Limit $lim_{xto 0^-}frac{1}{1+e^{1/x}}$












-1












$begingroup$


I have a small question about evaluating $$lim_{xto 0^-}frac{1}{1+e^{1/x}}$$



If I look at $e^{1/x}$, as $xto 0^-$, it approaches $frac{1}{-infty}$.



How does $$lim_{xto 0^-}frac{1}{1+e^{1/x}}=1?$$



Some insight would be great.
Thanks!






calculus limits







share|cite|improve this question











$endgroup$












  • $begingroup$
    $lim_{tto-infty}e^t=0$
    $endgroup$
    – Nosrati
    Nov 24 '18 at 11:24












  • $begingroup$
    Just to point out, a positive value raised to an exponent can never give a negative value.
    $endgroup$
    – KM101
    Nov 24 '18 at 11:25










  • $begingroup$
    @KM101 just edited that.
    $endgroup$
    – user472288
    Nov 24 '18 at 11:26










  • $begingroup$
    “Approaches $frac{1}{-infty}$” is a different way to say that the limit is $0$ (with a bit more information that in this case is irrelevant).
    $endgroup$
    – egreg
    Nov 24 '18 at 11:59


















-1












$begingroup$


I have a small question about evaluating $$lim_{xto 0^-}frac{1}{1+e^{1/x}}$$



If I look at $e^{1/x}$, as $xto 0^-$, it approaches $frac{1}{-infty}$.



How does $$lim_{xto 0^-}frac{1}{1+e^{1/x}}=1?$$



Some insight would be great.
Thanks!






calculus limits







share|cite|improve this question











$endgroup$












  • $begingroup$
    $lim_{tto-infty}e^t=0$
    $endgroup$
    – Nosrati
    Nov 24 '18 at 11:24












  • $begingroup$
    Just to point out, a positive value raised to an exponent can never give a negative value.
    $endgroup$
    – KM101
    Nov 24 '18 at 11:25










  • $begingroup$
    @KM101 just edited that.
    $endgroup$
    – user472288
    Nov 24 '18 at 11:26










  • $begingroup$
    “Approaches $frac{1}{-infty}$” is a different way to say that the limit is $0$ (with a bit more information that in this case is irrelevant).
    $endgroup$
    – egreg
    Nov 24 '18 at 11:59
















-1












-1








-1





$begingroup$


I have a small question about evaluating $$lim_{xto 0^-}frac{1}{1+e^{1/x}}$$



If I look at $e^{1/x}$, as $xto 0^-$, it approaches $frac{1}{-infty}$.



How does $$lim_{xto 0^-}frac{1}{1+e^{1/x}}=1?$$



Some insight would be great.
Thanks!






calculus limits







share|cite|improve this question











$endgroup$




I have a small question about evaluating $$lim_{xto 0^-}frac{1}{1+e^{1/x}}$$



If I look at $e^{1/x}$, as $xto 0^-$, it approaches $frac{1}{-infty}$.



How does $$lim_{xto 0^-}frac{1}{1+e^{1/x}}=1?$$



Some insight would be great.
Thanks!






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 24 '18 at 11:35









amWhy

192k28225439




192k28225439










asked Nov 24 '18 at 11:19









user472288user472288

457211




457211












  • $begingroup$
    $lim_{tto-infty}e^t=0$
    $endgroup$
    – Nosrati
    Nov 24 '18 at 11:24












  • $begingroup$
    Just to point out, a positive value raised to an exponent can never give a negative value.
    $endgroup$
    – KM101
    Nov 24 '18 at 11:25










  • $begingroup$
    @KM101 just edited that.
    $endgroup$
    – user472288
    Nov 24 '18 at 11:26










  • $begingroup$
    “Approaches $frac{1}{-infty}$” is a different way to say that the limit is $0$ (with a bit more information that in this case is irrelevant).
    $endgroup$
    – egreg
    Nov 24 '18 at 11:59




















  • $begingroup$
    $lim_{tto-infty}e^t=0$
    $endgroup$
    – Nosrati
    Nov 24 '18 at 11:24












  • $begingroup$
    Just to point out, a positive value raised to an exponent can never give a negative value.
    $endgroup$
    – KM101
    Nov 24 '18 at 11:25










  • $begingroup$
    @KM101 just edited that.
    $endgroup$
    – user472288
    Nov 24 '18 at 11:26










  • $begingroup$
    “Approaches $frac{1}{-infty}$” is a different way to say that the limit is $0$ (with a bit more information that in this case is irrelevant).
    $endgroup$
    – egreg
    Nov 24 '18 at 11:59


















$begingroup$
$lim_{tto-infty}e^t=0$
$endgroup$
– Nosrati
Nov 24 '18 at 11:24






$begingroup$
$lim_{tto-infty}e^t=0$
$endgroup$
– Nosrati
Nov 24 '18 at 11:24














$begingroup$
Just to point out, a positive value raised to an exponent can never give a negative value.
$endgroup$
– KM101
Nov 24 '18 at 11:25




$begingroup$
Just to point out, a positive value raised to an exponent can never give a negative value.
$endgroup$
– KM101
Nov 24 '18 at 11:25












$begingroup$
@KM101 just edited that.
$endgroup$
– user472288
Nov 24 '18 at 11:26




$begingroup$
@KM101 just edited that.
$endgroup$
– user472288
Nov 24 '18 at 11:26












$begingroup$
“Approaches $frac{1}{-infty}$” is a different way to say that the limit is $0$ (with a bit more information that in this case is irrelevant).
$endgroup$
– egreg
Nov 24 '18 at 11:59






$begingroup$
“Approaches $frac{1}{-infty}$” is a different way to say that the limit is $0$ (with a bit more information that in this case is irrelevant).
$endgroup$
– egreg
Nov 24 '18 at 11:59












4 Answers
4






active

oldest

votes


















2












$begingroup$

Note that :



$$lim_{x to 0^-} frac{1}{1+exp(1/x)} = frac{1}{lim_{x to 0^-} (1+exp(1/x))}= frac{1}{1+lim_{x to 0^-}exp(1/x)}$$



Now, since $x to 0^-$, this means it approaches zero by the negative side, which yields that :



$$lim_{x to 0^-} exp(1/x) = lim_{xto 0^-} exp(lim_{x to 0^-} 1/x) = exp(-infty) = 0$$



It is also easy to spot that, using the substitution $t = 1/x$ and noting that when $x to 0^- Rightarrow t to -infty$.



Thus, finally :



$$lim_{xto 0^-} frac{1}{1+e^frac{1}{x}} = 1$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    $lim_{xto0^-}frac1x=-infty$.



    So that $lim_{xto0^-}e^{frac1x}=lim_{yto-infty}e^{y}=lim_{zto+infty}e^{-z}=lim_{zto+infty}frac1{e^z}=0$.



    So that $lim_{xto0^-}frac1{1+e^{frac1x}}=frac1{1+0}=1$.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      Since $displaystylelim_{xto0^-}e^{frac1x}=e^{lim_{xto0^-}frac1x}=e^{-infty}=0$,$$lim_{xto0^-}frac1{1+e^{frac1x}}=1.$$






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        We have that $frac1x to -infty$ then by $y=-frac1x to infty$ we have



        $$frac{1}{1+e^frac{1}{x}}=frac{1}{1+e^{-y}}to 1$$



        since $e^{-y}=frac1 {e^y}to 0$.






        share|cite|improve this answer









        $endgroup$













          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%2f3011429%2flimit-lim-x-to-0-frac11e1-x%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          Note that :



          $$lim_{x to 0^-} frac{1}{1+exp(1/x)} = frac{1}{lim_{x to 0^-} (1+exp(1/x))}= frac{1}{1+lim_{x to 0^-}exp(1/x)}$$



          Now, since $x to 0^-$, this means it approaches zero by the negative side, which yields that :



          $$lim_{x to 0^-} exp(1/x) = lim_{xto 0^-} exp(lim_{x to 0^-} 1/x) = exp(-infty) = 0$$



          It is also easy to spot that, using the substitution $t = 1/x$ and noting that when $x to 0^- Rightarrow t to -infty$.



          Thus, finally :



          $$lim_{xto 0^-} frac{1}{1+e^frac{1}{x}} = 1$$






          share|cite|improve this answer









          $endgroup$


















            2












            $begingroup$

            Note that :



            $$lim_{x to 0^-} frac{1}{1+exp(1/x)} = frac{1}{lim_{x to 0^-} (1+exp(1/x))}= frac{1}{1+lim_{x to 0^-}exp(1/x)}$$



            Now, since $x to 0^-$, this means it approaches zero by the negative side, which yields that :



            $$lim_{x to 0^-} exp(1/x) = lim_{xto 0^-} exp(lim_{x to 0^-} 1/x) = exp(-infty) = 0$$



            It is also easy to spot that, using the substitution $t = 1/x$ and noting that when $x to 0^- Rightarrow t to -infty$.



            Thus, finally :



            $$lim_{xto 0^-} frac{1}{1+e^frac{1}{x}} = 1$$






            share|cite|improve this answer









            $endgroup$
















              2












              2








              2





              $begingroup$

              Note that :



              $$lim_{x to 0^-} frac{1}{1+exp(1/x)} = frac{1}{lim_{x to 0^-} (1+exp(1/x))}= frac{1}{1+lim_{x to 0^-}exp(1/x)}$$



              Now, since $x to 0^-$, this means it approaches zero by the negative side, which yields that :



              $$lim_{x to 0^-} exp(1/x) = lim_{xto 0^-} exp(lim_{x to 0^-} 1/x) = exp(-infty) = 0$$



              It is also easy to spot that, using the substitution $t = 1/x$ and noting that when $x to 0^- Rightarrow t to -infty$.



              Thus, finally :



              $$lim_{xto 0^-} frac{1}{1+e^frac{1}{x}} = 1$$






              share|cite|improve this answer









              $endgroup$



              Note that :



              $$lim_{x to 0^-} frac{1}{1+exp(1/x)} = frac{1}{lim_{x to 0^-} (1+exp(1/x))}= frac{1}{1+lim_{x to 0^-}exp(1/x)}$$



              Now, since $x to 0^-$, this means it approaches zero by the negative side, which yields that :



              $$lim_{x to 0^-} exp(1/x) = lim_{xto 0^-} exp(lim_{x to 0^-} 1/x) = exp(-infty) = 0$$



              It is also easy to spot that, using the substitution $t = 1/x$ and noting that when $x to 0^- Rightarrow t to -infty$.



              Thus, finally :



              $$lim_{xto 0^-} frac{1}{1+e^frac{1}{x}} = 1$$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 24 '18 at 11:26









              RebellosRebellos

              14.5k31246




              14.5k31246























                  1












                  $begingroup$

                  $lim_{xto0^-}frac1x=-infty$.



                  So that $lim_{xto0^-}e^{frac1x}=lim_{yto-infty}e^{y}=lim_{zto+infty}e^{-z}=lim_{zto+infty}frac1{e^z}=0$.



                  So that $lim_{xto0^-}frac1{1+e^{frac1x}}=frac1{1+0}=1$.






                  share|cite|improve this answer









                  $endgroup$


















                    1












                    $begingroup$

                    $lim_{xto0^-}frac1x=-infty$.



                    So that $lim_{xto0^-}e^{frac1x}=lim_{yto-infty}e^{y}=lim_{zto+infty}e^{-z}=lim_{zto+infty}frac1{e^z}=0$.



                    So that $lim_{xto0^-}frac1{1+e^{frac1x}}=frac1{1+0}=1$.






                    share|cite|improve this answer









                    $endgroup$
















                      1












                      1








                      1





                      $begingroup$

                      $lim_{xto0^-}frac1x=-infty$.



                      So that $lim_{xto0^-}e^{frac1x}=lim_{yto-infty}e^{y}=lim_{zto+infty}e^{-z}=lim_{zto+infty}frac1{e^z}=0$.



                      So that $lim_{xto0^-}frac1{1+e^{frac1x}}=frac1{1+0}=1$.






                      share|cite|improve this answer









                      $endgroup$



                      $lim_{xto0^-}frac1x=-infty$.



                      So that $lim_{xto0^-}e^{frac1x}=lim_{yto-infty}e^{y}=lim_{zto+infty}e^{-z}=lim_{zto+infty}frac1{e^z}=0$.



                      So that $lim_{xto0^-}frac1{1+e^{frac1x}}=frac1{1+0}=1$.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Nov 24 '18 at 11:26









                      drhabdrhab

                      98.7k544129




                      98.7k544129























                          0












                          $begingroup$

                          Since $displaystylelim_{xto0^-}e^{frac1x}=e^{lim_{xto0^-}frac1x}=e^{-infty}=0$,$$lim_{xto0^-}frac1{1+e^{frac1x}}=1.$$






                          share|cite|improve this answer









                          $endgroup$


















                            0












                            $begingroup$

                            Since $displaystylelim_{xto0^-}e^{frac1x}=e^{lim_{xto0^-}frac1x}=e^{-infty}=0$,$$lim_{xto0^-}frac1{1+e^{frac1x}}=1.$$






                            share|cite|improve this answer









                            $endgroup$
















                              0












                              0








                              0





                              $begingroup$

                              Since $displaystylelim_{xto0^-}e^{frac1x}=e^{lim_{xto0^-}frac1x}=e^{-infty}=0$,$$lim_{xto0^-}frac1{1+e^{frac1x}}=1.$$






                              share|cite|improve this answer









                              $endgroup$



                              Since $displaystylelim_{xto0^-}e^{frac1x}=e^{lim_{xto0^-}frac1x}=e^{-infty}=0$,$$lim_{xto0^-}frac1{1+e^{frac1x}}=1.$$







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Nov 24 '18 at 11:26









                              José Carlos SantosJosé Carlos Santos

                              153k22123225




                              153k22123225























                                  0












                                  $begingroup$

                                  We have that $frac1x to -infty$ then by $y=-frac1x to infty$ we have



                                  $$frac{1}{1+e^frac{1}{x}}=frac{1}{1+e^{-y}}to 1$$



                                  since $e^{-y}=frac1 {e^y}to 0$.






                                  share|cite|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    We have that $frac1x to -infty$ then by $y=-frac1x to infty$ we have



                                    $$frac{1}{1+e^frac{1}{x}}=frac{1}{1+e^{-y}}to 1$$



                                    since $e^{-y}=frac1 {e^y}to 0$.






                                    share|cite|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      We have that $frac1x to -infty$ then by $y=-frac1x to infty$ we have



                                      $$frac{1}{1+e^frac{1}{x}}=frac{1}{1+e^{-y}}to 1$$



                                      since $e^{-y}=frac1 {e^y}to 0$.






                                      share|cite|improve this answer









                                      $endgroup$



                                      We have that $frac1x to -infty$ then by $y=-frac1x to infty$ we have



                                      $$frac{1}{1+e^frac{1}{x}}=frac{1}{1+e^{-y}}to 1$$



                                      since $e^{-y}=frac1 {e^y}to 0$.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Nov 24 '18 at 11:28









                                      gimusigimusi

                                      1




                                      1






























                                          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.




                                          draft saved


                                          draft discarded














                                          StackExchange.ready(
                                          function () {
                                          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3011429%2flimit-lim-x-to-0-frac11e1-x%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

                                          How to change which sound is reproduced for terminal bell?

                                          Image
                                          3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question ...
                                          Read more

                                          Can I use Tabulator js library in my java Spring + Thymeleaf project?

                                          Image
                                          0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"...
                                          Read more

                                          Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

                                          Image
                                          up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx...
                                          Read more