Why isn't L Hospital Rule applicable here?












2














$lim_{x to 0} frac{e^frac{1}{x}}{e^frac{1}{x}+1}$



I'm trying to use this form of LH Rule. enter image description here






calculus







share|cite|improve this question






















  • Why not? I think it is allright
    – José Alejandro Aburto Araneda
    4 hours ago






  • 6




    The problem is that $e^{1/x}$ doesn't approach a limit as $xto 0$. It approaches $+infty$ as $x$ goes to $0$ from the right, and it goes to $0$ as $x$ goes to $0$ from the left. Thus your expression goes to $1$ on the right, and $0$ on the left.
    – lulu
    4 hours ago












  • You can use L'hopital to find the right side limit but it's not necessary. You can't apply it to find the left side limit as that is neither $infty/infty$ nor $0/0$.
    – fleablood
    3 hours ago
















2














$lim_{x to 0} frac{e^frac{1}{x}}{e^frac{1}{x}+1}$



I'm trying to use this form of LH Rule. enter image description here






calculus







share|cite|improve this question






















  • Why not? I think it is allright
    – José Alejandro Aburto Araneda
    4 hours ago






  • 6




    The problem is that $e^{1/x}$ doesn't approach a limit as $xto 0$. It approaches $+infty$ as $x$ goes to $0$ from the right, and it goes to $0$ as $x$ goes to $0$ from the left. Thus your expression goes to $1$ on the right, and $0$ on the left.
    – lulu
    4 hours ago












  • You can use L'hopital to find the right side limit but it's not necessary. You can't apply it to find the left side limit as that is neither $infty/infty$ nor $0/0$.
    – fleablood
    3 hours ago














2












2








2







$lim_{x to 0} frac{e^frac{1}{x}}{e^frac{1}{x}+1}$



I'm trying to use this form of LH Rule. enter image description here






calculus







share|cite|improve this question













$lim_{x to 0} frac{e^frac{1}{x}}{e^frac{1}{x}+1}$



I'm trying to use this form of LH Rule. enter image description here






calculus




calculus






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 4 hours ago









user1752323

304




304












  • Why not? I think it is allright
    – José Alejandro Aburto Araneda
    4 hours ago






  • 6




    The problem is that $e^{1/x}$ doesn't approach a limit as $xto 0$. It approaches $+infty$ as $x$ goes to $0$ from the right, and it goes to $0$ as $x$ goes to $0$ from the left. Thus your expression goes to $1$ on the right, and $0$ on the left.
    – lulu
    4 hours ago












  • You can use L'hopital to find the right side limit but it's not necessary. You can't apply it to find the left side limit as that is neither $infty/infty$ nor $0/0$.
    – fleablood
    3 hours ago


















  • Why not? I think it is allright
    – José Alejandro Aburto Araneda
    4 hours ago






  • 6




    The problem is that $e^{1/x}$ doesn't approach a limit as $xto 0$. It approaches $+infty$ as $x$ goes to $0$ from the right, and it goes to $0$ as $x$ goes to $0$ from the left. Thus your expression goes to $1$ on the right, and $0$ on the left.
    – lulu
    4 hours ago












  • You can use L'hopital to find the right side limit but it's not necessary. You can't apply it to find the left side limit as that is neither $infty/infty$ nor $0/0$.
    – fleablood
    3 hours ago
















Why not? I think it is allright
– José Alejandro Aburto Araneda
4 hours ago




Why not? I think it is allright
– José Alejandro Aburto Araneda
4 hours ago




6




6




The problem is that $e^{1/x}$ doesn't approach a limit as $xto 0$. It approaches $+infty$ as $x$ goes to $0$ from the right, and it goes to $0$ as $x$ goes to $0$ from the left. Thus your expression goes to $1$ on the right, and $0$ on the left.
– lulu
4 hours ago






The problem is that $e^{1/x}$ doesn't approach a limit as $xto 0$. It approaches $+infty$ as $x$ goes to $0$ from the right, and it goes to $0$ as $x$ goes to $0$ from the left. Thus your expression goes to $1$ on the right, and $0$ on the left.
– lulu
4 hours ago














You can use L'hopital to find the right side limit but it's not necessary. You can't apply it to find the left side limit as that is neither $infty/infty$ nor $0/0$.
– fleablood
3 hours ago




You can use L'hopital to find the right side limit but it's not necessary. You can't apply it to find the left side limit as that is neither $infty/infty$ nor $0/0$.
– fleablood
3 hours ago










3 Answers
3






active

oldest

votes


















3














L'Hopital's rule does not apply because the limit
$$
lim_{xto 0}e^{1/x}
$$
does not exist, since
$$
lim_{xto 0^-}e^{1/x}=0quadtext{while}quad lim_{xto 0^+}e^{1/x}=infty
$$
So, consequently
$$
lim_{xto 0^-}frac{e^{1/x}}{e^{1/x}+1}=0quadtext{while}quad lim_{xto 0^+}frac{e^{1/x}}{e^{1/x}+1}=1
$$






share|cite|improve this answer



















  • 1




    This is not the reason why l'Hôpital cannot be applied.
    – egreg
    3 hours ago



















1














You can apply l’Hôpital for the limit from the right, because it is in the form $infty/infty$.



The limit from the left is instead in the form $0/1$, which is not indeterminate. In order to use that form of l'Hôpital for a two sided limit, you need $infty/infty$ on both sides.



More generally you need $infty/infty$ or $0/0$ on either side.



If you blindly apply l’Hôpital, you end up with
$$
lim_{xto0}frac{-e^{1/x}/x^2}{-e^{1/x}/x^2}=1
$$
but this would be incorrect, because
$$
lim_{xto0^-}frac{e^{1/x}}{e^{1/x}+1}=0
$$
For the limit from the right the limit is correct, but using l’Hôpital is of course not necessary:
$$
lim_{xto0^+}frac{e^{1/x}}{e^{1/x}+1}=lim_{xto0^+}frac{1}{1+e^{-1/x}}=
frac{1}{1+0}=1
$$






share|cite|improve this answer































    0














    here, $$when~~~xto 0^+,e^{frac{1}{x}}to infty$$
    So,$$lim_{x to 0^+} frac{e^frac{1}{x}}{e^frac{1}{x}+1}=lim_{x to 0^+}frac{e^frac{1}{x}}{e^frac{1}{x}left(1+frac{1}{e^frac{1}{x}} right)}=lim_{x to 0^+}frac{1}{left(1+frac{1}{e^frac{1}{x}} right)}=frac{1}{1+0}=1$$
    $$when~~~xto 0^-,e^{frac{1}{x}} to 0$$
    So,
    $$lim_{x to 0^-} frac{e^frac{1}{x}}{e^frac{1}{x}+1}=frac{0}{0+1}=0$$



    so,simply limit doesn't exist.so we can't find the limit when $x to 0$ no matter what procedure you follow.






    share|cite|improve this answer





















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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3














      L'Hopital's rule does not apply because the limit
      $$
      lim_{xto 0}e^{1/x}
      $$
      does not exist, since
      $$
      lim_{xto 0^-}e^{1/x}=0quadtext{while}quad lim_{xto 0^+}e^{1/x}=infty
      $$
      So, consequently
      $$
      lim_{xto 0^-}frac{e^{1/x}}{e^{1/x}+1}=0quadtext{while}quad lim_{xto 0^+}frac{e^{1/x}}{e^{1/x}+1}=1
      $$






      share|cite|improve this answer



















      • 1




        This is not the reason why l'Hôpital cannot be applied.
        – egreg
        3 hours ago
















      3














      L'Hopital's rule does not apply because the limit
      $$
      lim_{xto 0}e^{1/x}
      $$
      does not exist, since
      $$
      lim_{xto 0^-}e^{1/x}=0quadtext{while}quad lim_{xto 0^+}e^{1/x}=infty
      $$
      So, consequently
      $$
      lim_{xto 0^-}frac{e^{1/x}}{e^{1/x}+1}=0quadtext{while}quad lim_{xto 0^+}frac{e^{1/x}}{e^{1/x}+1}=1
      $$






      share|cite|improve this answer



















      • 1




        This is not the reason why l'Hôpital cannot be applied.
        – egreg
        3 hours ago














      3












      3








      3






      L'Hopital's rule does not apply because the limit
      $$
      lim_{xto 0}e^{1/x}
      $$
      does not exist, since
      $$
      lim_{xto 0^-}e^{1/x}=0quadtext{while}quad lim_{xto 0^+}e^{1/x}=infty
      $$
      So, consequently
      $$
      lim_{xto 0^-}frac{e^{1/x}}{e^{1/x}+1}=0quadtext{while}quad lim_{xto 0^+}frac{e^{1/x}}{e^{1/x}+1}=1
      $$






      share|cite|improve this answer














      L'Hopital's rule does not apply because the limit
      $$
      lim_{xto 0}e^{1/x}
      $$
      does not exist, since
      $$
      lim_{xto 0^-}e^{1/x}=0quadtext{while}quad lim_{xto 0^+}e^{1/x}=infty
      $$
      So, consequently
      $$
      lim_{xto 0^-}frac{e^{1/x}}{e^{1/x}+1}=0quadtext{while}quad lim_{xto 0^+}frac{e^{1/x}}{e^{1/x}+1}=1
      $$







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited 3 hours ago









      Szeto

      6,4362926




      6,4362926










      answered 4 hours ago









      Yiorgos S. Smyrlis

      62.6k1383163




      62.6k1383163








      • 1




        This is not the reason why l'Hôpital cannot be applied.
        – egreg
        3 hours ago














      • 1




        This is not the reason why l'Hôpital cannot be applied.
        – egreg
        3 hours ago








      1




      1




      This is not the reason why l'Hôpital cannot be applied.
      – egreg
      3 hours ago




      This is not the reason why l'Hôpital cannot be applied.
      – egreg
      3 hours ago











      1














      You can apply l’Hôpital for the limit from the right, because it is in the form $infty/infty$.



      The limit from the left is instead in the form $0/1$, which is not indeterminate. In order to use that form of l'Hôpital for a two sided limit, you need $infty/infty$ on both sides.



      More generally you need $infty/infty$ or $0/0$ on either side.



      If you blindly apply l’Hôpital, you end up with
      $$
      lim_{xto0}frac{-e^{1/x}/x^2}{-e^{1/x}/x^2}=1
      $$
      but this would be incorrect, because
      $$
      lim_{xto0^-}frac{e^{1/x}}{e^{1/x}+1}=0
      $$
      For the limit from the right the limit is correct, but using l’Hôpital is of course not necessary:
      $$
      lim_{xto0^+}frac{e^{1/x}}{e^{1/x}+1}=lim_{xto0^+}frac{1}{1+e^{-1/x}}=
      frac{1}{1+0}=1
      $$






      share|cite|improve this answer




























        1














        You can apply l’Hôpital for the limit from the right, because it is in the form $infty/infty$.



        The limit from the left is instead in the form $0/1$, which is not indeterminate. In order to use that form of l'Hôpital for a two sided limit, you need $infty/infty$ on both sides.



        More generally you need $infty/infty$ or $0/0$ on either side.



        If you blindly apply l’Hôpital, you end up with
        $$
        lim_{xto0}frac{-e^{1/x}/x^2}{-e^{1/x}/x^2}=1
        $$
        but this would be incorrect, because
        $$
        lim_{xto0^-}frac{e^{1/x}}{e^{1/x}+1}=0
        $$
        For the limit from the right the limit is correct, but using l’Hôpital is of course not necessary:
        $$
        lim_{xto0^+}frac{e^{1/x}}{e^{1/x}+1}=lim_{xto0^+}frac{1}{1+e^{-1/x}}=
        frac{1}{1+0}=1
        $$






        share|cite|improve this answer


























          1












          1








          1






          You can apply l’Hôpital for the limit from the right, because it is in the form $infty/infty$.



          The limit from the left is instead in the form $0/1$, which is not indeterminate. In order to use that form of l'Hôpital for a two sided limit, you need $infty/infty$ on both sides.



          More generally you need $infty/infty$ or $0/0$ on either side.



          If you blindly apply l’Hôpital, you end up with
          $$
          lim_{xto0}frac{-e^{1/x}/x^2}{-e^{1/x}/x^2}=1
          $$
          but this would be incorrect, because
          $$
          lim_{xto0^-}frac{e^{1/x}}{e^{1/x}+1}=0
          $$
          For the limit from the right the limit is correct, but using l’Hôpital is of course not necessary:
          $$
          lim_{xto0^+}frac{e^{1/x}}{e^{1/x}+1}=lim_{xto0^+}frac{1}{1+e^{-1/x}}=
          frac{1}{1+0}=1
          $$






          share|cite|improve this answer














          You can apply l’Hôpital for the limit from the right, because it is in the form $infty/infty$.



          The limit from the left is instead in the form $0/1$, which is not indeterminate. In order to use that form of l'Hôpital for a two sided limit, you need $infty/infty$ on both sides.



          More generally you need $infty/infty$ or $0/0$ on either side.



          If you blindly apply l’Hôpital, you end up with
          $$
          lim_{xto0}frac{-e^{1/x}/x^2}{-e^{1/x}/x^2}=1
          $$
          but this would be incorrect, because
          $$
          lim_{xto0^-}frac{e^{1/x}}{e^{1/x}+1}=0
          $$
          For the limit from the right the limit is correct, but using l’Hôpital is of course not necessary:
          $$
          lim_{xto0^+}frac{e^{1/x}}{e^{1/x}+1}=lim_{xto0^+}frac{1}{1+e^{-1/x}}=
          frac{1}{1+0}=1
          $$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 2 hours ago

























          answered 3 hours ago









          egreg

          178k1484201




          178k1484201























              0














              here, $$when~~~xto 0^+,e^{frac{1}{x}}to infty$$
              So,$$lim_{x to 0^+} frac{e^frac{1}{x}}{e^frac{1}{x}+1}=lim_{x to 0^+}frac{e^frac{1}{x}}{e^frac{1}{x}left(1+frac{1}{e^frac{1}{x}} right)}=lim_{x to 0^+}frac{1}{left(1+frac{1}{e^frac{1}{x}} right)}=frac{1}{1+0}=1$$
              $$when~~~xto 0^-,e^{frac{1}{x}} to 0$$
              So,
              $$lim_{x to 0^-} frac{e^frac{1}{x}}{e^frac{1}{x}+1}=frac{0}{0+1}=0$$



              so,simply limit doesn't exist.so we can't find the limit when $x to 0$ no matter what procedure you follow.






              share|cite|improve this answer


























                0














                here, $$when~~~xto 0^+,e^{frac{1}{x}}to infty$$
                So,$$lim_{x to 0^+} frac{e^frac{1}{x}}{e^frac{1}{x}+1}=lim_{x to 0^+}frac{e^frac{1}{x}}{e^frac{1}{x}left(1+frac{1}{e^frac{1}{x}} right)}=lim_{x to 0^+}frac{1}{left(1+frac{1}{e^frac{1}{x}} right)}=frac{1}{1+0}=1$$
                $$when~~~xto 0^-,e^{frac{1}{x}} to 0$$
                So,
                $$lim_{x to 0^-} frac{e^frac{1}{x}}{e^frac{1}{x}+1}=frac{0}{0+1}=0$$



                so,simply limit doesn't exist.so we can't find the limit when $x to 0$ no matter what procedure you follow.






                share|cite|improve this answer
























                  0












                  0








                  0






                  here, $$when~~~xto 0^+,e^{frac{1}{x}}to infty$$
                  So,$$lim_{x to 0^+} frac{e^frac{1}{x}}{e^frac{1}{x}+1}=lim_{x to 0^+}frac{e^frac{1}{x}}{e^frac{1}{x}left(1+frac{1}{e^frac{1}{x}} right)}=lim_{x to 0^+}frac{1}{left(1+frac{1}{e^frac{1}{x}} right)}=frac{1}{1+0}=1$$
                  $$when~~~xto 0^-,e^{frac{1}{x}} to 0$$
                  So,
                  $$lim_{x to 0^-} frac{e^frac{1}{x}}{e^frac{1}{x}+1}=frac{0}{0+1}=0$$



                  so,simply limit doesn't exist.so we can't find the limit when $x to 0$ no matter what procedure you follow.






                  share|cite|improve this answer












                  here, $$when~~~xto 0^+,e^{frac{1}{x}}to infty$$
                  So,$$lim_{x to 0^+} frac{e^frac{1}{x}}{e^frac{1}{x}+1}=lim_{x to 0^+}frac{e^frac{1}{x}}{e^frac{1}{x}left(1+frac{1}{e^frac{1}{x}} right)}=lim_{x to 0^+}frac{1}{left(1+frac{1}{e^frac{1}{x}} right)}=frac{1}{1+0}=1$$
                  $$when~~~xto 0^-,e^{frac{1}{x}} to 0$$
                  So,
                  $$lim_{x to 0^-} frac{e^frac{1}{x}}{e^frac{1}{x}+1}=frac{0}{0+1}=0$$



                  so,simply limit doesn't exist.so we can't find the limit when $x to 0$ no matter what procedure you follow.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 3 hours ago









                  Rakibul Islam Prince

                  952211




                  952211






























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