Flaw in proof that a differentiable function has continuous derivative












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












    $begingroup$


    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
          3






          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





















          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













          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3135150%2fflaw-in-proof-that-a-differentiable-function-has-continuous-derivative%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          3 Answers
          3






          active

          oldest

          votes








          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




















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3135150%2fflaw-in-proof-that-a-differentiable-function-has-continuous-derivative%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          -

          Popular posts from this blog

          If I really need a card on my start hand, how many mulligans make sense? [duplicate]

          Image
          5 This question already has an answer here: How do you calculate the likelihood of drawing certain cards in your opening hand? 3 answers There is a funny modern deck where you play Treasure Hunt, but you really need to have it on the starting hand. How many mulligans should I take, if my deck contains 60 cards and contains 4 copies of Treasure Hunt? magic-the-gathering probability share | improve this question asked Nov 23 '18 at 12:12 dan dan 152 4
          Read more

          Alcedinidae

          Image
          Alcedinidae Martin-pêcheur à dos bleu ( Alcedo azurea ) Classification (COI) Règne Animalia Embranchement Chordata Sous-embr. Vertebrata Classe Aves Ordre Coraciiformes Famille Alcedinidae Rafinesque, 1815 Les Alcédinidés ou Alcedinidae forment une famille d'oiseaux plus connus sous les noms de martins-pêcheurs et martins-chasseurs. Sommaire 1 Description 2 Répartition et habitat 3 Systématique 3.1 Liste alphabétique des genres 3.2 Liste des espèces 4 Chez les Anciens : l'alcyon 4.1 Étymologie (Littré) 4.2 Oiseau « fabuleux » ? 5 Notes 6 Articles connexes 7 Liens externes 7.1 Bibliographie Description | Ce sont des oiseaux compacts de taille petite à moyenne (10 à 46 cm), au bec droit et long, en forme de poignard. Leurs pattes sont courtes, et ils portent un plumage aux couleurs vives. Répartition et habitat | Cosmopolites, ils fréquentent surtout
          Read more

          Can an atomic nucleus contain both particles and antiparticles? [duplicate]

          Image
          9 $begingroup$ This question already has an answer here: What happens if we put together a proton and an antineutron? 2 answers Is it theoretically possible to make a "deuterium" atom containing a proton and an antineutron in its nucleus? Would the strong nuclear force cause attraction between a proton and an antineutron? Would such a nucleus be stable, or would the proton somehow annihilate the antineutron when close enough? nuclear-physics antimatter share | cite | improve this question asked yesterday
          Read more