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










  • @fleablood Oh okay! The theorem is only about the right hand limit. Thanks!
    – user1752323
    12 mins 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










  • @fleablood Oh okay! The theorem is only about the right hand limit. Thanks!
    – user1752323
    12 mins 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

324




324












  • 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










  • @fleablood Oh okay! The theorem is only about the right hand limit. Thanks!
    – user1752323
    12 mins 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










  • @fleablood Oh okay! The theorem is only about the right hand limit. Thanks!
    – user1752323
    12 mins 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












@fleablood Oh okay! The theorem is only about the right hand limit. Thanks!
– user1752323
12 mins ago




@fleablood Oh okay! The theorem is only about the right hand limit. Thanks!
– user1752323
12 mins ago










3 Answers
3






active

oldest

votes


















2














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































    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



















    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









      2














      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




























        2














        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


























          2












          2








          2






          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 3 hours ago

























          answered 3 hours ago









          egreg

          178k1484201




          178k1484201























              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











              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 4 hours ago









                  Rakibul Islam Prince

                  952211




                  952211






























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                      Ida-Boy-Ed-Garten

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                      Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich ...
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