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







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  • $lim_{tto-infty}e^t=0$
    – 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






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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






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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





















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          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






























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