Limit to compare growth of function











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I wanted to compare growth of two functions



$F_1:n^{,lg,lg n}$



$F_2:(3/2)^n$



$lim_{n to infty} frac{n^{lglg n}}{(3/2)^n}$



After differentiating it $lg , lg n$ times I get



$lim_{n to infty} frac{(lglg n)!}{(lg(3/2))^{lglg n}(3/2)^n}$



How do I proceed forward?






limits







share|cite|improve this question




























    up vote
    2
    down vote

    favorite












    I wanted to compare growth of two functions



    $F_1:n^{,lg,lg n}$



    $F_2:(3/2)^n$



    $lim_{n to infty} frac{n^{lglg n}}{(3/2)^n}$



    After differentiating it $lg , lg n$ times I get



    $lim_{n to infty} frac{(lglg n)!}{(lg(3/2))^{lglg n}(3/2)^n}$



    How do I proceed forward?






    limits







    share|cite|improve this question


























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      I wanted to compare growth of two functions



      $F_1:n^{,lg,lg n}$



      $F_2:(3/2)^n$



      $lim_{n to infty} frac{n^{lglg n}}{(3/2)^n}$



      After differentiating it $lg , lg n$ times I get



      $lim_{n to infty} frac{(lglg n)!}{(lg(3/2))^{lglg n}(3/2)^n}$



      How do I proceed forward?






      limits







      share|cite|improve this question















      I wanted to compare growth of two functions



      $F_1:n^{,lg,lg n}$



      $F_2:(3/2)^n$



      $lim_{n to infty} frac{n^{lglg n}}{(3/2)^n}$



      After differentiating it $lg , lg n$ times I get



      $lim_{n to infty} frac{(lglg n)!}{(lg(3/2))^{lglg n}(3/2)^n}$



      How do I proceed forward?






      limits




      limits






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




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      edited 9 hours ago









      user376343

      2,1041716




      2,1041716










      asked 13 hours ago









      user3767495

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      1348






















          2 Answers
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          $(ln ln n) (ln n) - n ln (3/2)=n[frac {(ln ln n) (ln n)} n - ln (3/2)] to -infty$ because $frac {(ln ln n) (ln n)} n to 0$. [ Use L'Hopital's Rule for this]. Taking exponential we get $e^{(ln ln n) (ln n) } /(3/2)^{n} to 0$. This is same as $frac {n^{ln ln n}} {(3/2)^{n}} to 0$






          share|cite|improve this answer





















          • Same time, same answer !
            – Claude Leibovici
            13 hours ago


















          up vote
          3
          down vote













          Consider
          $$y=frac{F_1}{F_2}=left(frac{3}{2}right)^{-n} n^{log (log (n))}$$ and take logarithms
          $$log(y)={log (log (n))}times log(n)-nlog left(frac{3}{2}right)=nleft({log (log (n))}times frac {log(n)}n-log left(frac{3}{2}right) right)$$ When $n to infty$, since $frac {log(n)}n to0 $, you have
          $$log(y) sim -n log left(frac{3}{2}right) to -inftyimplies y=e^{log(n)} to 0$$






          share|cite|improve this answer





















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            2 Answers
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            active

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            2 Answers
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            active

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            active

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            up vote
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            down vote













            $(ln ln n) (ln n) - n ln (3/2)=n[frac {(ln ln n) (ln n)} n - ln (3/2)] to -infty$ because $frac {(ln ln n) (ln n)} n to 0$. [ Use L'Hopital's Rule for this]. Taking exponential we get $e^{(ln ln n) (ln n) } /(3/2)^{n} to 0$. This is same as $frac {n^{ln ln n}} {(3/2)^{n}} to 0$






            share|cite|improve this answer





















            • Same time, same answer !
              – Claude Leibovici
              13 hours ago















            up vote
            3
            down vote













            $(ln ln n) (ln n) - n ln (3/2)=n[frac {(ln ln n) (ln n)} n - ln (3/2)] to -infty$ because $frac {(ln ln n) (ln n)} n to 0$. [ Use L'Hopital's Rule for this]. Taking exponential we get $e^{(ln ln n) (ln n) } /(3/2)^{n} to 0$. This is same as $frac {n^{ln ln n}} {(3/2)^{n}} to 0$






            share|cite|improve this answer





















            • Same time, same answer !
              – Claude Leibovici
              13 hours ago













            up vote
            3
            down vote










            up vote
            3
            down vote









            $(ln ln n) (ln n) - n ln (3/2)=n[frac {(ln ln n) (ln n)} n - ln (3/2)] to -infty$ because $frac {(ln ln n) (ln n)} n to 0$. [ Use L'Hopital's Rule for this]. Taking exponential we get $e^{(ln ln n) (ln n) } /(3/2)^{n} to 0$. This is same as $frac {n^{ln ln n}} {(3/2)^{n}} to 0$






            share|cite|improve this answer












            $(ln ln n) (ln n) - n ln (3/2)=n[frac {(ln ln n) (ln n)} n - ln (3/2)] to -infty$ because $frac {(ln ln n) (ln n)} n to 0$. [ Use L'Hopital's Rule for this]. Taking exponential we get $e^{(ln ln n) (ln n) } /(3/2)^{n} to 0$. This is same as $frac {n^{ln ln n}} {(3/2)^{n}} to 0$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 13 hours ago









            Kavi Rama Murthy

            39k31748




            39k31748












            • Same time, same answer !
              – Claude Leibovici
              13 hours ago


















            • Same time, same answer !
              – Claude Leibovici
              13 hours ago
















            Same time, same answer !
            – Claude Leibovici
            13 hours ago




            Same time, same answer !
            – Claude Leibovici
            13 hours ago










            up vote
            3
            down vote













            Consider
            $$y=frac{F_1}{F_2}=left(frac{3}{2}right)^{-n} n^{log (log (n))}$$ and take logarithms
            $$log(y)={log (log (n))}times log(n)-nlog left(frac{3}{2}right)=nleft({log (log (n))}times frac {log(n)}n-log left(frac{3}{2}right) right)$$ When $n to infty$, since $frac {log(n)}n to0 $, you have
            $$log(y) sim -n log left(frac{3}{2}right) to -inftyimplies y=e^{log(n)} to 0$$






            share|cite|improve this answer

























              up vote
              3
              down vote













              Consider
              $$y=frac{F_1}{F_2}=left(frac{3}{2}right)^{-n} n^{log (log (n))}$$ and take logarithms
              $$log(y)={log (log (n))}times log(n)-nlog left(frac{3}{2}right)=nleft({log (log (n))}times frac {log(n)}n-log left(frac{3}{2}right) right)$$ When $n to infty$, since $frac {log(n)}n to0 $, you have
              $$log(y) sim -n log left(frac{3}{2}right) to -inftyimplies y=e^{log(n)} to 0$$






              share|cite|improve this answer























                up vote
                3
                down vote










                up vote
                3
                down vote









                Consider
                $$y=frac{F_1}{F_2}=left(frac{3}{2}right)^{-n} n^{log (log (n))}$$ and take logarithms
                $$log(y)={log (log (n))}times log(n)-nlog left(frac{3}{2}right)=nleft({log (log (n))}times frac {log(n)}n-log left(frac{3}{2}right) right)$$ When $n to infty$, since $frac {log(n)}n to0 $, you have
                $$log(y) sim -n log left(frac{3}{2}right) to -inftyimplies y=e^{log(n)} to 0$$






                share|cite|improve this answer












                Consider
                $$y=frac{F_1}{F_2}=left(frac{3}{2}right)^{-n} n^{log (log (n))}$$ and take logarithms
                $$log(y)={log (log (n))}times log(n)-nlog left(frac{3}{2}right)=nleft({log (log (n))}times frac {log(n)}n-log left(frac{3}{2}right) right)$$ When $n to infty$, since $frac {log(n)}n to0 $, you have
                $$log(y) sim -n log left(frac{3}{2}right) to -inftyimplies y=e^{log(n)} to 0$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 13 hours ago









                Claude Leibovici

                116k1156131




                116k1156131






























                     

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