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












6














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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  • 1




    Limit operator is not a linear transformation, that is why.
    – Bertrand Wittgenstein's Ghost
    Nov 29 at 0:20








  • 1




    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
    Nov 29 at 0:54
















6














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







share|cite|improve this question




















  • 1




    Limit operator is not a linear transformation, that is why.
    – Bertrand Wittgenstein's Ghost
    Nov 29 at 0:20








  • 1




    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
    Nov 29 at 0:54














6












6








6


1





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







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




calculus limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 at 4:27









snorepion

31




31










asked Nov 29 at 0:16









Danielle

2179




2179








  • 1




    Limit operator is not a linear transformation, that is why.
    – Bertrand Wittgenstein's Ghost
    Nov 29 at 0:20








  • 1




    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
    Nov 29 at 0:54














  • 1




    Limit operator is not a linear transformation, that is why.
    – Bertrand Wittgenstein's Ghost
    Nov 29 at 0:20








  • 1




    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
    Nov 29 at 0:54








1




1




Limit operator is not a linear transformation, that is why.
– Bertrand Wittgenstein's Ghost
Nov 29 at 0:20






Limit operator is not a linear transformation, that is why.
– Bertrand Wittgenstein's Ghost
Nov 29 at 0:20






1




1




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
Nov 29 at 0:54




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
Nov 29 at 0:54










4 Answers
4






active

oldest

votes


















13















I was told that I cannot apply L'Hôpital's rule to each individual limit and then join the limits as ...




This has nothing to do with the L'Hôpital's rule itself.



The rule that you cannot use is:




$limlimits_{xto...}f(x)+limlimits_{xto...}g(x)=limlimits_{xto...}(f(x)+g(x))$




And you can see from JDMan4444's answer that there are situations where this rule does not work.



However, if you are sure that $limlimits_{xto...}f(x)$ and $limlimits_{xto...}g(x)$ exist (and are not $pminfty$), you can apply that rule. (And of course you can apply L'Hôpital's rule to the sum in this case.)



This is important because in some cases it is possible to prove that both limits exist but calculating the limits directly is very difficult or even impossible. In such cases calculating the limit of the sum may be easier.






share|cite|improve this answer





















  • The OP is related to THAT and the main doubt is on the application of HR.
    – gimusi
    Nov 29 at 7:42





















10














Consider the following example:



$$
lim_{xrightarrowinfty}frac{x^2}{x}+lim_{xrightarrowinfty}frac{-x^2}{x} = infty - infty quad text{(which is undefined)}
$$
$$
lim_{xrightarrowinfty}frac{x^2}{x}+frac{-x^2}{x} = 0
$$






share|cite|improve this answer























  • Ypu example is fine and correct but note that the main doubt was in the application of l'HR and is related to this other OP.
    – gimusi
    Nov 29 at 7:44






  • 1




    Your second line "$lim_{xrightarrowinfty}frac{x^2}{x}+frac{-x^2}{x} = 0$" is wrong. Addition has lower precedence than the limit operator, by convention.
    – user21820
    Nov 29 at 13:51






  • 1




    @user21820 indeed; round brackets are needed around both fractions together, otherwise it's +∞ because the limit applies only to the left fraction, the right remains -x^2/x. But that was obvious from the context. More importantly, this applies to the original question as much as here, so comment should be there.
    – user3445853
    Nov 29 at 15:17












  • @user21820 It could be convention but since the expression $frac{x^2}{x}+frac{-x^2}{x}$ is identically equal to zero for all $xneq 0$ I think that it is a noce assumption read that as $$lim_{xrightarrowinfty} left(frac{x^2}{x}+frac{-x^2}{x}right) =lim_{xrightarrowinfty} 0= 0$$
    – gimusi
    Nov 29 at 15:29










  • @user3445853: No. Teachers making errors are one big reason why students don't learn. I've nothing more to say here.
    – user21820
    Nov 29 at 16:47



















3














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





























    1














    It seems to me you got some bad, or incomplete, advice. It's not incorrect if both of those limits exist and are finite. This doesn't have much to do with L'Hopital. The $infty - infty$ or $-infty + infty$ cases are problematic whether you're contemplating L'Hopital or not.






    share|cite|improve this answer





















    • The advice was given HERE for that specific example and it was a completely different case to the one you are referring to.
      – gimusi
      Nov 29 at 7:41










    • @gimusi I responded to the OP's question here. She wrote "I was told that I cannot apply L'Hôpital's rule to each individual limit and then join the limits" and then asked why this is incorrect. I answered that. Why are you sending me to some other question?
      – zhw.
      Nov 29 at 19:19










    • My aim was solely give you a better context for the question posed.
      – gimusi
      Nov 29 at 19:23











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    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    13















    I was told that I cannot apply L'Hôpital's rule to each individual limit and then join the limits as ...




    This has nothing to do with the L'Hôpital's rule itself.



    The rule that you cannot use is:




    $limlimits_{xto...}f(x)+limlimits_{xto...}g(x)=limlimits_{xto...}(f(x)+g(x))$




    And you can see from JDMan4444's answer that there are situations where this rule does not work.



    However, if you are sure that $limlimits_{xto...}f(x)$ and $limlimits_{xto...}g(x)$ exist (and are not $pminfty$), you can apply that rule. (And of course you can apply L'Hôpital's rule to the sum in this case.)



    This is important because in some cases it is possible to prove that both limits exist but calculating the limits directly is very difficult or even impossible. In such cases calculating the limit of the sum may be easier.






    share|cite|improve this answer





















    • The OP is related to THAT and the main doubt is on the application of HR.
      – gimusi
      Nov 29 at 7:42


















    13















    I was told that I cannot apply L'Hôpital's rule to each individual limit and then join the limits as ...




    This has nothing to do with the L'Hôpital's rule itself.



    The rule that you cannot use is:




    $limlimits_{xto...}f(x)+limlimits_{xto...}g(x)=limlimits_{xto...}(f(x)+g(x))$




    And you can see from JDMan4444's answer that there are situations where this rule does not work.



    However, if you are sure that $limlimits_{xto...}f(x)$ and $limlimits_{xto...}g(x)$ exist (and are not $pminfty$), you can apply that rule. (And of course you can apply L'Hôpital's rule to the sum in this case.)



    This is important because in some cases it is possible to prove that both limits exist but calculating the limits directly is very difficult or even impossible. In such cases calculating the limit of the sum may be easier.






    share|cite|improve this answer





















    • The OP is related to THAT and the main doubt is on the application of HR.
      – gimusi
      Nov 29 at 7:42
















    13












    13








    13







    I was told that I cannot apply L'Hôpital's rule to each individual limit and then join the limits as ...




    This has nothing to do with the L'Hôpital's rule itself.



    The rule that you cannot use is:




    $limlimits_{xto...}f(x)+limlimits_{xto...}g(x)=limlimits_{xto...}(f(x)+g(x))$




    And you can see from JDMan4444's answer that there are situations where this rule does not work.



    However, if you are sure that $limlimits_{xto...}f(x)$ and $limlimits_{xto...}g(x)$ exist (and are not $pminfty$), you can apply that rule. (And of course you can apply L'Hôpital's rule to the sum in this case.)



    This is important because in some cases it is possible to prove that both limits exist but calculating the limits directly is very difficult or even impossible. In such cases calculating the limit of the sum may be easier.






    share|cite|improve this answer













    I was told that I cannot apply L'Hôpital's rule to each individual limit and then join the limits as ...




    This has nothing to do with the L'Hôpital's rule itself.



    The rule that you cannot use is:




    $limlimits_{xto...}f(x)+limlimits_{xto...}g(x)=limlimits_{xto...}(f(x)+g(x))$




    And you can see from JDMan4444's answer that there are situations where this rule does not work.



    However, if you are sure that $limlimits_{xto...}f(x)$ and $limlimits_{xto...}g(x)$ exist (and are not $pminfty$), you can apply that rule. (And of course you can apply L'Hôpital's rule to the sum in this case.)



    This is important because in some cases it is possible to prove that both limits exist but calculating the limits directly is very difficult or even impossible. In such cases calculating the limit of the sum may be easier.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 29 at 7:09









    Martin Rosenau

    1,156139




    1,156139












    • The OP is related to THAT and the main doubt is on the application of HR.
      – gimusi
      Nov 29 at 7:42




















    • The OP is related to THAT and the main doubt is on the application of HR.
      – gimusi
      Nov 29 at 7:42


















    The OP is related to THAT and the main doubt is on the application of HR.
    – gimusi
    Nov 29 at 7:42






    The OP is related to THAT and the main doubt is on the application of HR.
    – gimusi
    Nov 29 at 7:42













    10














    Consider the following example:



    $$
    lim_{xrightarrowinfty}frac{x^2}{x}+lim_{xrightarrowinfty}frac{-x^2}{x} = infty - infty quad text{(which is undefined)}
    $$
    $$
    lim_{xrightarrowinfty}frac{x^2}{x}+frac{-x^2}{x} = 0
    $$






    share|cite|improve this answer























    • Ypu example is fine and correct but note that the main doubt was in the application of l'HR and is related to this other OP.
      – gimusi
      Nov 29 at 7:44






    • 1




      Your second line "$lim_{xrightarrowinfty}frac{x^2}{x}+frac{-x^2}{x} = 0$" is wrong. Addition has lower precedence than the limit operator, by convention.
      – user21820
      Nov 29 at 13:51






    • 1




      @user21820 indeed; round brackets are needed around both fractions together, otherwise it's +∞ because the limit applies only to the left fraction, the right remains -x^2/x. But that was obvious from the context. More importantly, this applies to the original question as much as here, so comment should be there.
      – user3445853
      Nov 29 at 15:17












    • @user21820 It could be convention but since the expression $frac{x^2}{x}+frac{-x^2}{x}$ is identically equal to zero for all $xneq 0$ I think that it is a noce assumption read that as $$lim_{xrightarrowinfty} left(frac{x^2}{x}+frac{-x^2}{x}right) =lim_{xrightarrowinfty} 0= 0$$
      – gimusi
      Nov 29 at 15:29










    • @user3445853: No. Teachers making errors are one big reason why students don't learn. I've nothing more to say here.
      – user21820
      Nov 29 at 16:47
















    10














    Consider the following example:



    $$
    lim_{xrightarrowinfty}frac{x^2}{x}+lim_{xrightarrowinfty}frac{-x^2}{x} = infty - infty quad text{(which is undefined)}
    $$
    $$
    lim_{xrightarrowinfty}frac{x^2}{x}+frac{-x^2}{x} = 0
    $$






    share|cite|improve this answer























    • Ypu example is fine and correct but note that the main doubt was in the application of l'HR and is related to this other OP.
      – gimusi
      Nov 29 at 7:44






    • 1




      Your second line "$lim_{xrightarrowinfty}frac{x^2}{x}+frac{-x^2}{x} = 0$" is wrong. Addition has lower precedence than the limit operator, by convention.
      – user21820
      Nov 29 at 13:51






    • 1




      @user21820 indeed; round brackets are needed around both fractions together, otherwise it's +∞ because the limit applies only to the left fraction, the right remains -x^2/x. But that was obvious from the context. More importantly, this applies to the original question as much as here, so comment should be there.
      – user3445853
      Nov 29 at 15:17












    • @user21820 It could be convention but since the expression $frac{x^2}{x}+frac{-x^2}{x}$ is identically equal to zero for all $xneq 0$ I think that it is a noce assumption read that as $$lim_{xrightarrowinfty} left(frac{x^2}{x}+frac{-x^2}{x}right) =lim_{xrightarrowinfty} 0= 0$$
      – gimusi
      Nov 29 at 15:29










    • @user3445853: No. Teachers making errors are one big reason why students don't learn. I've nothing more to say here.
      – user21820
      Nov 29 at 16:47














    10












    10








    10






    Consider the following example:



    $$
    lim_{xrightarrowinfty}frac{x^2}{x}+lim_{xrightarrowinfty}frac{-x^2}{x} = infty - infty quad text{(which is undefined)}
    $$
    $$
    lim_{xrightarrowinfty}frac{x^2}{x}+frac{-x^2}{x} = 0
    $$






    share|cite|improve this answer














    Consider the following example:



    $$
    lim_{xrightarrowinfty}frac{x^2}{x}+lim_{xrightarrowinfty}frac{-x^2}{x} = infty - infty quad text{(which is undefined)}
    $$
    $$
    lim_{xrightarrowinfty}frac{x^2}{x}+frac{-x^2}{x} = 0
    $$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 29 at 5:08









    Tanner Swett

    3,9841638




    3,9841638










    answered Nov 29 at 0:18









    JDMan4444

    23514




    23514












    • Ypu example is fine and correct but note that the main doubt was in the application of l'HR and is related to this other OP.
      – gimusi
      Nov 29 at 7:44






    • 1




      Your second line "$lim_{xrightarrowinfty}frac{x^2}{x}+frac{-x^2}{x} = 0$" is wrong. Addition has lower precedence than the limit operator, by convention.
      – user21820
      Nov 29 at 13:51






    • 1




      @user21820 indeed; round brackets are needed around both fractions together, otherwise it's +∞ because the limit applies only to the left fraction, the right remains -x^2/x. But that was obvious from the context. More importantly, this applies to the original question as much as here, so comment should be there.
      – user3445853
      Nov 29 at 15:17












    • @user21820 It could be convention but since the expression $frac{x^2}{x}+frac{-x^2}{x}$ is identically equal to zero for all $xneq 0$ I think that it is a noce assumption read that as $$lim_{xrightarrowinfty} left(frac{x^2}{x}+frac{-x^2}{x}right) =lim_{xrightarrowinfty} 0= 0$$
      – gimusi
      Nov 29 at 15:29










    • @user3445853: No. Teachers making errors are one big reason why students don't learn. I've nothing more to say here.
      – user21820
      Nov 29 at 16:47


















    • Ypu example is fine and correct but note that the main doubt was in the application of l'HR and is related to this other OP.
      – gimusi
      Nov 29 at 7:44






    • 1




      Your second line "$lim_{xrightarrowinfty}frac{x^2}{x}+frac{-x^2}{x} = 0$" is wrong. Addition has lower precedence than the limit operator, by convention.
      – user21820
      Nov 29 at 13:51






    • 1




      @user21820 indeed; round brackets are needed around both fractions together, otherwise it's +∞ because the limit applies only to the left fraction, the right remains -x^2/x. But that was obvious from the context. More importantly, this applies to the original question as much as here, so comment should be there.
      – user3445853
      Nov 29 at 15:17












    • @user21820 It could be convention but since the expression $frac{x^2}{x}+frac{-x^2}{x}$ is identically equal to zero for all $xneq 0$ I think that it is a noce assumption read that as $$lim_{xrightarrowinfty} left(frac{x^2}{x}+frac{-x^2}{x}right) =lim_{xrightarrowinfty} 0= 0$$
      – gimusi
      Nov 29 at 15:29










    • @user3445853: No. Teachers making errors are one big reason why students don't learn. I've nothing more to say here.
      – user21820
      Nov 29 at 16:47
















    Ypu example is fine and correct but note that the main doubt was in the application of l'HR and is related to this other OP.
    – gimusi
    Nov 29 at 7:44




    Ypu example is fine and correct but note that the main doubt was in the application of l'HR and is related to this other OP.
    – gimusi
    Nov 29 at 7:44




    1




    1




    Your second line "$lim_{xrightarrowinfty}frac{x^2}{x}+frac{-x^2}{x} = 0$" is wrong. Addition has lower precedence than the limit operator, by convention.
    – user21820
    Nov 29 at 13:51




    Your second line "$lim_{xrightarrowinfty}frac{x^2}{x}+frac{-x^2}{x} = 0$" is wrong. Addition has lower precedence than the limit operator, by convention.
    – user21820
    Nov 29 at 13:51




    1




    1




    @user21820 indeed; round brackets are needed around both fractions together, otherwise it's +∞ because the limit applies only to the left fraction, the right remains -x^2/x. But that was obvious from the context. More importantly, this applies to the original question as much as here, so comment should be there.
    – user3445853
    Nov 29 at 15:17






    @user21820 indeed; round brackets are needed around both fractions together, otherwise it's +∞ because the limit applies only to the left fraction, the right remains -x^2/x. But that was obvious from the context. More importantly, this applies to the original question as much as here, so comment should be there.
    – user3445853
    Nov 29 at 15:17














    @user21820 It could be convention but since the expression $frac{x^2}{x}+frac{-x^2}{x}$ is identically equal to zero for all $xneq 0$ I think that it is a noce assumption read that as $$lim_{xrightarrowinfty} left(frac{x^2}{x}+frac{-x^2}{x}right) =lim_{xrightarrowinfty} 0= 0$$
    – gimusi
    Nov 29 at 15:29




    @user21820 It could be convention but since the expression $frac{x^2}{x}+frac{-x^2}{x}$ is identically equal to zero for all $xneq 0$ I think that it is a noce assumption read that as $$lim_{xrightarrowinfty} left(frac{x^2}{x}+frac{-x^2}{x}right) =lim_{xrightarrowinfty} 0= 0$$
    – gimusi
    Nov 29 at 15:29












    @user3445853: No. Teachers making errors are one big reason why students don't learn. I've nothing more to say here.
    – user21820
    Nov 29 at 16:47




    @user3445853: No. Teachers making errors are one big reason why students don't learn. I've nothing more to say here.
    – user21820
    Nov 29 at 16:47











    3














    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


























      3














      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
























        3












        3








        3






        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 Nov 29 at 0:23









        gimusi

        1




        1























            1














            It seems to me you got some bad, or incomplete, advice. It's not incorrect if both of those limits exist and are finite. This doesn't have much to do with L'Hopital. The $infty - infty$ or $-infty + infty$ cases are problematic whether you're contemplating L'Hopital or not.






            share|cite|improve this answer





















            • The advice was given HERE for that specific example and it was a completely different case to the one you are referring to.
              – gimusi
              Nov 29 at 7:41










            • @gimusi I responded to the OP's question here. She wrote "I was told that I cannot apply L'Hôpital's rule to each individual limit and then join the limits" and then asked why this is incorrect. I answered that. Why are you sending me to some other question?
              – zhw.
              Nov 29 at 19:19










            • My aim was solely give you a better context for the question posed.
              – gimusi
              Nov 29 at 19:23
















            1














            It seems to me you got some bad, or incomplete, advice. It's not incorrect if both of those limits exist and are finite. This doesn't have much to do with L'Hopital. The $infty - infty$ or $-infty + infty$ cases are problematic whether you're contemplating L'Hopital or not.






            share|cite|improve this answer





















            • The advice was given HERE for that specific example and it was a completely different case to the one you are referring to.
              – gimusi
              Nov 29 at 7:41










            • @gimusi I responded to the OP's question here. She wrote "I was told that I cannot apply L'Hôpital's rule to each individual limit and then join the limits" and then asked why this is incorrect. I answered that. Why are you sending me to some other question?
              – zhw.
              Nov 29 at 19:19










            • My aim was solely give you a better context for the question posed.
              – gimusi
              Nov 29 at 19:23














            1












            1








            1






            It seems to me you got some bad, or incomplete, advice. It's not incorrect if both of those limits exist and are finite. This doesn't have much to do with L'Hopital. The $infty - infty$ or $-infty + infty$ cases are problematic whether you're contemplating L'Hopital or not.






            share|cite|improve this answer












            It seems to me you got some bad, or incomplete, advice. It's not incorrect if both of those limits exist and are finite. This doesn't have much to do with L'Hopital. The $infty - infty$ or $-infty + infty$ cases are problematic whether you're contemplating L'Hopital or not.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 29 at 7:28









            zhw.

            71.5k43075




            71.5k43075












            • The advice was given HERE for that specific example and it was a completely different case to the one you are referring to.
              – gimusi
              Nov 29 at 7:41










            • @gimusi I responded to the OP's question here. She wrote "I was told that I cannot apply L'Hôpital's rule to each individual limit and then join the limits" and then asked why this is incorrect. I answered that. Why are you sending me to some other question?
              – zhw.
              Nov 29 at 19:19










            • My aim was solely give you a better context for the question posed.
              – gimusi
              Nov 29 at 19:23


















            • The advice was given HERE for that specific example and it was a completely different case to the one you are referring to.
              – gimusi
              Nov 29 at 7:41










            • @gimusi I responded to the OP's question here. She wrote "I was told that I cannot apply L'Hôpital's rule to each individual limit and then join the limits" and then asked why this is incorrect. I answered that. Why are you sending me to some other question?
              – zhw.
              Nov 29 at 19:19










            • My aim was solely give you a better context for the question posed.
              – gimusi
              Nov 29 at 19:23
















            The advice was given HERE for that specific example and it was a completely different case to the one you are referring to.
            – gimusi
            Nov 29 at 7:41




            The advice was given HERE for that specific example and it was a completely different case to the one you are referring to.
            – gimusi
            Nov 29 at 7:41












            @gimusi I responded to the OP's question here. She wrote "I was told that I cannot apply L'Hôpital's rule to each individual limit and then join the limits" and then asked why this is incorrect. I answered that. Why are you sending me to some other question?
            – zhw.
            Nov 29 at 19:19




            @gimusi I responded to the OP's question here. She wrote "I was told that I cannot apply L'Hôpital's rule to each individual limit and then join the limits" and then asked why this is incorrect. I answered that. Why are you sending me to some other question?
            – zhw.
            Nov 29 at 19:19












            My aim was solely give you a better context for the question posed.
            – gimusi
            Nov 29 at 19:23




            My aim was solely give you a better context for the question posed.
            – gimusi
            Nov 29 at 19:23


















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