Limit involving logarithms and exponentials

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How do I calculate:



$$lim limits_n to inftyleft(1-e^left(fract+ log^2(n)1-log(n)right)right)^n$$



It's from a Statistics exercise, I haven't done analysis for a long time.
I think what I have to do is try to get it to look something like



$$lim limits_n to inftyleft(1+fracAnright)^n$$ for a suitable $A$ and apply the standard limit $$lim limits_n to inftyleft(1+fracAnright)^n = e^A$$ I am stuck though.




limits analysis




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    up vote
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    down vote

    favorite












    How do I calculate:



    $$lim limits_n to inftyleft(1-e^left(fract+ log^2(n)1-log(n)right)right)^n$$



    It's from a Statistics exercise, I haven't done analysis for a long time.
    I think what I have to do is try to get it to look something like



    $$lim limits_n to inftyleft(1+fracAnright)^n$$ for a suitable $A$ and apply the standard limit $$lim limits_n to inftyleft(1+fracAnright)^n = e^A$$ I am stuck though.




    limits analysis




    share|cite|improve this question























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      How do I calculate:



      $$lim limits_n to inftyleft(1-e^left(fract+ log^2(n)1-log(n)right)right)^n$$



      It's from a Statistics exercise, I haven't done analysis for a long time.
      I think what I have to do is try to get it to look something like



      $$lim limits_n to inftyleft(1+fracAnright)^n$$ for a suitable $A$ and apply the standard limit $$lim limits_n to inftyleft(1+fracAnright)^n = e^A$$ I am stuck though.




      limits analysis




      share|cite|improve this question













      How do I calculate:



      $$lim limits_n to inftyleft(1-e^left(fract+ log^2(n)1-log(n)right)right)^n$$



      It's from a Statistics exercise, I haven't done analysis for a long time.
      I think what I have to do is try to get it to look something like



      $$lim limits_n to inftyleft(1+fracAnright)^n$$ for a suitable $A$ and apply the standard limit $$lim limits_n to inftyleft(1+fracAnright)^n = e^A$$ I am stuck though.





      limits analysis





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      edited 6 hours ago









      Davide Morgante

      1,566118




      1,566118









      asked 6 hours ago









      Noppawee Apichonpongpan

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          First note that as $ntoinfty$ we have
          beginalign
          fract+ log^2(n)1-log(n)
          &sim-log(n)\
          &to-infty
          endalign
          hence $e^fract+ log^2(n)1-log(n)to 0$.
          Since $log(1+varepsilon)simvarepsilon$ as $varepsilonto 0$, by taking logarithm, we have
          beginalign
          lim_n to inftylogleft(1-e^fract+ log^2(n)1-log(n)right)^n
          &=lim_n to inftynlogleft(1-e^fract+ log^2(n)1-log(n)right)\
          &sim -ne^fract+ log^2(n)1-log(n)
          endalign
          By taking logarithm again:
          beginalign
          lim_n to inftylogleft(ne^fract+ log^2(n)1-log(n)right)
          &=log n+fract+ log^2(n)1-log(n)\
          &=fract+log(n)1-log(n)\
          &to -1
          endalign
          Consequently,
          $$lim_ntoinftyleft(1-e^fract+ log^2(n)1-log(n)right)^n=e^-1/e$$






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            First note that as $ntoinfty$ we have
            beginalign
            fract+ log^2(n)1-log(n)
            &sim-log(n)\
            &to-infty
            endalign
            hence $e^fract+ log^2(n)1-log(n)to 0$.
            Since $log(1+varepsilon)simvarepsilon$ as $varepsilonto 0$, by taking logarithm, we have
            beginalign
            lim_n to inftylogleft(1-e^fract+ log^2(n)1-log(n)right)^n
            &=lim_n to inftynlogleft(1-e^fract+ log^2(n)1-log(n)right)\
            &sim -ne^fract+ log^2(n)1-log(n)
            endalign
            By taking logarithm again:
            beginalign
            lim_n to inftylogleft(ne^fract+ log^2(n)1-log(n)right)
            &=log n+fract+ log^2(n)1-log(n)\
            &=fract+log(n)1-log(n)\
            &to -1
            endalign
            Consequently,
            $$lim_ntoinftyleft(1-e^fract+ log^2(n)1-log(n)right)^n=e^-1/e$$






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              First note that as $ntoinfty$ we have
              beginalign
              fract+ log^2(n)1-log(n)
              &sim-log(n)\
              &to-infty
              endalign
              hence $e^fract+ log^2(n)1-log(n)to 0$.
              Since $log(1+varepsilon)simvarepsilon$ as $varepsilonto 0$, by taking logarithm, we have
              beginalign
              lim_n to inftylogleft(1-e^fract+ log^2(n)1-log(n)right)^n
              &=lim_n to inftynlogleft(1-e^fract+ log^2(n)1-log(n)right)\
              &sim -ne^fract+ log^2(n)1-log(n)
              endalign
              By taking logarithm again:
              beginalign
              lim_n to inftylogleft(ne^fract+ log^2(n)1-log(n)right)
              &=log n+fract+ log^2(n)1-log(n)\
              &=fract+log(n)1-log(n)\
              &to -1
              endalign
              Consequently,
              $$lim_ntoinftyleft(1-e^fract+ log^2(n)1-log(n)right)^n=e^-1/e$$






              share|cite|improve this answer























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                up vote
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                First note that as $ntoinfty$ we have
                beginalign
                fract+ log^2(n)1-log(n)
                &sim-log(n)\
                &to-infty
                endalign
                hence $e^fract+ log^2(n)1-log(n)to 0$.
                Since $log(1+varepsilon)simvarepsilon$ as $varepsilonto 0$, by taking logarithm, we have
                beginalign
                lim_n to inftylogleft(1-e^fract+ log^2(n)1-log(n)right)^n
                &=lim_n to inftynlogleft(1-e^fract+ log^2(n)1-log(n)right)\
                &sim -ne^fract+ log^2(n)1-log(n)
                endalign
                By taking logarithm again:
                beginalign
                lim_n to inftylogleft(ne^fract+ log^2(n)1-log(n)right)
                &=log n+fract+ log^2(n)1-log(n)\
                &=fract+log(n)1-log(n)\
                &to -1
                endalign
                Consequently,
                $$lim_ntoinftyleft(1-e^fract+ log^2(n)1-log(n)right)^n=e^-1/e$$






                share|cite|improve this answer













                First note that as $ntoinfty$ we have
                beginalign
                fract+ log^2(n)1-log(n)
                &sim-log(n)\
                &to-infty
                endalign
                hence $e^fract+ log^2(n)1-log(n)to 0$.
                Since $log(1+varepsilon)simvarepsilon$ as $varepsilonto 0$, by taking logarithm, we have
                beginalign
                lim_n to inftylogleft(1-e^fract+ log^2(n)1-log(n)right)^n
                &=lim_n to inftynlogleft(1-e^fract+ log^2(n)1-log(n)right)\
                &sim -ne^fract+ log^2(n)1-log(n)
                endalign
                By taking logarithm again:
                beginalign
                lim_n to inftylogleft(ne^fract+ log^2(n)1-log(n)right)
                &=log n+fract+ log^2(n)1-log(n)\
                &=fract+log(n)1-log(n)\
                &to -1
                endalign
                Consequently,
                $$lim_ntoinftyleft(1-e^fract+ log^2(n)1-log(n)right)^n=e^-1/e$$







                share|cite|improve this answer













                share|cite|improve this answer



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                answered 5 hours ago









                Fabio Lucchini

                5,55411025




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