Flaw in proof that a differentiable function has continuous derivative












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Let f be a function differentiable on $(a,b)$ and continuous on $cin(a,b)$. If $c+h in (a,b)$ then by the mean value theorem $frac{f(c+h)-f(c)}{h}=f'(c+theta h)$ for $theta in [0,1]$. Let $h xrightarrow{}0$, then $f'(c+theta h) xrightarrow{} f'(c)$ by the above.



My reasoning is the following: $theta$ is a function of $h$, so in fact it is not true that that $theta(h_n)h_n$ represents any arbitrary sequence that tends to 0, so using this definition of limits(the sequence definition), what we have does not follow. EDIT: I am aware of the explicit counterexample $x^2sin(1/x)$ but that doesn't capture the full meat of the question.






real-analysis limits analysis derivatives convergence







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$endgroup$

















    1












    $begingroup$


    Let f be a function differentiable on $(a,b)$ and continuous on $cin(a,b)$. If $c+h in (a,b)$ then by the mean value theorem $frac{f(c+h)-f(c)}{h}=f'(c+theta h)$ for $theta in [0,1]$. Let $h xrightarrow{}0$, then $f'(c+theta h) xrightarrow{} f'(c)$ by the above.



    My reasoning is the following: $theta$ is a function of $h$, so in fact it is not true that that $theta(h_n)h_n$ represents any arbitrary sequence that tends to 0, so using this definition of limits(the sequence definition), what we have does not follow. EDIT: I am aware of the explicit counterexample $x^2sin(1/x)$ but that doesn't capture the full meat of the question.






    real-analysis limits analysis derivatives convergence







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Let f be a function differentiable on $(a,b)$ and continuous on $cin(a,b)$. If $c+h in (a,b)$ then by the mean value theorem $frac{f(c+h)-f(c)}{h}=f'(c+theta h)$ for $theta in [0,1]$. Let $h xrightarrow{}0$, then $f'(c+theta h) xrightarrow{} f'(c)$ by the above.



      My reasoning is the following: $theta$ is a function of $h$, so in fact it is not true that that $theta(h_n)h_n$ represents any arbitrary sequence that tends to 0, so using this definition of limits(the sequence definition), what we have does not follow. EDIT: I am aware of the explicit counterexample $x^2sin(1/x)$ but that doesn't capture the full meat of the question.






      real-analysis limits analysis derivatives convergence







      share|cite|improve this question











      $endgroup$




      Let f be a function differentiable on $(a,b)$ and continuous on $cin(a,b)$. If $c+h in (a,b)$ then by the mean value theorem $frac{f(c+h)-f(c)}{h}=f'(c+theta h)$ for $theta in [0,1]$. Let $h xrightarrow{}0$, then $f'(c+theta h) xrightarrow{} f'(c)$ by the above.



      My reasoning is the following: $theta$ is a function of $h$, so in fact it is not true that that $theta(h_n)h_n$ represents any arbitrary sequence that tends to 0, so using this definition of limits(the sequence definition), what we have does not follow. EDIT: I am aware of the explicit counterexample $x^2sin(1/x)$ but that doesn't capture the full meat of the question.






      real-analysis limits analysis derivatives convergence




      real-analysis limits analysis derivatives convergence






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 13 hours ago







      user3184807

















      asked 13 hours ago









      user3184807user3184807

      357110




      357110






















          3 Answers
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          active

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          4












          $begingroup$

          It is not true that $theta$ need be a function of $h,$ because it may not be uniquely specified. There could be many $theta$'s that work for a given $h.$



          We only know $f'(y)to f'(c)$ as $yto c$ within the set of $y=x+theta h$ that arise in the MVT process you described. But the set of such $y$ may not equal any deleted neighborhood of $c,$ as the example $x^2sin(1/x)$ shows.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            So we can make $theta$ a function by choosing a value of $h$ and then say the union of all the sequences $theta(h_n)h_n$ where $h_n xrightarrow{}0$ over all functions $theta$ need not be all sequences tending to 0?
            $endgroup$
            – user3184807
            13 hours ago








          • 1




            $begingroup$
            That's a lot of choosing, and the axiom of choice may even come in here. But yes, the sequences we get through the MVT may not be all necessary sequences. In the case of $x^2sin(1/x)$ the sequence $1/(2npi)to 0,$ but this sequence does not arise as $theta(h_n)h_n$ in the MVT process.
            $endgroup$
            – zhw.
            13 hours ago





















          3












          $begingroup$

          If you look at $f(x)=x^2 sin(1/x)$ on $(0,1]$ and $f(0)=0$, you have $f'(x)=2xsin(1/x)-cos(1/x)$ on $(0,1]$ and $f'(0)=0$. Thus MVT tells you that for every $h in (0,1)$ there exists $theta(h) in (0,1)$ with



          $$hsin(1/h)=2htheta(h)sin(1/(htheta(h)))-cos(1/(htheta(h))).$$



          Notice that once $h$ is small enough, $htheta(h)$ is forced to stay relatively close to the zeros of $cos(1/x)$, since



          $$|cos(1/(htheta(h)))|=|hsin(1/h)-2htheta(h)sin(1/(htheta(h)))|<3h.$$



          This means that the MVT is only providing you with information about $f'(x)$ approaching $f'(0)$ along sequences which stay sufficiently close to the zeros of $cos(1/x)$. If you instead look at something like $f' left ( frac{1}{pi n} right )$ you find that it doesn't go to zero. There is no contradiction because the MVT is never giving you information about those points.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            So does this essentially come down to the idea that $theta(h_n)h_n$ cannot always be made to be any arbitrary sequence tending to 0?
            $endgroup$
            – user3184807
            13 hours ago










          • $begingroup$
            @user3184807 Yes, that is essentially what happens.
            $endgroup$
            – Ian
            13 hours ago



















          0












          $begingroup$

          You are assuming what you want to prove: that $f’$ is continuous at $c$ when you say that $f'(c+theta h) to f'(c)$ as $hto0$.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            No, because OP defined $theta(h)$ through the MVT in the first place. The point is that you know that $lim_{h to 0} f'(c+htheta(h))=f'(c)$, but why doesn't this imply $lim_{h to 0} f'(c+h)=f'(c)$? Presumably this is because of some very bad property of $theta(h)$.
            $endgroup$
            – Ian
            13 hours ago








          • 1




            $begingroup$
            That has nothing to do with muy answer.
            $endgroup$
            – Julián Aguirre
            13 hours ago










          • $begingroup$
            Let me spell it out, then. By definition $lim_{h to 0} frac{f(c+h)-f(c)}{h}=f'(c)$, and by MVT $frac{f(c+h)-f(c)}{h}=f'(c+htheta(h))$ for some $theta(h) in (0,1)$. That's all OP assumed. Then they send $h to 0$ in the MVT relation and find $lim_{h to 0} f'(c+h theta(h))=f'(c)$. Now they're wondering why they can't conclude $lim_{h to 0} f'(c+h)=f'(c)$.
            $endgroup$
            – Ian
            13 hours ago












          • $begingroup$
            @Ian: The problem is that the limit may not exist. If it exists, then it must equal $f'(c)$ (as has been covered many times on this site already, for example here: math.stackexchange.com/questions/257907/…).
            $endgroup$
            – Hans Lundmark
            13 hours ago










          • $begingroup$
            @HansLundmark OP knows that already as well, the question is about intuition. The answer, for the case of something like $f(x)=x^2sin(1/x)$, is that $htheta(h)$ has numerous discontinuities in a vicinity of $h=0$, so you don't get information about entire neighborhoods of $h=0$ from the MVT.
            $endgroup$
            – Ian
            13 hours ago













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          3 Answers
          3






          active

          oldest

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          3 Answers
          3






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          4












          $begingroup$

          It is not true that $theta$ need be a function of $h,$ because it may not be uniquely specified. There could be many $theta$'s that work for a given $h.$



          We only know $f'(y)to f'(c)$ as $yto c$ within the set of $y=x+theta h$ that arise in the MVT process you described. But the set of such $y$ may not equal any deleted neighborhood of $c,$ as the example $x^2sin(1/x)$ shows.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            So we can make $theta$ a function by choosing a value of $h$ and then say the union of all the sequences $theta(h_n)h_n$ where $h_n xrightarrow{}0$ over all functions $theta$ need not be all sequences tending to 0?
            $endgroup$
            – user3184807
            13 hours ago








          • 1




            $begingroup$
            That's a lot of choosing, and the axiom of choice may even come in here. But yes, the sequences we get through the MVT may not be all necessary sequences. In the case of $x^2sin(1/x)$ the sequence $1/(2npi)to 0,$ but this sequence does not arise as $theta(h_n)h_n$ in the MVT process.
            $endgroup$
            – zhw.
            13 hours ago


















          4












          $begingroup$

          It is not true that $theta$ need be a function of $h,$ because it may not be uniquely specified. There could be many $theta$'s that work for a given $h.$



          We only know $f'(y)to f'(c)$ as $yto c$ within the set of $y=x+theta h$ that arise in the MVT process you described. But the set of such $y$ may not equal any deleted neighborhood of $c,$ as the example $x^2sin(1/x)$ shows.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            So we can make $theta$ a function by choosing a value of $h$ and then say the union of all the sequences $theta(h_n)h_n$ where $h_n xrightarrow{}0$ over all functions $theta$ need not be all sequences tending to 0?
            $endgroup$
            – user3184807
            13 hours ago








          • 1




            $begingroup$
            That's a lot of choosing, and the axiom of choice may even come in here. But yes, the sequences we get through the MVT may not be all necessary sequences. In the case of $x^2sin(1/x)$ the sequence $1/(2npi)to 0,$ but this sequence does not arise as $theta(h_n)h_n$ in the MVT process.
            $endgroup$
            – zhw.
            13 hours ago
















          4












          4








          4





          $begingroup$

          It is not true that $theta$ need be a function of $h,$ because it may not be uniquely specified. There could be many $theta$'s that work for a given $h.$



          We only know $f'(y)to f'(c)$ as $yto c$ within the set of $y=x+theta h$ that arise in the MVT process you described. But the set of such $y$ may not equal any deleted neighborhood of $c,$ as the example $x^2sin(1/x)$ shows.






          share|cite|improve this answer











          $endgroup$



          It is not true that $theta$ need be a function of $h,$ because it may not be uniquely specified. There could be many $theta$'s that work for a given $h.$



          We only know $f'(y)to f'(c)$ as $yto c$ within the set of $y=x+theta h$ that arise in the MVT process you described. But the set of such $y$ may not equal any deleted neighborhood of $c,$ as the example $x^2sin(1/x)$ shows.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 9 hours ago

























          answered 13 hours ago









          zhw.zhw.

          73.7k43175




          73.7k43175












          • $begingroup$
            So we can make $theta$ a function by choosing a value of $h$ and then say the union of all the sequences $theta(h_n)h_n$ where $h_n xrightarrow{}0$ over all functions $theta$ need not be all sequences tending to 0?
            $endgroup$
            – user3184807
            13 hours ago








          • 1




            $begingroup$
            That's a lot of choosing, and the axiom of choice may even come in here. But yes, the sequences we get through the MVT may not be all necessary sequences. In the case of $x^2sin(1/x)$ the sequence $1/(2npi)to 0,$ but this sequence does not arise as $theta(h_n)h_n$ in the MVT process.
            $endgroup$
            – zhw.
            13 hours ago




















          • $begingroup$
            So we can make $theta$ a function by choosing a value of $h$ and then say the union of all the sequences $theta(h_n)h_n$ where $h_n xrightarrow{}0$ over all functions $theta$ need not be all sequences tending to 0?
            $endgroup$
            – user3184807
            13 hours ago








          • 1




            $begingroup$
            That's a lot of choosing, and the axiom of choice may even come in here. But yes, the sequences we get through the MVT may not be all necessary sequences. In the case of $x^2sin(1/x)$ the sequence $1/(2npi)to 0,$ but this sequence does not arise as $theta(h_n)h_n$ in the MVT process.
            $endgroup$
            – zhw.
            13 hours ago


















          $begingroup$
          So we can make $theta$ a function by choosing a value of $h$ and then say the union of all the sequences $theta(h_n)h_n$ where $h_n xrightarrow{}0$ over all functions $theta$ need not be all sequences tending to 0?
          $endgroup$
          – user3184807
          13 hours ago






          $begingroup$
          So we can make $theta$ a function by choosing a value of $h$ and then say the union of all the sequences $theta(h_n)h_n$ where $h_n xrightarrow{}0$ over all functions $theta$ need not be all sequences tending to 0?
          $endgroup$
          – user3184807
          13 hours ago






          1




          1




          $begingroup$
          That's a lot of choosing, and the axiom of choice may even come in here. But yes, the sequences we get through the MVT may not be all necessary sequences. In the case of $x^2sin(1/x)$ the sequence $1/(2npi)to 0,$ but this sequence does not arise as $theta(h_n)h_n$ in the MVT process.
          $endgroup$
          – zhw.
          13 hours ago






          $begingroup$
          That's a lot of choosing, and the axiom of choice may even come in here. But yes, the sequences we get through the MVT may not be all necessary sequences. In the case of $x^2sin(1/x)$ the sequence $1/(2npi)to 0,$ but this sequence does not arise as $theta(h_n)h_n$ in the MVT process.
          $endgroup$
          – zhw.
          13 hours ago













          3












          $begingroup$

          If you look at $f(x)=x^2 sin(1/x)$ on $(0,1]$ and $f(0)=0$, you have $f'(x)=2xsin(1/x)-cos(1/x)$ on $(0,1]$ and $f'(0)=0$. Thus MVT tells you that for every $h in (0,1)$ there exists $theta(h) in (0,1)$ with



          $$hsin(1/h)=2htheta(h)sin(1/(htheta(h)))-cos(1/(htheta(h))).$$



          Notice that once $h$ is small enough, $htheta(h)$ is forced to stay relatively close to the zeros of $cos(1/x)$, since



          $$|cos(1/(htheta(h)))|=|hsin(1/h)-2htheta(h)sin(1/(htheta(h)))|<3h.$$



          This means that the MVT is only providing you with information about $f'(x)$ approaching $f'(0)$ along sequences which stay sufficiently close to the zeros of $cos(1/x)$. If you instead look at something like $f' left ( frac{1}{pi n} right )$ you find that it doesn't go to zero. There is no contradiction because the MVT is never giving you information about those points.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            So does this essentially come down to the idea that $theta(h_n)h_n$ cannot always be made to be any arbitrary sequence tending to 0?
            $endgroup$
            – user3184807
            13 hours ago










          • $begingroup$
            @user3184807 Yes, that is essentially what happens.
            $endgroup$
            – Ian
            13 hours ago
















          3












          $begingroup$

          If you look at $f(x)=x^2 sin(1/x)$ on $(0,1]$ and $f(0)=0$, you have $f'(x)=2xsin(1/x)-cos(1/x)$ on $(0,1]$ and $f'(0)=0$. Thus MVT tells you that for every $h in (0,1)$ there exists $theta(h) in (0,1)$ with



          $$hsin(1/h)=2htheta(h)sin(1/(htheta(h)))-cos(1/(htheta(h))).$$



          Notice that once $h$ is small enough, $htheta(h)$ is forced to stay relatively close to the zeros of $cos(1/x)$, since



          $$|cos(1/(htheta(h)))|=|hsin(1/h)-2htheta(h)sin(1/(htheta(h)))|<3h.$$



          This means that the MVT is only providing you with information about $f'(x)$ approaching $f'(0)$ along sequences which stay sufficiently close to the zeros of $cos(1/x)$. If you instead look at something like $f' left ( frac{1}{pi n} right )$ you find that it doesn't go to zero. There is no contradiction because the MVT is never giving you information about those points.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            So does this essentially come down to the idea that $theta(h_n)h_n$ cannot always be made to be any arbitrary sequence tending to 0?
            $endgroup$
            – user3184807
            13 hours ago










          • $begingroup$
            @user3184807 Yes, that is essentially what happens.
            $endgroup$
            – Ian
            13 hours ago














          3












          3








          3





          $begingroup$

          If you look at $f(x)=x^2 sin(1/x)$ on $(0,1]$ and $f(0)=0$, you have $f'(x)=2xsin(1/x)-cos(1/x)$ on $(0,1]$ and $f'(0)=0$. Thus MVT tells you that for every $h in (0,1)$ there exists $theta(h) in (0,1)$ with



          $$hsin(1/h)=2htheta(h)sin(1/(htheta(h)))-cos(1/(htheta(h))).$$



          Notice that once $h$ is small enough, $htheta(h)$ is forced to stay relatively close to the zeros of $cos(1/x)$, since



          $$|cos(1/(htheta(h)))|=|hsin(1/h)-2htheta(h)sin(1/(htheta(h)))|<3h.$$



          This means that the MVT is only providing you with information about $f'(x)$ approaching $f'(0)$ along sequences which stay sufficiently close to the zeros of $cos(1/x)$. If you instead look at something like $f' left ( frac{1}{pi n} right )$ you find that it doesn't go to zero. There is no contradiction because the MVT is never giving you information about those points.






          share|cite|improve this answer











          $endgroup$



          If you look at $f(x)=x^2 sin(1/x)$ on $(0,1]$ and $f(0)=0$, you have $f'(x)=2xsin(1/x)-cos(1/x)$ on $(0,1]$ and $f'(0)=0$. Thus MVT tells you that for every $h in (0,1)$ there exists $theta(h) in (0,1)$ with



          $$hsin(1/h)=2htheta(h)sin(1/(htheta(h)))-cos(1/(htheta(h))).$$



          Notice that once $h$ is small enough, $htheta(h)$ is forced to stay relatively close to the zeros of $cos(1/x)$, since



          $$|cos(1/(htheta(h)))|=|hsin(1/h)-2htheta(h)sin(1/(htheta(h)))|<3h.$$



          This means that the MVT is only providing you with information about $f'(x)$ approaching $f'(0)$ along sequences which stay sufficiently close to the zeros of $cos(1/x)$. If you instead look at something like $f' left ( frac{1}{pi n} right )$ you find that it doesn't go to zero. There is no contradiction because the MVT is never giving you information about those points.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 13 hours ago

























          answered 13 hours ago









          IanIan

          68.7k25389




          68.7k25389












          • $begingroup$
            So does this essentially come down to the idea that $theta(h_n)h_n$ cannot always be made to be any arbitrary sequence tending to 0?
            $endgroup$
            – user3184807
            13 hours ago










          • $begingroup$
            @user3184807 Yes, that is essentially what happens.
            $endgroup$
            – Ian
            13 hours ago


















          • $begingroup$
            So does this essentially come down to the idea that $theta(h_n)h_n$ cannot always be made to be any arbitrary sequence tending to 0?
            $endgroup$
            – user3184807
            13 hours ago










          • $begingroup$
            @user3184807 Yes, that is essentially what happens.
            $endgroup$
            – Ian
            13 hours ago
















          $begingroup$
          So does this essentially come down to the idea that $theta(h_n)h_n$ cannot always be made to be any arbitrary sequence tending to 0?
          $endgroup$
          – user3184807
          13 hours ago




          $begingroup$
          So does this essentially come down to the idea that $theta(h_n)h_n$ cannot always be made to be any arbitrary sequence tending to 0?
          $endgroup$
          – user3184807
          13 hours ago












          $begingroup$
          @user3184807 Yes, that is essentially what happens.
          $endgroup$
          – Ian
          13 hours ago




          $begingroup$
          @user3184807 Yes, that is essentially what happens.
          $endgroup$
          – Ian
          13 hours ago











          0












          $begingroup$

          You are assuming what you want to prove: that $f’$ is continuous at $c$ when you say that $f'(c+theta h) to f'(c)$ as $hto0$.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            No, because OP defined $theta(h)$ through the MVT in the first place. The point is that you know that $lim_{h to 0} f'(c+htheta(h))=f'(c)$, but why doesn't this imply $lim_{h to 0} f'(c+h)=f'(c)$? Presumably this is because of some very bad property of $theta(h)$.
            $endgroup$
            – Ian
            13 hours ago








          • 1




            $begingroup$
            That has nothing to do with muy answer.
            $endgroup$
            – Julián Aguirre
            13 hours ago










          • $begingroup$
            Let me spell it out, then. By definition $lim_{h to 0} frac{f(c+h)-f(c)}{h}=f'(c)$, and by MVT $frac{f(c+h)-f(c)}{h}=f'(c+htheta(h))$ for some $theta(h) in (0,1)$. That's all OP assumed. Then they send $h to 0$ in the MVT relation and find $lim_{h to 0} f'(c+h theta(h))=f'(c)$. Now they're wondering why they can't conclude $lim_{h to 0} f'(c+h)=f'(c)$.
            $endgroup$
            – Ian
            13 hours ago












          • $begingroup$
            @Ian: The problem is that the limit may not exist. If it exists, then it must equal $f'(c)$ (as has been covered many times on this site already, for example here: math.stackexchange.com/questions/257907/…).
            $endgroup$
            – Hans Lundmark
            13 hours ago










          • $begingroup$
            @HansLundmark OP knows that already as well, the question is about intuition. The answer, for the case of something like $f(x)=x^2sin(1/x)$, is that $htheta(h)$ has numerous discontinuities in a vicinity of $h=0$, so you don't get information about entire neighborhoods of $h=0$ from the MVT.
            $endgroup$
            – Ian
            13 hours ago


















          0












          $begingroup$

          You are assuming what you want to prove: that $f’$ is continuous at $c$ when you say that $f'(c+theta h) to f'(c)$ as $hto0$.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            No, because OP defined $theta(h)$ through the MVT in the first place. The point is that you know that $lim_{h to 0} f'(c+htheta(h))=f'(c)$, but why doesn't this imply $lim_{h to 0} f'(c+h)=f'(c)$? Presumably this is because of some very bad property of $theta(h)$.
            $endgroup$
            – Ian
            13 hours ago








          • 1




            $begingroup$
            That has nothing to do with muy answer.
            $endgroup$
            – Julián Aguirre
            13 hours ago










          • $begingroup$
            Let me spell it out, then. By definition $lim_{h to 0} frac{f(c+h)-f(c)}{h}=f'(c)$, and by MVT $frac{f(c+h)-f(c)}{h}=f'(c+htheta(h))$ for some $theta(h) in (0,1)$. That's all OP assumed. Then they send $h to 0$ in the MVT relation and find $lim_{h to 0} f'(c+h theta(h))=f'(c)$. Now they're wondering why they can't conclude $lim_{h to 0} f'(c+h)=f'(c)$.
            $endgroup$
            – Ian
            13 hours ago












          • $begingroup$
            @Ian: The problem is that the limit may not exist. If it exists, then it must equal $f'(c)$ (as has been covered many times on this site already, for example here: math.stackexchange.com/questions/257907/…).
            $endgroup$
            – Hans Lundmark
            13 hours ago










          • $begingroup$
            @HansLundmark OP knows that already as well, the question is about intuition. The answer, for the case of something like $f(x)=x^2sin(1/x)$, is that $htheta(h)$ has numerous discontinuities in a vicinity of $h=0$, so you don't get information about entire neighborhoods of $h=0$ from the MVT.
            $endgroup$
            – Ian
            13 hours ago
















          0












          0








          0





          $begingroup$

          You are assuming what you want to prove: that $f’$ is continuous at $c$ when you say that $f'(c+theta h) to f'(c)$ as $hto0$.






          share|cite|improve this answer









          $endgroup$



          You are assuming what you want to prove: that $f’$ is continuous at $c$ when you say that $f'(c+theta h) to f'(c)$ as $hto0$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 13 hours ago









          Julián AguirreJulián Aguirre

          69.2k24096




          69.2k24096








          • 1




            $begingroup$
            No, because OP defined $theta(h)$ through the MVT in the first place. The point is that you know that $lim_{h to 0} f'(c+htheta(h))=f'(c)$, but why doesn't this imply $lim_{h to 0} f'(c+h)=f'(c)$? Presumably this is because of some very bad property of $theta(h)$.
            $endgroup$
            – Ian
            13 hours ago








          • 1




            $begingroup$
            That has nothing to do with muy answer.
            $endgroup$
            – Julián Aguirre
            13 hours ago










          • $begingroup$
            Let me spell it out, then. By definition $lim_{h to 0} frac{f(c+h)-f(c)}{h}=f'(c)$, and by MVT $frac{f(c+h)-f(c)}{h}=f'(c+htheta(h))$ for some $theta(h) in (0,1)$. That's all OP assumed. Then they send $h to 0$ in the MVT relation and find $lim_{h to 0} f'(c+h theta(h))=f'(c)$. Now they're wondering why they can't conclude $lim_{h to 0} f'(c+h)=f'(c)$.
            $endgroup$
            – Ian
            13 hours ago












          • $begingroup$
            @Ian: The problem is that the limit may not exist. If it exists, then it must equal $f'(c)$ (as has been covered many times on this site already, for example here: math.stackexchange.com/questions/257907/…).
            $endgroup$
            – Hans Lundmark
            13 hours ago










          • $begingroup$
            @HansLundmark OP knows that already as well, the question is about intuition. The answer, for the case of something like $f(x)=x^2sin(1/x)$, is that $htheta(h)$ has numerous discontinuities in a vicinity of $h=0$, so you don't get information about entire neighborhoods of $h=0$ from the MVT.
            $endgroup$
            – Ian
            13 hours ago
















          • 1




            $begingroup$
            No, because OP defined $theta(h)$ through the MVT in the first place. The point is that you know that $lim_{h to 0} f'(c+htheta(h))=f'(c)$, but why doesn't this imply $lim_{h to 0} f'(c+h)=f'(c)$? Presumably this is because of some very bad property of $theta(h)$.
            $endgroup$
            – Ian
            13 hours ago








          • 1




            $begingroup$
            That has nothing to do with muy answer.
            $endgroup$
            – Julián Aguirre
            13 hours ago










          • $begingroup$
            Let me spell it out, then. By definition $lim_{h to 0} frac{f(c+h)-f(c)}{h}=f'(c)$, and by MVT $frac{f(c+h)-f(c)}{h}=f'(c+htheta(h))$ for some $theta(h) in (0,1)$. That's all OP assumed. Then they send $h to 0$ in the MVT relation and find $lim_{h to 0} f'(c+h theta(h))=f'(c)$. Now they're wondering why they can't conclude $lim_{h to 0} f'(c+h)=f'(c)$.
            $endgroup$
            – Ian
            13 hours ago












          • $begingroup$
            @Ian: The problem is that the limit may not exist. If it exists, then it must equal $f'(c)$ (as has been covered many times on this site already, for example here: math.stackexchange.com/questions/257907/…).
            $endgroup$
            – Hans Lundmark
            13 hours ago










          • $begingroup$
            @HansLundmark OP knows that already as well, the question is about intuition. The answer, for the case of something like $f(x)=x^2sin(1/x)$, is that $htheta(h)$ has numerous discontinuities in a vicinity of $h=0$, so you don't get information about entire neighborhoods of $h=0$ from the MVT.
            $endgroup$
            – Ian
            13 hours ago










          1




          1




          $begingroup$
          No, because OP defined $theta(h)$ through the MVT in the first place. The point is that you know that $lim_{h to 0} f'(c+htheta(h))=f'(c)$, but why doesn't this imply $lim_{h to 0} f'(c+h)=f'(c)$? Presumably this is because of some very bad property of $theta(h)$.
          $endgroup$
          – Ian
          13 hours ago






          $begingroup$
          No, because OP defined $theta(h)$ through the MVT in the first place. The point is that you know that $lim_{h to 0} f'(c+htheta(h))=f'(c)$, but why doesn't this imply $lim_{h to 0} f'(c+h)=f'(c)$? Presumably this is because of some very bad property of $theta(h)$.
          $endgroup$
          – Ian
          13 hours ago






          1




          1




          $begingroup$
          That has nothing to do with muy answer.
          $endgroup$
          – Julián Aguirre
          13 hours ago




          $begingroup$
          That has nothing to do with muy answer.
          $endgroup$
          – Julián Aguirre
          13 hours ago












          $begingroup$
          Let me spell it out, then. By definition $lim_{h to 0} frac{f(c+h)-f(c)}{h}=f'(c)$, and by MVT $frac{f(c+h)-f(c)}{h}=f'(c+htheta(h))$ for some $theta(h) in (0,1)$. That's all OP assumed. Then they send $h to 0$ in the MVT relation and find $lim_{h to 0} f'(c+h theta(h))=f'(c)$. Now they're wondering why they can't conclude $lim_{h to 0} f'(c+h)=f'(c)$.
          $endgroup$
          – Ian
          13 hours ago






          $begingroup$
          Let me spell it out, then. By definition $lim_{h to 0} frac{f(c+h)-f(c)}{h}=f'(c)$, and by MVT $frac{f(c+h)-f(c)}{h}=f'(c+htheta(h))$ for some $theta(h) in (0,1)$. That's all OP assumed. Then they send $h to 0$ in the MVT relation and find $lim_{h to 0} f'(c+h theta(h))=f'(c)$. Now they're wondering why they can't conclude $lim_{h to 0} f'(c+h)=f'(c)$.
          $endgroup$
          – Ian
          13 hours ago














          $begingroup$
          @Ian: The problem is that the limit may not exist. If it exists, then it must equal $f'(c)$ (as has been covered many times on this site already, for example here: math.stackexchange.com/questions/257907/…).
          $endgroup$
          – Hans Lundmark
          13 hours ago




          $begingroup$
          @Ian: The problem is that the limit may not exist. If it exists, then it must equal $f'(c)$ (as has been covered many times on this site already, for example here: math.stackexchange.com/questions/257907/…).
          $endgroup$
          – Hans Lundmark
          13 hours ago












          $begingroup$
          @HansLundmark OP knows that already as well, the question is about intuition. The answer, for the case of something like $f(x)=x^2sin(1/x)$, is that $htheta(h)$ has numerous discontinuities in a vicinity of $h=0$, so you don't get information about entire neighborhoods of $h=0$ from the MVT.
          $endgroup$
          – Ian
          13 hours ago






          $begingroup$
          @HansLundmark OP knows that already as well, the question is about intuition. The answer, for the case of something like $f(x)=x^2sin(1/x)$, is that $htheta(h)$ has numerous discontinuities in a vicinity of $h=0$, so you don't get information about entire neighborhoods of $h=0$ from the MVT.
          $endgroup$
          – Ian
          13 hours ago




















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