What is the limit of $(1-x)^frac{1}{x}$ as x approaches 0?












1












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Basically, if something has a probability of 1/5, doing it 5 times gives a probability of it not happening that's close to that formula. Doing something with a probability of 1/100 100 times makes it even closer. I'm wondering if this is a well-known named constant, the way these probabilities converge is interesting to me.






limits







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  • 2




    $begingroup$
    If $x=1/n$ then this becomes $$left(1-frac1nright)^n$$ which has a ver well-known limit as $ntoinfty$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 31 at 20:57












  • $begingroup$
    Seee (sorry), See e en.wikipedia.org/wiki/E_(mathematical_constant)#History
    $endgroup$
    – Tito Eliatron
    Jan 31 at 20:58










  • $begingroup$
    This is about the second most famous mathematical constant.
    $endgroup$
    – Yves Daoust
    Jan 31 at 21:01










  • $begingroup$
    wolframalpha.com/input/?i=series+expansion+of+(1-x)%5E(1%2Fx)
    $endgroup$
    – Yves Daoust
    Jan 31 at 21:04
















1












$begingroup$


Basically, if something has a probability of 1/5, doing it 5 times gives a probability of it not happening that's close to that formula. Doing something with a probability of 1/100 100 times makes it even closer. I'm wondering if this is a well-known named constant, the way these probabilities converge is interesting to me.






limits







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    If $x=1/n$ then this becomes $$left(1-frac1nright)^n$$ which has a ver well-known limit as $ntoinfty$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 31 at 20:57












  • $begingroup$
    Seee (sorry), See e en.wikipedia.org/wiki/E_(mathematical_constant)#History
    $endgroup$
    – Tito Eliatron
    Jan 31 at 20:58










  • $begingroup$
    This is about the second most famous mathematical constant.
    $endgroup$
    – Yves Daoust
    Jan 31 at 21:01










  • $begingroup$
    wolframalpha.com/input/?i=series+expansion+of+(1-x)%5E(1%2Fx)
    $endgroup$
    – Yves Daoust
    Jan 31 at 21:04














1












1








1





$begingroup$


Basically, if something has a probability of 1/5, doing it 5 times gives a probability of it not happening that's close to that formula. Doing something with a probability of 1/100 100 times makes it even closer. I'm wondering if this is a well-known named constant, the way these probabilities converge is interesting to me.






limits







share|cite|improve this question









$endgroup$




Basically, if something has a probability of 1/5, doing it 5 times gives a probability of it not happening that's close to that formula. Doing something with a probability of 1/100 100 times makes it even closer. I'm wondering if this is a well-known named constant, the way these probabilities converge is interesting to me.






limits




limits






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 31 at 20:56









VyvianVyvian

82




82








  • 2




    $begingroup$
    If $x=1/n$ then this becomes $$left(1-frac1nright)^n$$ which has a ver well-known limit as $ntoinfty$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 31 at 20:57












  • $begingroup$
    Seee (sorry), See e en.wikipedia.org/wiki/E_(mathematical_constant)#History
    $endgroup$
    – Tito Eliatron
    Jan 31 at 20:58










  • $begingroup$
    This is about the second most famous mathematical constant.
    $endgroup$
    – Yves Daoust
    Jan 31 at 21:01










  • $begingroup$
    wolframalpha.com/input/?i=series+expansion+of+(1-x)%5E(1%2Fx)
    $endgroup$
    – Yves Daoust
    Jan 31 at 21:04














  • 2




    $begingroup$
    If $x=1/n$ then this becomes $$left(1-frac1nright)^n$$ which has a ver well-known limit as $ntoinfty$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 31 at 20:57












  • $begingroup$
    Seee (sorry), See e en.wikipedia.org/wiki/E_(mathematical_constant)#History
    $endgroup$
    – Tito Eliatron
    Jan 31 at 20:58










  • $begingroup$
    This is about the second most famous mathematical constant.
    $endgroup$
    – Yves Daoust
    Jan 31 at 21:01










  • $begingroup$
    wolframalpha.com/input/?i=series+expansion+of+(1-x)%5E(1%2Fx)
    $endgroup$
    – Yves Daoust
    Jan 31 at 21:04








2




2




$begingroup$
If $x=1/n$ then this becomes $$left(1-frac1nright)^n$$ which has a ver well-known limit as $ntoinfty$.
$endgroup$
– Lord Shark the Unknown
Jan 31 at 20:57






$begingroup$
If $x=1/n$ then this becomes $$left(1-frac1nright)^n$$ which has a ver well-known limit as $ntoinfty$.
$endgroup$
– Lord Shark the Unknown
Jan 31 at 20:57














$begingroup$
Seee (sorry), See e en.wikipedia.org/wiki/E_(mathematical_constant)#History
$endgroup$
– Tito Eliatron
Jan 31 at 20:58




$begingroup$
Seee (sorry), See e en.wikipedia.org/wiki/E_(mathematical_constant)#History
$endgroup$
– Tito Eliatron
Jan 31 at 20:58












$begingroup$
This is about the second most famous mathematical constant.
$endgroup$
– Yves Daoust
Jan 31 at 21:01




$begingroup$
This is about the second most famous mathematical constant.
$endgroup$
– Yves Daoust
Jan 31 at 21:01












$begingroup$
wolframalpha.com/input/?i=series+expansion+of+(1-x)%5E(1%2Fx)
$endgroup$
– Yves Daoust
Jan 31 at 21:04




$begingroup$
wolframalpha.com/input/?i=series+expansion+of+(1-x)%5E(1%2Fx)
$endgroup$
– Yves Daoust
Jan 31 at 21:04










3 Answers
3






active

oldest

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0












$begingroup$

Let $n=frac1x$. You get $lim_{ntoinfty}(1-frac1n)^n=frac1e$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I didn't identify that it was just the reciprocal of e, thanks
    $endgroup$
    – Vyvian
    Jan 31 at 21:30



















1












$begingroup$

Hint:



Determine first the limit of the logarithm: $;dfrac1x,ln (1-x)=dfrac{ln(1-x)-ln 1}x;$ is a rate of variation.






share|cite|improve this answer









$endgroup$





















    -1












    $begingroup$

    By the Binomial theorem,



    $$left(1-frac1nright)^n=sum_{k=0}^nbinom nkleft(-frac1nright)^k
    \=1-frac nn+frac{n(n-1)}{2n^2}-frac{n(n-1)(n-2)}{3!n^3}+frac{n(n-1)(n-2)(n-3)}{4!n^4}-cdotsleft(-frac1nright)^n
    \=frac12left(1-frac1nright)-frac1{3!}left(1-frac1nright)left(1-frac2nright)+frac1{4!}left(1-frac1nright)left(1-frac2nright)left(1-frac3nright)-cdotsleft(-frac1nright)^n$$



    The constant term of this expression is, for unbounded $n$,



    $$frac12-frac1{3!}+frac1{4!}-cdotsto e^{-1}$$



    Then coefficient of $dfrac1n$ is



    $$-frac12+frac{1+2}{3!}-frac{1+2+3}{4!}-cdots frac{(n-1)n}{2n!}cdotsto -frac{e^{-1}}2$$



    So the first order approximation is



    $$left(1-xright)^x=frac1e-frac x{2e}.$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Undue downvote.
      $endgroup$
      – Yves Daoust
      Jan 31 at 21:22












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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    Let $n=frac1x$. You get $lim_{ntoinfty}(1-frac1n)^n=frac1e$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I didn't identify that it was just the reciprocal of e, thanks
      $endgroup$
      – Vyvian
      Jan 31 at 21:30
















    0












    $begingroup$

    Let $n=frac1x$. You get $lim_{ntoinfty}(1-frac1n)^n=frac1e$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I didn't identify that it was just the reciprocal of e, thanks
      $endgroup$
      – Vyvian
      Jan 31 at 21:30














    0












    0








    0





    $begingroup$

    Let $n=frac1x$. You get $lim_{ntoinfty}(1-frac1n)^n=frac1e$.






    share|cite|improve this answer









    $endgroup$



    Let $n=frac1x$. You get $lim_{ntoinfty}(1-frac1n)^n=frac1e$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 31 at 21:02









    Chris CusterChris Custer

    14.3k3827




    14.3k3827












    • $begingroup$
      I didn't identify that it was just the reciprocal of e, thanks
      $endgroup$
      – Vyvian
      Jan 31 at 21:30


















    • $begingroup$
      I didn't identify that it was just the reciprocal of e, thanks
      $endgroup$
      – Vyvian
      Jan 31 at 21:30
















    $begingroup$
    I didn't identify that it was just the reciprocal of e, thanks
    $endgroup$
    – Vyvian
    Jan 31 at 21:30




    $begingroup$
    I didn't identify that it was just the reciprocal of e, thanks
    $endgroup$
    – Vyvian
    Jan 31 at 21:30











    1












    $begingroup$

    Hint:



    Determine first the limit of the logarithm: $;dfrac1x,ln (1-x)=dfrac{ln(1-x)-ln 1}x;$ is a rate of variation.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Hint:



      Determine first the limit of the logarithm: $;dfrac1x,ln (1-x)=dfrac{ln(1-x)-ln 1}x;$ is a rate of variation.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Hint:



        Determine first the limit of the logarithm: $;dfrac1x,ln (1-x)=dfrac{ln(1-x)-ln 1}x;$ is a rate of variation.






        share|cite|improve this answer









        $endgroup$



        Hint:



        Determine first the limit of the logarithm: $;dfrac1x,ln (1-x)=dfrac{ln(1-x)-ln 1}x;$ is a rate of variation.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 31 at 21:00









        BernardBernard

        124k741117




        124k741117























            -1












            $begingroup$

            By the Binomial theorem,



            $$left(1-frac1nright)^n=sum_{k=0}^nbinom nkleft(-frac1nright)^k
            \=1-frac nn+frac{n(n-1)}{2n^2}-frac{n(n-1)(n-2)}{3!n^3}+frac{n(n-1)(n-2)(n-3)}{4!n^4}-cdotsleft(-frac1nright)^n
            \=frac12left(1-frac1nright)-frac1{3!}left(1-frac1nright)left(1-frac2nright)+frac1{4!}left(1-frac1nright)left(1-frac2nright)left(1-frac3nright)-cdotsleft(-frac1nright)^n$$



            The constant term of this expression is, for unbounded $n$,



            $$frac12-frac1{3!}+frac1{4!}-cdotsto e^{-1}$$



            Then coefficient of $dfrac1n$ is



            $$-frac12+frac{1+2}{3!}-frac{1+2+3}{4!}-cdots frac{(n-1)n}{2n!}cdotsto -frac{e^{-1}}2$$



            So the first order approximation is



            $$left(1-xright)^x=frac1e-frac x{2e}.$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Undue downvote.
              $endgroup$
              – Yves Daoust
              Jan 31 at 21:22
















            -1












            $begingroup$

            By the Binomial theorem,



            $$left(1-frac1nright)^n=sum_{k=0}^nbinom nkleft(-frac1nright)^k
            \=1-frac nn+frac{n(n-1)}{2n^2}-frac{n(n-1)(n-2)}{3!n^3}+frac{n(n-1)(n-2)(n-3)}{4!n^4}-cdotsleft(-frac1nright)^n
            \=frac12left(1-frac1nright)-frac1{3!}left(1-frac1nright)left(1-frac2nright)+frac1{4!}left(1-frac1nright)left(1-frac2nright)left(1-frac3nright)-cdotsleft(-frac1nright)^n$$



            The constant term of this expression is, for unbounded $n$,



            $$frac12-frac1{3!}+frac1{4!}-cdotsto e^{-1}$$



            Then coefficient of $dfrac1n$ is



            $$-frac12+frac{1+2}{3!}-frac{1+2+3}{4!}-cdots frac{(n-1)n}{2n!}cdotsto -frac{e^{-1}}2$$



            So the first order approximation is



            $$left(1-xright)^x=frac1e-frac x{2e}.$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Undue downvote.
              $endgroup$
              – Yves Daoust
              Jan 31 at 21:22














            -1












            -1








            -1





            $begingroup$

            By the Binomial theorem,



            $$left(1-frac1nright)^n=sum_{k=0}^nbinom nkleft(-frac1nright)^k
            \=1-frac nn+frac{n(n-1)}{2n^2}-frac{n(n-1)(n-2)}{3!n^3}+frac{n(n-1)(n-2)(n-3)}{4!n^4}-cdotsleft(-frac1nright)^n
            \=frac12left(1-frac1nright)-frac1{3!}left(1-frac1nright)left(1-frac2nright)+frac1{4!}left(1-frac1nright)left(1-frac2nright)left(1-frac3nright)-cdotsleft(-frac1nright)^n$$



            The constant term of this expression is, for unbounded $n$,



            $$frac12-frac1{3!}+frac1{4!}-cdotsto e^{-1}$$



            Then coefficient of $dfrac1n$ is



            $$-frac12+frac{1+2}{3!}-frac{1+2+3}{4!}-cdots frac{(n-1)n}{2n!}cdotsto -frac{e^{-1}}2$$



            So the first order approximation is



            $$left(1-xright)^x=frac1e-frac x{2e}.$$






            share|cite|improve this answer









            $endgroup$



            By the Binomial theorem,



            $$left(1-frac1nright)^n=sum_{k=0}^nbinom nkleft(-frac1nright)^k
            \=1-frac nn+frac{n(n-1)}{2n^2}-frac{n(n-1)(n-2)}{3!n^3}+frac{n(n-1)(n-2)(n-3)}{4!n^4}-cdotsleft(-frac1nright)^n
            \=frac12left(1-frac1nright)-frac1{3!}left(1-frac1nright)left(1-frac2nright)+frac1{4!}left(1-frac1nright)left(1-frac2nright)left(1-frac3nright)-cdotsleft(-frac1nright)^n$$



            The constant term of this expression is, for unbounded $n$,



            $$frac12-frac1{3!}+frac1{4!}-cdotsto e^{-1}$$



            Then coefficient of $dfrac1n$ is



            $$-frac12+frac{1+2}{3!}-frac{1+2+3}{4!}-cdots frac{(n-1)n}{2n!}cdotsto -frac{e^{-1}}2$$



            So the first order approximation is



            $$left(1-xright)^x=frac1e-frac x{2e}.$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 31 at 21:21









            Yves DaoustYves Daoust

            132k676230




            132k676230












            • $begingroup$
              Undue downvote.
              $endgroup$
              – Yves Daoust
              Jan 31 at 21:22


















            • $begingroup$
              Undue downvote.
              $endgroup$
              – Yves Daoust
              Jan 31 at 21:22
















            $begingroup$
            Undue downvote.
            $endgroup$
            – Yves Daoust
            Jan 31 at 21:22




            $begingroup$
            Undue downvote.
            $endgroup$
            – Yves Daoust
            Jan 31 at 21:22


















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