Can we apply L'Hospital's rule?












4












$begingroup$


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 it's derivative may not be continuous at some point. One of many counter examples is:
$$f(x) = begin{cases} 0 text{ ; if x=0} \ x^2 sin frac{1}{x} text{; if x $neq$ 0 } end{cases}$$
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







share|cite|improve this question









$endgroup$

















    4












    $begingroup$


    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 it's derivative may not be continuous at some point. One of many counter examples is:
    $$f(x) = begin{cases} 0 text{ ; if x=0} \ x^2 sin frac{1}{x} text{; if x $neq$ 0 } end{cases}$$
    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







    share|cite|improve this question









    $endgroup$















      4












      4








      4


      0



      $begingroup$


      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 it's derivative may not be continuous at some point. One of many counter examples is:
      $$f(x) = begin{cases} 0 text{ ; if x=0} \ x^2 sin frac{1}{x} text{; if x $neq$ 0 } end{cases}$$
      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







      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 it's derivative may not be continuous at some point. One of many counter examples is:
      $$f(x) = begin{cases} 0 text{ ; if x=0} \ x^2 sin frac{1}{x} text{; if x $neq$ 0 } end{cases}$$
      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




      limits derivatives






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 1 hour ago









      DhvanitDhvanit

      1088




      1088






















          2 Answers
          2






          active

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          4












          $begingroup$

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






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            @Dhvanit when $lim_{hto0}frac{f'(h)}{g'(h)}$ is one of $pminfty$, which some people classify as not existing, the implication also holds.
            $endgroup$
            – user647486
            1 hour ago












          • $begingroup$
            Basically, the condition mentioned in this answer means that applying L'Hospital's rule assumes that $lim_{hto 0}f'(a+h)$ 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
            52 mins ago



















          1












          $begingroup$

          In this case - yes, you need derivative to be continuous. In general, you need $lim frac{f'(x)}{g'(x)}$ 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.






          $endgroup$














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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4












            $begingroup$

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






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              @Dhvanit when $lim_{hto0}frac{f'(h)}{g'(h)}$ is one of $pminfty$, which some people classify as not existing, the implication also holds.
              $endgroup$
              – user647486
              1 hour ago












            • $begingroup$
              Basically, the condition mentioned in this answer means that applying L'Hospital's rule assumes that $lim_{hto 0}f'(a+h)$ 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
              52 mins ago
















            4












            $begingroup$

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






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              @Dhvanit when $lim_{hto0}frac{f'(h)}{g'(h)}$ is one of $pminfty$, which some people classify as not existing, the implication also holds.
              $endgroup$
              – user647486
              1 hour ago












            • $begingroup$
              Basically, the condition mentioned in this answer means that applying L'Hospital's rule assumes that $lim_{hto 0}f'(a+h)$ 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
              52 mins ago














            4












            4








            4





            $begingroup$

            L'Hospital's rule says under certain conditions: IF $lim_{hto 0} frac{f'(h)}{g'(h)}=c$ exists, then also $lim_{hto 0} frac{f(h)}{g(h)}=c$. 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 $lim_{hto 0} frac{f'(h)}{g'(h)}=c$ exists, then also $lim_{hto 0} frac{f(h)}{g(h)}=c$. 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 1 hour ago









            HelmutHelmut

            739128




            739128












            • $begingroup$
              @Dhvanit when $lim_{hto0}frac{f'(h)}{g'(h)}$ is one of $pminfty$, which some people classify as not existing, the implication also holds.
              $endgroup$
              – user647486
              1 hour ago












            • $begingroup$
              Basically, the condition mentioned in this answer means that applying L'Hospital's rule assumes that $lim_{hto 0}f'(a+h)$ 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
              52 mins ago


















            • $begingroup$
              @Dhvanit when $lim_{hto0}frac{f'(h)}{g'(h)}$ is one of $pminfty$, which some people classify as not existing, the implication also holds.
              $endgroup$
              – user647486
              1 hour ago












            • $begingroup$
              Basically, the condition mentioned in this answer means that applying L'Hospital's rule assumes that $lim_{hto 0}f'(a+h)$ 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
              52 mins ago
















            $begingroup$
            @Dhvanit when $lim_{hto0}frac{f'(h)}{g'(h)}$ is one of $pminfty$, which some people classify as not existing, the implication also holds.
            $endgroup$
            – user647486
            1 hour ago






            $begingroup$
            @Dhvanit when $lim_{hto0}frac{f'(h)}{g'(h)}$ is one of $pminfty$, which some people classify as not existing, the implication also holds.
            $endgroup$
            – user647486
            1 hour ago














            $begingroup$
            Basically, the condition mentioned in this answer means that applying L'Hospital's rule assumes that $lim_{hto 0}f'(a+h)$ 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
            52 mins ago




            $begingroup$
            Basically, the condition mentioned in this answer means that applying L'Hospital's rule assumes that $lim_{hto 0}f'(a+h)$ 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
            52 mins ago











            1












            $begingroup$

            In this case - yes, you need derivative to be continuous. In general, you need $lim frac{f'(x)}{g'(x)}$ 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 $lim frac{f'(x)}{g'(x)}$ 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 $lim frac{f'(x)}{g'(x)}$ 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 $lim frac{f'(x)}{g'(x)}$ 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 1 hour ago









                mihaildmihaild

                5689




                5689




                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.






























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