How to evaluate $lim_{nto infty }(frac{n^{2}+1}{n-1})^{-2n^{2}}$?











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How to evaluate the following limit?
$$
lim_{nto infty }(frac{n^{2}+1}{n-1})^{-2n^{2}}
$$




I have tried the following:
$$
begin{align}
lim_{nrightarrow infty }(frac{n^{2}+1+2n-2n}{n-1})^{-2n^{2}}
&= lim_{nrightarrow infty}(frac{(n-1)^{2}+2n}{n-1})^{-2n^{2}}\
&= lim_{nrightarrow infty}(frac{(n-1)^{2}}{n-1}+frac{2n}{n-1})^{-2n^{2}}\
&= lim_{nto infty }({n-1}+frac{2n}{n-1})^{-2n^{2}}
end{align}
$$


I know that:
$$
lim_{nrightarrow infty }(1+frac{1}{n})^{n}=e
$$
But I can't come to that form in my equation.
Can you help me, please?






calculus sequences-and-series limits







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    up vote
    0
    down vote

    favorite













    How to evaluate the following limit?
    $$
    lim_{nto infty }(frac{n^{2}+1}{n-1})^{-2n^{2}}
    $$




    I have tried the following:
    $$
    begin{align}
    lim_{nrightarrow infty }(frac{n^{2}+1+2n-2n}{n-1})^{-2n^{2}}
    &= lim_{nrightarrow infty}(frac{(n-1)^{2}+2n}{n-1})^{-2n^{2}}\
    &= lim_{nrightarrow infty}(frac{(n-1)^{2}}{n-1}+frac{2n}{n-1})^{-2n^{2}}\
    &= lim_{nto infty }({n-1}+frac{2n}{n-1})^{-2n^{2}}
    end{align}
    $$


    I know that:
    $$
    lim_{nrightarrow infty }(1+frac{1}{n})^{n}=e
    $$
    But I can't come to that form in my equation.
    Can you help me, please?






    calculus sequences-and-series limits







    share|cite|improve this question


























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite












      How to evaluate the following limit?
      $$
      lim_{nto infty }(frac{n^{2}+1}{n-1})^{-2n^{2}}
      $$




      I have tried the following:
      $$
      begin{align}
      lim_{nrightarrow infty }(frac{n^{2}+1+2n-2n}{n-1})^{-2n^{2}}
      &= lim_{nrightarrow infty}(frac{(n-1)^{2}+2n}{n-1})^{-2n^{2}}\
      &= lim_{nrightarrow infty}(frac{(n-1)^{2}}{n-1}+frac{2n}{n-1})^{-2n^{2}}\
      &= lim_{nto infty }({n-1}+frac{2n}{n-1})^{-2n^{2}}
      end{align}
      $$


      I know that:
      $$
      lim_{nrightarrow infty }(1+frac{1}{n})^{n}=e
      $$
      But I can't come to that form in my equation.
      Can you help me, please?






      calculus sequences-and-series limits







      share|cite|improve this question
















      How to evaluate the following limit?
      $$
      lim_{nto infty }(frac{n^{2}+1}{n-1})^{-2n^{2}}
      $$




      I have tried the following:
      $$
      begin{align}
      lim_{nrightarrow infty }(frac{n^{2}+1+2n-2n}{n-1})^{-2n^{2}}
      &= lim_{nrightarrow infty}(frac{(n-1)^{2}+2n}{n-1})^{-2n^{2}}\
      &= lim_{nrightarrow infty}(frac{(n-1)^{2}}{n-1}+frac{2n}{n-1})^{-2n^{2}}\
      &= lim_{nto infty }({n-1}+frac{2n}{n-1})^{-2n^{2}}
      end{align}
      $$


      I know that:
      $$
      lim_{nrightarrow infty }(1+frac{1}{n})^{n}=e
      $$
      But I can't come to that form in my equation.
      Can you help me, please?






      calculus sequences-and-series limits




      calculus sequences-and-series limits






      share|cite|improve this question















      share|cite|improve this question













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      edited Nov 23 at 14:02









      user587192

      1,488112




      1,488112










      asked Nov 23 at 12:54









      violettagold

      216




      216






















          3 Answers
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          $$lim_{ntoinfty}left(dfrac{n^2+1}{n-1}right)^{-2n^2}=dfrac1{lim_{ntoinfty}left(n+1+dfrac2{n-1}right)^{2n^2}}=?$$






          share|cite|improve this answer




























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            Note that:
            $$left(frac{n^{2}+1}{n-1}right)^{-2n^{2}}=left(frac{n-1}{n^{2}+1}right)^{2n^{2}}<left(frac1nright)^{2n^2}to 0.$$






            share|cite|improve this answer




























              up vote
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              We have that



              $$left(frac{n^{2}+1}{n-1}right)^{-2n^{2}}=n^{-2n^2}left(frac{n+frac1n}{n-1}right)^{-2n^{2}}=n^{-2n^2}left(frac{n-1+1+frac1n}{n-1}right)^{-2n^{2}}=$$



              $$=n^{-2n^2}left(1+frac{n+1}{n(n-1)}right)^{-2n^{2}}=n^{-2n^2}left[left(1+frac{n+1}{n(n-1)}right)^{frac{n(n-1)}{n+1}}right]^{frac{-2n^{2}(n+1)}{n(n-1)}}=$$$$=frac1{n^{2n^2}left[left(1+frac{n+1}{n(n-1)}right)^{frac{n(n-1)}{n+1}}right]^{frac{2n^{2}(n+1)}{n(n-1)}}}sim frac{1}{n^{2n^2}e^{2n}}$$






              share|cite|improve this answer





















              • Great solution, thank you very much.
                – violettagold
                Nov 23 at 13:16










              • @violettagold It can be simplified a lot, I did in that way in order to use the knwn limit for $e$ but it is not necessary of course.
                – gimusi
                Nov 23 at 13:50












              • Yes, I have noticed. I had simplified my answer at the end.
                – violettagold
                Nov 23 at 15:27











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              3 Answers
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              up vote
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              $$lim_{ntoinfty}left(dfrac{n^2+1}{n-1}right)^{-2n^2}=dfrac1{lim_{ntoinfty}left(n+1+dfrac2{n-1}right)^{2n^2}}=?$$






              share|cite|improve this answer

























                up vote
                4
                down vote













                $$lim_{ntoinfty}left(dfrac{n^2+1}{n-1}right)^{-2n^2}=dfrac1{lim_{ntoinfty}left(n+1+dfrac2{n-1}right)^{2n^2}}=?$$






                share|cite|improve this answer























                  up vote
                  4
                  down vote










                  up vote
                  4
                  down vote









                  $$lim_{ntoinfty}left(dfrac{n^2+1}{n-1}right)^{-2n^2}=dfrac1{lim_{ntoinfty}left(n+1+dfrac2{n-1}right)^{2n^2}}=?$$






                  share|cite|improve this answer












                  $$lim_{ntoinfty}left(dfrac{n^2+1}{n-1}right)^{-2n^2}=dfrac1{lim_{ntoinfty}left(n+1+dfrac2{n-1}right)^{2n^2}}=?$$







                  share|cite|improve this answer












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                  answered Nov 23 at 13:01









                  lab bhattacharjee

                  222k15155273




                  222k15155273






















                      up vote
                      2
                      down vote













                      Note that:
                      $$left(frac{n^{2}+1}{n-1}right)^{-2n^{2}}=left(frac{n-1}{n^{2}+1}right)^{2n^{2}}<left(frac1nright)^{2n^2}to 0.$$






                      share|cite|improve this answer

























                        up vote
                        2
                        down vote













                        Note that:
                        $$left(frac{n^{2}+1}{n-1}right)^{-2n^{2}}=left(frac{n-1}{n^{2}+1}right)^{2n^{2}}<left(frac1nright)^{2n^2}to 0.$$






                        share|cite|improve this answer























                          up vote
                          2
                          down vote










                          up vote
                          2
                          down vote









                          Note that:
                          $$left(frac{n^{2}+1}{n-1}right)^{-2n^{2}}=left(frac{n-1}{n^{2}+1}right)^{2n^{2}}<left(frac1nright)^{2n^2}to 0.$$






                          share|cite|improve this answer












                          Note that:
                          $$left(frac{n^{2}+1}{n-1}right)^{-2n^{2}}=left(frac{n-1}{n^{2}+1}right)^{2n^{2}}<left(frac1nright)^{2n^2}to 0.$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 23 at 13:22









                          farruhota

                          18.8k2736




                          18.8k2736






















                              up vote
                              0
                              down vote













                              We have that



                              $$left(frac{n^{2}+1}{n-1}right)^{-2n^{2}}=n^{-2n^2}left(frac{n+frac1n}{n-1}right)^{-2n^{2}}=n^{-2n^2}left(frac{n-1+1+frac1n}{n-1}right)^{-2n^{2}}=$$



                              $$=n^{-2n^2}left(1+frac{n+1}{n(n-1)}right)^{-2n^{2}}=n^{-2n^2}left[left(1+frac{n+1}{n(n-1)}right)^{frac{n(n-1)}{n+1}}right]^{frac{-2n^{2}(n+1)}{n(n-1)}}=$$$$=frac1{n^{2n^2}left[left(1+frac{n+1}{n(n-1)}right)^{frac{n(n-1)}{n+1}}right]^{frac{2n^{2}(n+1)}{n(n-1)}}}sim frac{1}{n^{2n^2}e^{2n}}$$






                              share|cite|improve this answer





















                              • Great solution, thank you very much.
                                – violettagold
                                Nov 23 at 13:16










                              • @violettagold It can be simplified a lot, I did in that way in order to use the knwn limit for $e$ but it is not necessary of course.
                                – gimusi
                                Nov 23 at 13:50












                              • Yes, I have noticed. I had simplified my answer at the end.
                                – violettagold
                                Nov 23 at 15:27















                              up vote
                              0
                              down vote













                              We have that



                              $$left(frac{n^{2}+1}{n-1}right)^{-2n^{2}}=n^{-2n^2}left(frac{n+frac1n}{n-1}right)^{-2n^{2}}=n^{-2n^2}left(frac{n-1+1+frac1n}{n-1}right)^{-2n^{2}}=$$



                              $$=n^{-2n^2}left(1+frac{n+1}{n(n-1)}right)^{-2n^{2}}=n^{-2n^2}left[left(1+frac{n+1}{n(n-1)}right)^{frac{n(n-1)}{n+1}}right]^{frac{-2n^{2}(n+1)}{n(n-1)}}=$$$$=frac1{n^{2n^2}left[left(1+frac{n+1}{n(n-1)}right)^{frac{n(n-1)}{n+1}}right]^{frac{2n^{2}(n+1)}{n(n-1)}}}sim frac{1}{n^{2n^2}e^{2n}}$$






                              share|cite|improve this answer





















                              • Great solution, thank you very much.
                                – violettagold
                                Nov 23 at 13:16










                              • @violettagold It can be simplified a lot, I did in that way in order to use the knwn limit for $e$ but it is not necessary of course.
                                – gimusi
                                Nov 23 at 13:50












                              • Yes, I have noticed. I had simplified my answer at the end.
                                – violettagold
                                Nov 23 at 15:27













                              up vote
                              0
                              down vote










                              up vote
                              0
                              down vote









                              We have that



                              $$left(frac{n^{2}+1}{n-1}right)^{-2n^{2}}=n^{-2n^2}left(frac{n+frac1n}{n-1}right)^{-2n^{2}}=n^{-2n^2}left(frac{n-1+1+frac1n}{n-1}right)^{-2n^{2}}=$$



                              $$=n^{-2n^2}left(1+frac{n+1}{n(n-1)}right)^{-2n^{2}}=n^{-2n^2}left[left(1+frac{n+1}{n(n-1)}right)^{frac{n(n-1)}{n+1}}right]^{frac{-2n^{2}(n+1)}{n(n-1)}}=$$$$=frac1{n^{2n^2}left[left(1+frac{n+1}{n(n-1)}right)^{frac{n(n-1)}{n+1}}right]^{frac{2n^{2}(n+1)}{n(n-1)}}}sim frac{1}{n^{2n^2}e^{2n}}$$






                              share|cite|improve this answer












                              We have that



                              $$left(frac{n^{2}+1}{n-1}right)^{-2n^{2}}=n^{-2n^2}left(frac{n+frac1n}{n-1}right)^{-2n^{2}}=n^{-2n^2}left(frac{n-1+1+frac1n}{n-1}right)^{-2n^{2}}=$$



                              $$=n^{-2n^2}left(1+frac{n+1}{n(n-1)}right)^{-2n^{2}}=n^{-2n^2}left[left(1+frac{n+1}{n(n-1)}right)^{frac{n(n-1)}{n+1}}right]^{frac{-2n^{2}(n+1)}{n(n-1)}}=$$$$=frac1{n^{2n^2}left[left(1+frac{n+1}{n(n-1)}right)^{frac{n(n-1)}{n+1}}right]^{frac{2n^{2}(n+1)}{n(n-1)}}}sim frac{1}{n^{2n^2}e^{2n}}$$







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Nov 23 at 13:09









                              gimusi

                              93k94495




                              93k94495












                              • Great solution, thank you very much.
                                – violettagold
                                Nov 23 at 13:16










                              • @violettagold It can be simplified a lot, I did in that way in order to use the knwn limit for $e$ but it is not necessary of course.
                                – gimusi
                                Nov 23 at 13:50












                              • Yes, I have noticed. I had simplified my answer at the end.
                                – violettagold
                                Nov 23 at 15:27


















                              • Great solution, thank you very much.
                                – violettagold
                                Nov 23 at 13:16










                              • @violettagold It can be simplified a lot, I did in that way in order to use the knwn limit for $e$ but it is not necessary of course.
                                – gimusi
                                Nov 23 at 13:50












                              • Yes, I have noticed. I had simplified my answer at the end.
                                – violettagold
                                Nov 23 at 15:27
















                              Great solution, thank you very much.
                              – violettagold
                              Nov 23 at 13:16




                              Great solution, thank you very much.
                              – violettagold
                              Nov 23 at 13:16












                              @violettagold It can be simplified a lot, I did in that way in order to use the knwn limit for $e$ but it is not necessary of course.
                              – gimusi
                              Nov 23 at 13:50






                              @violettagold It can be simplified a lot, I did in that way in order to use the knwn limit for $e$ but it is not necessary of course.
                              – gimusi
                              Nov 23 at 13:50














                              Yes, I have noticed. I had simplified my answer at the end.
                              – violettagold
                              Nov 23 at 15:27




                              Yes, I have noticed. I had simplified my answer at the end.
                              – violettagold
                              Nov 23 at 15:27


















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