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












-1














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
























  • $lim_{tto-infty}e^t=0$
    – Nosrati
    Nov 24 at 11:24












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










  • @KM101 just edited that.
    – user472288
    Nov 24 at 11:26










  • “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).
    – egreg
    Nov 24 at 11:59


















-1














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
























  • $lim_{tto-infty}e^t=0$
    – Nosrati
    Nov 24 at 11:24












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










  • @KM101 just edited that.
    – user472288
    Nov 24 at 11:26










  • “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).
    – egreg
    Nov 24 at 11:59
















-1












-1








-1







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















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 at 11:35









amWhy

191k28224439




191k28224439










asked Nov 24 at 11:19









user472288

457211




457211












  • $lim_{tto-infty}e^t=0$
    – Nosrati
    Nov 24 at 11:24












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










  • @KM101 just edited that.
    – user472288
    Nov 24 at 11:26










  • “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).
    – egreg
    Nov 24 at 11:59




















  • $lim_{tto-infty}e^t=0$
    – Nosrati
    Nov 24 at 11:24












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










  • @KM101 just edited that.
    – user472288
    Nov 24 at 11:26










  • “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).
    – egreg
    Nov 24 at 11:59


















$lim_{tto-infty}e^t=0$
– Nosrati
Nov 24 at 11:24






$lim_{tto-infty}e^t=0$
– Nosrati
Nov 24 at 11:24














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




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












@KM101 just edited that.
– user472288
Nov 24 at 11:26




@KM101 just edited that.
– user472288
Nov 24 at 11:26












“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).
– egreg
Nov 24 at 11:59






“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).
– egreg
Nov 24 at 11:59












4 Answers
4






active

oldest

votes


















2














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





























    1














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



    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





























      0














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






      share|cite|improve this answer





























        0














        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





















          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














          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


























            2














            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
























              2












              2








              2






              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












              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 at 11:26









              Rebellos

              14.2k31244




              14.2k31244























                  1














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



                  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


























                    1














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



                    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
























                      1












                      1








                      1






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



                      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












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



                      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 at 11:26









                      drhab

                      96.5k544127




                      96.5k544127























                          0














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






                          share|cite|improve this answer


























                            0














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






                            share|cite|improve this answer
























                              0












                              0








                              0






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






                              share|cite|improve this answer












                              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 at 11:26









                              José Carlos Santos

                              148k22117218




                              148k22117218























                                  0














                                  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


























                                    0














                                    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
























                                      0












                                      0








                                      0






                                      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












                                      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 at 11:28









                                      gimusi

                                      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.





                                          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%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

                                          Quarter-circle Tiles

                                          Image
                                          up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2
                                          Read more

                                          build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

                                          Image
                                          0 I think it's similiar to NFAs. I replace the finite states of the given automaton for start-states for my new automaton. I do it with $epsilon$ -transition from the start state to the actual finite states of the given automaton. The start state of the given automaton will be my accepting(finite) state of my new one. All transitions will be reversed of course to recognize the reversed word. But how do I have to change the stack of the pushdown automaton? That's tricky. formal-languages automata context-free-grammar share | cite | improve this question edited Nov 29 '18 at 14:20 joee
                                          Read more

                                          Mont Emei

                                          Image
                                          Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
                                          Read more