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Can we apply L'Hospital's rule where the derivative is not continuous?



The 2019 Stack Overflow Developer Survey Results Are InResilient L'Hospital's rule questionIs this a valid use of l'Hospital's Rule? Can it be used recursively?Simple Derivation of Functional Equation Question (L'Hospital's Rule)L'Hôpital's rule and Difference QuotientsL'Hôpital's rule does not apply?!Fake proof for “differentiability implies continuous derivative”: reviewL'Hospital's rule helpCan a function be differentiable on an interval, but not continuously differentiable somewhere besides an oscillating discontinuity in the derivative?Compute the limit without L'Hospital's ruleProof of L'Hospital's Rule










7












$begingroup$


My doubt arises due to the following :



We know that the definition of the derivative of a function at a point $x=a$, if it is differentiable at $a$, is:
$$f'(a) = lim_h rightarrow 0 frac f(a+h) - f(a)h$$



Suppose that the function $f(x)$ is differentiable in a finite interval $[c,d]$ and $a in (c,d) $



So, we can apply L'Hospital's rule. On differentiating numerator and denominator with respect to $h$, we get:
$$f'(a) = lim_h rightarrow 0 frac f(a+h) - f(a)h = lim_h rightarrow 0 frac f'(a+h)1$$
Which implies that
$$f'(a) = lim_h rightarrow 0 f'(a+h)$$
Which means that the function $f'(x)$ is continuous at $x=a$



But this not necessarily true. A function may have a derivative everywhere but its derivative may not be continuous at some point. One of many counterexamples is:
$$f(x) = begincases 0 text ; if x=0 \ x^2 sin frac1x text; if x $neq$ 0 endcases$$
Whose derivative isn't continuous at $0$



So, is something wrong with what I have done ? Or is it necessary that for applying L'Hospital's rule, the function's derivative must be a continuous function?



If the latter is true, why does that condition appear in the proof for L'Hospital's rule ?










share|cite|improve this question











$endgroup$
















    7












    $begingroup$


    My doubt arises due to the following :



    We know that the definition of the derivative of a function at a point $x=a$, if it is differentiable at $a$, is:
    $$f'(a) = lim_h rightarrow 0 frac f(a+h) - f(a)h$$



    Suppose that the function $f(x)$ is differentiable in a finite interval $[c,d]$ and $a in (c,d) $



    So, we can apply L'Hospital's rule. On differentiating numerator and denominator with respect to $h$, we get:
    $$f'(a) = lim_h rightarrow 0 frac f(a+h) - f(a)h = lim_h rightarrow 0 frac f'(a+h)1$$
    Which implies that
    $$f'(a) = lim_h rightarrow 0 f'(a+h)$$
    Which means that the function $f'(x)$ is continuous at $x=a$



    But this not necessarily true. A function may have a derivative everywhere but its derivative may not be continuous at some point. One of many counterexamples is:
    $$f(x) = begincases 0 text ; if x=0 \ x^2 sin frac1x text; if x $neq$ 0 endcases$$
    Whose derivative isn't continuous at $0$



    So, is something wrong with what I have done ? Or is it necessary that for applying L'Hospital's rule, the function's derivative must be a continuous function?



    If the latter is true, why does that condition appear in the proof for L'Hospital's rule ?










    share|cite|improve this question











    $endgroup$














      7












      7








      7


      1



      $begingroup$


      My doubt arises due to the following :



      We know that the definition of the derivative of a function at a point $x=a$, if it is differentiable at $a$, is:
      $$f'(a) = lim_h rightarrow 0 frac f(a+h) - f(a)h$$



      Suppose that the function $f(x)$ is differentiable in a finite interval $[c,d]$ and $a in (c,d) $



      So, we can apply L'Hospital's rule. On differentiating numerator and denominator with respect to $h$, we get:
      $$f'(a) = lim_h rightarrow 0 frac f(a+h) - f(a)h = lim_h rightarrow 0 frac f'(a+h)1$$
      Which implies that
      $$f'(a) = lim_h rightarrow 0 f'(a+h)$$
      Which means that the function $f'(x)$ is continuous at $x=a$



      But this not necessarily true. A function may have a derivative everywhere but its derivative may not be continuous at some point. One of many counterexamples is:
      $$f(x) = begincases 0 text ; if x=0 \ x^2 sin frac1x text; if x $neq$ 0 endcases$$
      Whose derivative isn't continuous at $0$



      So, is something wrong with what I have done ? Or is it necessary that for applying L'Hospital's rule, the function's derivative must be a continuous function?



      If the latter is true, why does that condition appear in the proof for L'Hospital's rule ?










      share|cite|improve this question











      $endgroup$




      My doubt arises due to the following :



      We know that the definition of the derivative of a function at a point $x=a$, if it is differentiable at $a$, is:
      $$f'(a) = lim_h rightarrow 0 frac f(a+h) - f(a)h$$



      Suppose that the function $f(x)$ is differentiable in a finite interval $[c,d]$ and $a in (c,d) $



      So, we can apply L'Hospital's rule. On differentiating numerator and denominator with respect to $h$, we get:
      $$f'(a) = lim_h rightarrow 0 frac f(a+h) - f(a)h = lim_h rightarrow 0 frac f'(a+h)1$$
      Which implies that
      $$f'(a) = lim_h rightarrow 0 f'(a+h)$$
      Which means that the function $f'(x)$ is continuous at $x=a$



      But this not necessarily true. A function may have a derivative everywhere but its derivative may not be continuous at some point. One of many counterexamples is:
      $$f(x) = begincases 0 text ; if x=0 \ x^2 sin frac1x text; if x $neq$ 0 endcases$$
      Whose derivative isn't continuous at $0$



      So, is something wrong with what I have done ? Or is it necessary that for applying L'Hospital's rule, the function's derivative must be a continuous function?



      If the latter is true, why does that condition appear in the proof for L'Hospital's rule ?







      limits derivatives continuity






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 4 hours ago









      200_success

      668515




      668515










      asked 12 hours ago









      DhvanitDhvanit

      1239




      1239




















          3 Answers
          3






          active

          oldest

          votes


















          8












          $begingroup$

          L'Hospital's rule says under certain conditions: IF $lim_hto 0 fracf'(h)g'(h)=c$ exists, then also $lim_hto 0 fracf(h)g(h)=c$. It does not say anything about the existence of the former limit.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            @Dhvanit when $lim_hto0fracf'(h)g'(h)$ is one of $pminfty$, which some people classify as not existing, the implication also holds.
            $endgroup$
            – user647486
            11 hours ago











          • $begingroup$
            Basically, the condition mentioned in this answer means that applying L'Hospital's rule assumes that $lim_hto 0f'(a+h)$ exists. And if it exists, it can be shown that it must be $f'(a)$ (because derivatives need not be continuous, but still must fulfill the intermdiate value condition).
            $endgroup$
            – Ingix
            11 hours ago


















          1












          $begingroup$

          In this case - yes, you need derivative to be continuous. In general, you need $lim fracf'(x)g'(x)$ to exist to apply L'Hospital's rule. As in your case $g'(x) = 1$, you proved that if there is a limit of $f'(a + h)$, then the limit is equal to $f'(a)$.






          share|cite|improve this answer








          New contributor




          mihaild is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          $endgroup$




















            0












            $begingroup$

            The derivative value exists if:
            1. the left-side (-0) derivative exists
            2. the right-side (+0) derivative exists
            3. and they are the same/identical .



            In your case they are not identical.






            share|cite|improve this answer








            New contributor




            user9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.






            $endgroup$













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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              8












              $begingroup$

              L'Hospital's rule says under certain conditions: IF $lim_hto 0 fracf'(h)g'(h)=c$ exists, then also $lim_hto 0 fracf(h)g(h)=c$. It does not say anything about the existence of the former limit.






              share|cite|improve this answer









              $endgroup$












              • $begingroup$
                @Dhvanit when $lim_hto0fracf'(h)g'(h)$ is one of $pminfty$, which some people classify as not existing, the implication also holds.
                $endgroup$
                – user647486
                11 hours ago











              • $begingroup$
                Basically, the condition mentioned in this answer means that applying L'Hospital's rule assumes that $lim_hto 0f'(a+h)$ exists. And if it exists, it can be shown that it must be $f'(a)$ (because derivatives need not be continuous, but still must fulfill the intermdiate value condition).
                $endgroup$
                – Ingix
                11 hours ago















              8












              $begingroup$

              L'Hospital's rule says under certain conditions: IF $lim_hto 0 fracf'(h)g'(h)=c$ exists, then also $lim_hto 0 fracf(h)g(h)=c$. It does not say anything about the existence of the former limit.






              share|cite|improve this answer









              $endgroup$












              • $begingroup$
                @Dhvanit when $lim_hto0fracf'(h)g'(h)$ is one of $pminfty$, which some people classify as not existing, the implication also holds.
                $endgroup$
                – user647486
                11 hours ago











              • $begingroup$
                Basically, the condition mentioned in this answer means that applying L'Hospital's rule assumes that $lim_hto 0f'(a+h)$ exists. And if it exists, it can be shown that it must be $f'(a)$ (because derivatives need not be continuous, but still must fulfill the intermdiate value condition).
                $endgroup$
                – Ingix
                11 hours ago













              8












              8








              8





              $begingroup$

              L'Hospital's rule says under certain conditions: IF $lim_hto 0 fracf'(h)g'(h)=c$ exists, then also $lim_hto 0 fracf(h)g(h)=c$. It does not say anything about the existence of the former limit.






              share|cite|improve this answer









              $endgroup$



              L'Hospital's rule says under certain conditions: IF $lim_hto 0 fracf'(h)g'(h)=c$ exists, then also $lim_hto 0 fracf(h)g(h)=c$. It does not say anything about the existence of the former limit.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered 12 hours ago









              HelmutHelmut

              789128




              789128











              • $begingroup$
                @Dhvanit when $lim_hto0fracf'(h)g'(h)$ is one of $pminfty$, which some people classify as not existing, the implication also holds.
                $endgroup$
                – user647486
                11 hours ago











              • $begingroup$
                Basically, the condition mentioned in this answer means that applying L'Hospital's rule assumes that $lim_hto 0f'(a+h)$ exists. And if it exists, it can be shown that it must be $f'(a)$ (because derivatives need not be continuous, but still must fulfill the intermdiate value condition).
                $endgroup$
                – Ingix
                11 hours ago
















              • $begingroup$
                @Dhvanit when $lim_hto0fracf'(h)g'(h)$ is one of $pminfty$, which some people classify as not existing, the implication also holds.
                $endgroup$
                – user647486
                11 hours ago











              • $begingroup$
                Basically, the condition mentioned in this answer means that applying L'Hospital's rule assumes that $lim_hto 0f'(a+h)$ exists. And if it exists, it can be shown that it must be $f'(a)$ (because derivatives need not be continuous, but still must fulfill the intermdiate value condition).
                $endgroup$
                – Ingix
                11 hours ago















              $begingroup$
              @Dhvanit when $lim_hto0fracf'(h)g'(h)$ is one of $pminfty$, which some people classify as not existing, the implication also holds.
              $endgroup$
              – user647486
              11 hours ago





              $begingroup$
              @Dhvanit when $lim_hto0fracf'(h)g'(h)$ is one of $pminfty$, which some people classify as not existing, the implication also holds.
              $endgroup$
              – user647486
              11 hours ago













              $begingroup$
              Basically, the condition mentioned in this answer means that applying L'Hospital's rule assumes that $lim_hto 0f'(a+h)$ exists. And if it exists, it can be shown that it must be $f'(a)$ (because derivatives need not be continuous, but still must fulfill the intermdiate value condition).
              $endgroup$
              – Ingix
              11 hours ago




              $begingroup$
              Basically, the condition mentioned in this answer means that applying L'Hospital's rule assumes that $lim_hto 0f'(a+h)$ exists. And if it exists, it can be shown that it must be $f'(a)$ (because derivatives need not be continuous, but still must fulfill the intermdiate value condition).
              $endgroup$
              – Ingix
              11 hours ago











              1












              $begingroup$

              In this case - yes, you need derivative to be continuous. In general, you need $lim fracf'(x)g'(x)$ to exist to apply L'Hospital's rule. As in your case $g'(x) = 1$, you proved that if there is a limit of $f'(a + h)$, then the limit is equal to $f'(a)$.






              share|cite|improve this answer








              New contributor




              mihaild is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.






              $endgroup$

















                1












                $begingroup$

                In this case - yes, you need derivative to be continuous. In general, you need $lim fracf'(x)g'(x)$ to exist to apply L'Hospital's rule. As in your case $g'(x) = 1$, you proved that if there is a limit of $f'(a + h)$, then the limit is equal to $f'(a)$.






                share|cite|improve this answer








                New contributor




                mihaild is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.






                $endgroup$















                  1












                  1








                  1





                  $begingroup$

                  In this case - yes, you need derivative to be continuous. In general, you need $lim fracf'(x)g'(x)$ to exist to apply L'Hospital's rule. As in your case $g'(x) = 1$, you proved that if there is a limit of $f'(a + h)$, then the limit is equal to $f'(a)$.






                  share|cite|improve this answer








                  New contributor




                  mihaild is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.






                  $endgroup$



                  In this case - yes, you need derivative to be continuous. In general, you need $lim fracf'(x)g'(x)$ to exist to apply L'Hospital's rule. As in your case $g'(x) = 1$, you proved that if there is a limit of $f'(a + h)$, then the limit is equal to $f'(a)$.







                  share|cite|improve this answer








                  New contributor




                  mihaild is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.









                  share|cite|improve this answer



                  share|cite|improve this answer






                  New contributor




                  mihaild is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.









                  answered 12 hours ago









                  mihaildmihaild

                  56810




                  56810




                  New contributor




                  mihaild is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.





                  New contributor





                  mihaild is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.






                  mihaild is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.





















                      0












                      $begingroup$

                      The derivative value exists if:
                      1. the left-side (-0) derivative exists
                      2. the right-side (+0) derivative exists
                      3. and they are the same/identical .



                      In your case they are not identical.






                      share|cite|improve this answer








                      New contributor




                      user9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.






                      $endgroup$

















                        0












                        $begingroup$

                        The derivative value exists if:
                        1. the left-side (-0) derivative exists
                        2. the right-side (+0) derivative exists
                        3. and they are the same/identical .



                        In your case they are not identical.






                        share|cite|improve this answer








                        New contributor




                        user9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                        Check out our Code of Conduct.






                        $endgroup$















                          0












                          0








                          0





                          $begingroup$

                          The derivative value exists if:
                          1. the left-side (-0) derivative exists
                          2. the right-side (+0) derivative exists
                          3. and they are the same/identical .



                          In your case they are not identical.






                          share|cite|improve this answer








                          New contributor




                          user9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.






                          $endgroup$



                          The derivative value exists if:
                          1. the left-side (-0) derivative exists
                          2. the right-side (+0) derivative exists
                          3. and they are the same/identical .



                          In your case they are not identical.







                          share|cite|improve this answer








                          New contributor




                          user9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.









                          share|cite|improve this answer



                          share|cite|improve this answer






                          New contributor




                          user9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.









                          answered 9 hours ago









                          user9user9

                          1




                          1




                          New contributor




                          user9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.





                          New contributor





                          user9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.






                          user9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.



























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