9 at 19:13




    $begingroup$
    +1 for pointing out the exact flaw. Also the argument used by asker proves that if the limit of derivative exists then derivative is continuous. This is very special of derivatives that they can't have what is usually called simple / jump discontinuity.
    $endgroup$
    – Paramanand Singh
    Jan 9 at 19:13











    0












    $begingroup$

    Let $a=0$, $h(x) = begin{cases} x^2 sin {1 over x} , & x neq 0 \
    0, & x = 0end{cases}$     on $K=(-1,1)$. Then $h$ is differentiable on $K$ but ${h(x) over x}$ has no limit as $x to 0$.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Let $a=0$, $h(x) = begin{cases} x^2 sin {1 over x} , & x neq 0 \
      0, & x = 0end{cases}$       on $K=(-1,1)$. Then $h$ is differentiable on $K$ but ${h(x) over x}$ has no limit as $x to 0$.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Let $a=0$, $h(x) = begin{cases} x^2 sin {1 over x} , & x neq 0 \
        0, & x = 0end{cases}$         on $K=(-1,1)$. Then $h$ is differentiable on $K$ but ${h(x) over x}$ has no limit as $x to 0$.






        share|cite|improve this answer











        $endgroup$



        Let $a=0$, $h(x) = begin{cases} x^2 sin {1 over x} , & x neq 0 \
        0, & x = 0end{cases}$         on $K=(-1,1)$. Then $h$ is differentiable on $K$ but ${h(x) over x}$ has no limit as $x to 0$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 8 at 20:55

























        answered Jan 8 at 20:49









        copper.hatcopper.hat

        126k559160




        126k559160






























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