What is the limit of $ (1 - frac{1}{n})^{2n}$ as n goes to infinity?












0












$begingroup$


$ (1 - frac{1}{n})^{2n} = ((1-frac{1}{n})^2)^n = (1-frac{2}{n} + frac{1}{n^2})^n$.



If I ignore the $frac{1}{n^2}$ term inside the parenthesis I have that the limit is $e^{-2}$ which agrees with Mathematica. Can I ignore this term? Why?






calculus







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$endgroup$








  • 2




    $begingroup$
    Hint. It's cleaner if you use $(X^n)^2$ rather than $(X^2)^n$.
    $endgroup$
    – Ethan Bolker
    Jan 7 at 15:54






  • 2




    $begingroup$
    $$((1-1/n)^n)^2 to (1/e)^2$$
    $endgroup$
    – Crostul
    Jan 7 at 15:54






  • 1




    $begingroup$
    write it as $lim_limits{ntoinfty} [(1+frac1{-n})^{-n}]^{-2}=cdots$
    $endgroup$
    – farruhota
    Jan 7 at 15:55
















0












$begingroup$


$ (1 - frac{1}{n})^{2n} = ((1-frac{1}{n})^2)^n = (1-frac{2}{n} + frac{1}{n^2})^n$.



If I ignore the $frac{1}{n^2}$ term inside the parenthesis I have that the limit is $e^{-2}$ which agrees with Mathematica. Can I ignore this term? Why?






calculus







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Hint. It's cleaner if you use $(X^n)^2$ rather than $(X^2)^n$.
    $endgroup$
    – Ethan Bolker
    Jan 7 at 15:54






  • 2




    $begingroup$
    $$((1-1/n)^n)^2 to (1/e)^2$$
    $endgroup$
    – Crostul
    Jan 7 at 15:54






  • 1




    $begingroup$
    write it as $lim_limits{ntoinfty} [(1+frac1{-n})^{-n}]^{-2}=cdots$
    $endgroup$
    – farruhota
    Jan 7 at 15:55














0












0








0





$begingroup$


$ (1 - frac{1}{n})^{2n} = ((1-frac{1}{n})^2)^n = (1-frac{2}{n} + frac{1}{n^2})^n$.



If I ignore the $frac{1}{n^2}$ term inside the parenthesis I have that the limit is $e^{-2}$ which agrees with Mathematica. Can I ignore this term? Why?






calculus







share|cite|improve this question









$endgroup$




$ (1 - frac{1}{n})^{2n} = ((1-frac{1}{n})^2)^n = (1-frac{2}{n} + frac{1}{n^2})^n$.



If I ignore the $frac{1}{n^2}$ term inside the parenthesis I have that the limit is $e^{-2}$ which agrees with Mathematica. Can I ignore this term? Why?






calculus




calculus






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 7 at 15:51









Geoffrey CritzerGeoffrey Critzer

1,5121128




1,5121128








  • 2




    $begingroup$
    Hint. It's cleaner if you use $(X^n)^2$ rather than $(X^2)^n$.
    $endgroup$
    – Ethan Bolker
    Jan 7 at 15:54






  • 2




    $begingroup$
    $$((1-1/n)^n)^2 to (1/e)^2$$
    $endgroup$
    – Crostul
    Jan 7 at 15:54






  • 1




    $begingroup$
    write it as $lim_limits{ntoinfty} [(1+frac1{-n})^{-n}]^{-2}=cdots$
    $endgroup$
    – farruhota
    Jan 7 at 15:55














  • 2




    $begingroup$
    Hint. It's cleaner if you use $(X^n)^2$ rather than $(X^2)^n$.
    $endgroup$
    – Ethan Bolker
    Jan 7 at 15:54






  • 2




    $begingroup$
    $$((1-1/n)^n)^2 to (1/e)^2$$
    $endgroup$
    – Crostul
    Jan 7 at 15:54






  • 1




    $begingroup$
    write it as $lim_limits{ntoinfty} [(1+frac1{-n})^{-n}]^{-2}=cdots$
    $endgroup$
    – farruhota
    Jan 7 at 15:55








2




2




$begingroup$
Hint. It's cleaner if you use $(X^n)^2$ rather than $(X^2)^n$.
$endgroup$
– Ethan Bolker
Jan 7 at 15:54




$begingroup$
Hint. It's cleaner if you use $(X^n)^2$ rather than $(X^2)^n$.
$endgroup$
– Ethan Bolker
Jan 7 at 15:54




2




2




$begingroup$
$$((1-1/n)^n)^2 to (1/e)^2$$
$endgroup$
– Crostul
Jan 7 at 15:54




$begingroup$
$$((1-1/n)^n)^2 to (1/e)^2$$
$endgroup$
– Crostul
Jan 7 at 15:54




1




1




$begingroup$
write it as $lim_limits{ntoinfty} [(1+frac1{-n})^{-n}]^{-2}=cdots$
$endgroup$
– farruhota
Jan 7 at 15:55




$begingroup$
write it as $lim_limits{ntoinfty} [(1+frac1{-n})^{-n}]^{-2}=cdots$
$endgroup$
– farruhota
Jan 7 at 15:55










2 Answers
2






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4












$begingroup$

You can ignore the term if you take a logarithm and expand $ln(1+x)$ as $x rightarrow 0$ (up to $O(x^2)$).



A better reason here is that $left(1-frac{1}{n}right)^{2n}=left(left(1-frac{1}{n}right)^{n}right)^2 longrightarrow (e^{-1})^2=e^{-2}$.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    $lim left(1-frac2n+frac1{n^2}right)^n=lim left(1+frac1nleft(-2+frac1{n}right)right)^n=e^{lim left(-2+frac1{n}right)}=e^{-2}.$



    Or you could just do $lim left(1-frac1nright)^{2n}=e^{2times (-1)}=e^-2.$
    The rule here is $lim (1+a_n)^{b_n}=e^{lim (na_nb_n)}.$






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

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      active

      oldest

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      4












      $begingroup$

      You can ignore the term if you take a logarithm and expand $ln(1+x)$ as $x rightarrow 0$ (up to $O(x^2)$).



      A better reason here is that $left(1-frac{1}{n}right)^{2n}=left(left(1-frac{1}{n}right)^{n}right)^2 longrightarrow (e^{-1})^2=e^{-2}$.






      share|cite|improve this answer









      $endgroup$


















        4












        $begingroup$

        You can ignore the term if you take a logarithm and expand $ln(1+x)$ as $x rightarrow 0$ (up to $O(x^2)$).



        A better reason here is that $left(1-frac{1}{n}right)^{2n}=left(left(1-frac{1}{n}right)^{n}right)^2 longrightarrow (e^{-1})^2=e^{-2}$.






        share|cite|improve this answer









        $endgroup$
















          4












          4








          4





          $begingroup$

          You can ignore the term if you take a logarithm and expand $ln(1+x)$ as $x rightarrow 0$ (up to $O(x^2)$).



          A better reason here is that $left(1-frac{1}{n}right)^{2n}=left(left(1-frac{1}{n}right)^{n}right)^2 longrightarrow (e^{-1})^2=e^{-2}$.






          share|cite|improve this answer









          $endgroup$



          You can ignore the term if you take a logarithm and expand $ln(1+x)$ as $x rightarrow 0$ (up to $O(x^2)$).



          A better reason here is that $left(1-frac{1}{n}right)^{2n}=left(left(1-frac{1}{n}right)^{n}right)^2 longrightarrow (e^{-1})^2=e^{-2}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 7 at 15:55









          MindlackMindlack

          2,70717




          2,70717























              2












              $begingroup$

              $lim left(1-frac2n+frac1{n^2}right)^n=lim left(1+frac1nleft(-2+frac1{n}right)right)^n=e^{lim left(-2+frac1{n}right)}=e^{-2}.$



              Or you could just do $lim left(1-frac1nright)^{2n}=e^{2times (-1)}=e^-2.$
              The rule here is $lim (1+a_n)^{b_n}=e^{lim (na_nb_n)}.$






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                $lim left(1-frac2n+frac1{n^2}right)^n=lim left(1+frac1nleft(-2+frac1{n}right)right)^n=e^{lim left(-2+frac1{n}right)}=e^{-2}.$



                Or you could just do $lim left(1-frac1nright)^{2n}=e^{2times (-1)}=e^-2.$
                The rule here is $lim (1+a_n)^{b_n}=e^{lim (na_nb_n)}.$






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  $lim left(1-frac2n+frac1{n^2}right)^n=lim left(1+frac1nleft(-2+frac1{n}right)right)^n=e^{lim left(-2+frac1{n}right)}=e^{-2}.$



                  Or you could just do $lim left(1-frac1nright)^{2n}=e^{2times (-1)}=e^-2.$
                  The rule here is $lim (1+a_n)^{b_n}=e^{lim (na_nb_n)}.$






                  share|cite|improve this answer









                  $endgroup$



                  $lim left(1-frac2n+frac1{n^2}right)^n=lim left(1+frac1nleft(-2+frac1{n}right)right)^n=e^{lim left(-2+frac1{n}right)}=e^{-2}.$



                  Or you could just do $lim left(1-frac1nright)^{2n}=e^{2times (-1)}=e^-2.$
                  The rule here is $lim (1+a_n)^{b_n}=e^{lim (na_nb_n)}.$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 7 at 15:57









                  John_WickJohn_Wick

                  1,486111




                  1,486111






























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