Why Can L'Hôpital's Rule Not be Applied to the Sum or Difference of Limits?











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1
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Consider $$lim_{xtoinfty}frac{f(x)}{g(x)} + lim_{xtoinfty}frac{h(x)}{i(x)}$$
I was told that I cannot apply L'Hôpital's rule to each individual limit and then join the limits as
$$lim_{xtoinfty}frac{f(x)}{g(x)} +frac{h(x)}{i(x)}$$
Why is this incorrect?










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  • 1




    Limit operator is not a linear transformation, that is why.
    – Bertrand Wittgenstein's Ghost
    4 hours ago












  • I think it only goes wrong when you will end up with something of the form $+infty - infty$. So, if you avoid these cases you can apply it if I am correct.
    – Stan Tendijck
    3 hours ago















up vote
1
down vote

favorite












Consider $$lim_{xtoinfty}frac{f(x)}{g(x)} + lim_{xtoinfty}frac{h(x)}{i(x)}$$
I was told that I cannot apply L'Hôpital's rule to each individual limit and then join the limits as
$$lim_{xtoinfty}frac{f(x)}{g(x)} +frac{h(x)}{i(x)}$$
Why is this incorrect?










share|cite|improve this question




















  • 1




    Limit operator is not a linear transformation, that is why.
    – Bertrand Wittgenstein's Ghost
    4 hours ago












  • I think it only goes wrong when you will end up with something of the form $+infty - infty$. So, if you avoid these cases you can apply it if I am correct.
    – Stan Tendijck
    3 hours ago













up vote
1
down vote

favorite









up vote
1
down vote

favorite











Consider $$lim_{xtoinfty}frac{f(x)}{g(x)} + lim_{xtoinfty}frac{h(x)}{i(x)}$$
I was told that I cannot apply L'Hôpital's rule to each individual limit and then join the limits as
$$lim_{xtoinfty}frac{f(x)}{g(x)} +frac{h(x)}{i(x)}$$
Why is this incorrect?










share|cite|improve this question















Consider $$lim_{xtoinfty}frac{f(x)}{g(x)} + lim_{xtoinfty}frac{h(x)}{i(x)}$$
I was told that I cannot apply L'Hôpital's rule to each individual limit and then join the limits as
$$lim_{xtoinfty}frac{f(x)}{g(x)} +frac{h(x)}{i(x)}$$
Why is this incorrect?







calculus limits






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edited 21 mins ago









snorepion

31




31










asked 4 hours ago









Danielle

1457




1457








  • 1




    Limit operator is not a linear transformation, that is why.
    – Bertrand Wittgenstein's Ghost
    4 hours ago












  • I think it only goes wrong when you will end up with something of the form $+infty - infty$. So, if you avoid these cases you can apply it if I am correct.
    – Stan Tendijck
    3 hours ago














  • 1




    Limit operator is not a linear transformation, that is why.
    – Bertrand Wittgenstein's Ghost
    4 hours ago












  • I think it only goes wrong when you will end up with something of the form $+infty - infty$. So, if you avoid these cases you can apply it if I am correct.
    – Stan Tendijck
    3 hours ago








1




1




Limit operator is not a linear transformation, that is why.
– Bertrand Wittgenstein's Ghost
4 hours ago






Limit operator is not a linear transformation, that is why.
– Bertrand Wittgenstein's Ghost
4 hours ago














I think it only goes wrong when you will end up with something of the form $+infty - infty$. So, if you avoid these cases you can apply it if I am correct.
– Stan Tendijck
3 hours ago




I think it only goes wrong when you will end up with something of the form $+infty - infty$. So, if you avoid these cases you can apply it if I am correct.
– Stan Tendijck
3 hours ago










2 Answers
2






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up vote
8
down vote



accepted










Consider the following example:



$$
lim_{xrightarrowinfty}frac{x^2}{x}+lim_{xrightarrowinfty}frac{-x^2}{x} = infty - infty = text{Undefined}
$$

$$
lim_{xrightarrowinfty}frac{x^2}{x}+frac{-x^2}{x} = 0
$$






share|cite|improve this answer








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JDMan4444 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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    up vote
    1
    down vote













    The following identity



    $$lim_{xto infty}left(frac{f(x)}{g(x)} +frac{h(x)}{i(x)}right)=lim_{xto infty}frac{f(x)}{g(x)} + lim_{xto infty}frac{h(x)}{i(x)}$$



    doesn't hold in general and to solve the LHS limit by l'Hopital, if necessary, we need to put it in the form



    $$lim_{xto infty}left(frac{f(x)}{g(x)} +frac{h(x)}{i(x)}right)=lim_{xto infty}frac{f(x)i(x)+h(x)g(x)}{g(x)i(x)} $$



    the reason is that the case you are referring to is not among the cases considered by l'Hopital theorem.






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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      8
      down vote



      accepted










      Consider the following example:



      $$
      lim_{xrightarrowinfty}frac{x^2}{x}+lim_{xrightarrowinfty}frac{-x^2}{x} = infty - infty = text{Undefined}
      $$

      $$
      lim_{xrightarrowinfty}frac{x^2}{x}+frac{-x^2}{x} = 0
      $$






      share|cite|improve this answer








      New contributor




      JDMan4444 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






















        up vote
        8
        down vote



        accepted










        Consider the following example:



        $$
        lim_{xrightarrowinfty}frac{x^2}{x}+lim_{xrightarrowinfty}frac{-x^2}{x} = infty - infty = text{Undefined}
        $$

        $$
        lim_{xrightarrowinfty}frac{x^2}{x}+frac{-x^2}{x} = 0
        $$






        share|cite|improve this answer








        New contributor




        JDMan4444 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.




















          up vote
          8
          down vote



          accepted







          up vote
          8
          down vote



          accepted






          Consider the following example:



          $$
          lim_{xrightarrowinfty}frac{x^2}{x}+lim_{xrightarrowinfty}frac{-x^2}{x} = infty - infty = text{Undefined}
          $$

          $$
          lim_{xrightarrowinfty}frac{x^2}{x}+frac{-x^2}{x} = 0
          $$






          share|cite|improve this answer








          New contributor




          JDMan4444 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.









          Consider the following example:



          $$
          lim_{xrightarrowinfty}frac{x^2}{x}+lim_{xrightarrowinfty}frac{-x^2}{x} = infty - infty = text{Undefined}
          $$

          $$
          lim_{xrightarrowinfty}frac{x^2}{x}+frac{-x^2}{x} = 0
          $$







          share|cite|improve this answer








          New contributor




          JDMan4444 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




          JDMan4444 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.









          answered 4 hours ago









          JDMan4444

          2143




          2143




          New contributor




          JDMan4444 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.





          New contributor





          JDMan4444 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          JDMan4444 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






















              up vote
              1
              down vote













              The following identity



              $$lim_{xto infty}left(frac{f(x)}{g(x)} +frac{h(x)}{i(x)}right)=lim_{xto infty}frac{f(x)}{g(x)} + lim_{xto infty}frac{h(x)}{i(x)}$$



              doesn't hold in general and to solve the LHS limit by l'Hopital, if necessary, we need to put it in the form



              $$lim_{xto infty}left(frac{f(x)}{g(x)} +frac{h(x)}{i(x)}right)=lim_{xto infty}frac{f(x)i(x)+h(x)g(x)}{g(x)i(x)} $$



              the reason is that the case you are referring to is not among the cases considered by l'Hopital theorem.






              share|cite|improve this answer

























                up vote
                1
                down vote













                The following identity



                $$lim_{xto infty}left(frac{f(x)}{g(x)} +frac{h(x)}{i(x)}right)=lim_{xto infty}frac{f(x)}{g(x)} + lim_{xto infty}frac{h(x)}{i(x)}$$



                doesn't hold in general and to solve the LHS limit by l'Hopital, if necessary, we need to put it in the form



                $$lim_{xto infty}left(frac{f(x)}{g(x)} +frac{h(x)}{i(x)}right)=lim_{xto infty}frac{f(x)i(x)+h(x)g(x)}{g(x)i(x)} $$



                the reason is that the case you are referring to is not among the cases considered by l'Hopital theorem.






                share|cite|improve this answer























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  The following identity



                  $$lim_{xto infty}left(frac{f(x)}{g(x)} +frac{h(x)}{i(x)}right)=lim_{xto infty}frac{f(x)}{g(x)} + lim_{xto infty}frac{h(x)}{i(x)}$$



                  doesn't hold in general and to solve the LHS limit by l'Hopital, if necessary, we need to put it in the form



                  $$lim_{xto infty}left(frac{f(x)}{g(x)} +frac{h(x)}{i(x)}right)=lim_{xto infty}frac{f(x)i(x)+h(x)g(x)}{g(x)i(x)} $$



                  the reason is that the case you are referring to is not among the cases considered by l'Hopital theorem.






                  share|cite|improve this answer












                  The following identity



                  $$lim_{xto infty}left(frac{f(x)}{g(x)} +frac{h(x)}{i(x)}right)=lim_{xto infty}frac{f(x)}{g(x)} + lim_{xto infty}frac{h(x)}{i(x)}$$



                  doesn't hold in general and to solve the LHS limit by l'Hopital, if necessary, we need to put it in the form



                  $$lim_{xto infty}left(frac{f(x)}{g(x)} +frac{h(x)}{i(x)}right)=lim_{xto infty}frac{f(x)i(x)+h(x)g(x)}{g(x)i(x)} $$



                  the reason is that the case you are referring to is not among the cases considered by l'Hopital theorem.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 4 hours ago









                  gimusi

                  88k74393




                  88k74393






























                       

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