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
    5 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
    5 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
    4 hours ago










  • @fleablood Oh okay! The theorem is only about the right hand limit. Thanks!
    – user1752323
    40 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
    5 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
    5 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
    4 hours ago










  • @fleablood Oh okay! The theorem is only about the right hand limit. Thanks!
    – user1752323
    40 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 5 hours ago









user1752323

324




324












  • Why not? I think it is allright
    – José Alejandro Aburto Araneda
    5 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
    5 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
    4 hours ago










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


















  • Why not? I think it is allright
    – José Alejandro Aburto Araneda
    5 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
    5 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
    4 hours ago










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
















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




Why not? I think it is allright
– José Alejandro Aburto Araneda
5 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
5 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
5 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
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
4 hours ago












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




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










3 Answers
3






active

oldest

votes


















3














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



















    • 2




      This is not the reason why l'Hôpital cannot be applied.
      – egreg
      4 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





















      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3058940%2fwhy-isnt-l-hospital-rule-applicable-here%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3














      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














        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












          3








          3






          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 4 hours ago

























          answered 4 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



















              • 2




                This is not the reason why l'Hôpital cannot be applied.
                – egreg
                4 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



















              • 2




                This is not the reason why l'Hôpital cannot be applied.
                – egreg
                4 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 4 hours ago









              Szeto

              6,4362926




              6,4362926










              answered 5 hours ago









              Yiorgos S. Smyrlis

              62.6k1383163




              62.6k1383163








              • 2




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














              • 2




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








              2




              2




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




              This is not the reason why l'Hôpital cannot be applied.
              – egreg
              4 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






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.





                      Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                      Please pay close attention to the following guidance:


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3058940%2fwhy-isnt-l-hospital-rule-applicable-here%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      Eastern Orthodox Church

                      Image
                      Not to be confused with Oriental Orthodoxy, Eastern Christianity, or Eastern Catholic Churches. "Eastern Orthodoxy", "Orthodox Church", "Orthodox Catholic Church", and "Orthodox Christian Church" redirect here. For other uses, see Orthodox (disambiguation). This article may be too long to read and navigate comfortably . The readable prose size is 132 kilobytes. Please consider splitting content into sub-articles, condensing it, or adding or removing subheadings. ( January 2018 ) Part of a series on the Eastern Orthodox Church Mosaic of Christ Pantocrator, Hagia Sophia Overview Structure Theology (History of theology) Liturgy Church history Holy Mysteries View of salvation View of Mary View of icons Background Crucifixion / Resurrection / Ascension of Jesus Christianity Christian Church Apostolic succession Four Marks of the Church Orthodoxy Organization Autocephaly Patriarchate ...
                      Read more

                      Zagreb

                      Image
                      This article is about the Croatian capital city. For other uses, see Zagreb (disambiguation). City in City of Zagreb, Croatia Zagreb City Grad Zagreb City of Zagreb Clockwise, from top: St. Mark's Square, Croatian State Archives, Zagreb Tram, Art Pavilion, Cibona & HOTO towers and Croatian National Theatre. Flag Coat of arms Zagreb Location of Zagreb in Croatia Show map of Croatia Zagreb Zagreb (Europe) Show map of Europe Coordinates: 45°49′N 15°59′E  /  45.817°N 15.983°E  / 45.817; 15.983 Coordinates: 45°49′N 15°59′E  /  45.817°N 15.983°E  / 45.817; 15.983 Country   Croatia County City of Zagreb RC diocese 1094 Free royal city 1242 Unified 1850 Subdivisions 17 city districts 218 local committees (70 settlements) Government  • Type Mayor-Council  • Mayor Milan Bandić (BM 365)  • City Assembly 51 members BM 365, ZL, NS-R (14) SDP, GLAS, HNS, HSS, NH-PS (13) ...
                      Read more

                      Understanding the information contained in the Deep Space Network XML data?

                      Image
                      up vote 1 down vote favorite NASA's Deep Space Network activity can be viewed conveniently at https://eyes.nasa.gov/dsn/dsn.html but if you would like data in a more raw XML form, it can be accessed as described in this answer Below is a small Python script that reads the XML into a list of dictionaries, and saves it to disk as a JSON. The dictionaries are organized at the higher level by dish, then by spacecraft per dish. Question: What are the most-likely meanings of all the data field contained in the Deep Space Network XML data? This is the example url from the other answer, used in the script below: https://eyes.nasa.gov/dsn/data/1365107113.xml def dictify(r,root=True): # https://stackoverflow.com/questions/2148119/how-to-convert-an-xml-string-to-a-dictionary-in-python/30923963#30923963 if root: ret...
                      Read more