Limit to compare growth of function

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$begingroup$


I wanted to compare growth of two functions



$F_1:n^,lg,lg n$



$F_2:(3/2)^n$



$lim_n to infty fracn^lglg n(3/2)^n$



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







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$endgroup$
















    1












    $begingroup$


    I wanted to compare growth of two functions



    $F_1:n^,lg,lg n$



    $F_2:(3/2)^n$



    $lim_n to infty fracn^lglg n(3/2)^n$



    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











    $endgroup$














      1












      1








      1





      $begingroup$


      I wanted to compare growth of two functions



      $F_1:n^,lg,lg n$



      $F_2:(3/2)^n$



      $lim_n to infty fracn^lglg n(3/2)^n$



      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











      $endgroup$




      I wanted to compare growth of two functions



      $F_1:n^,lg,lg n$



      $F_2:(3/2)^n$



      $lim_n to infty fracn^lglg n(3/2)^n$



      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




      share|cite|improve this question








      edited Nov 15 '18 at 9:51









      user376343

      3,8483829




      3,8483829










      asked Nov 15 '18 at 5:36









      user3767495user3767495

      4078




      4078




















          2 Answers
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          $begingroup$

          Consider
          $$y=fracF_1F_2=left(frac32right)^-n n^log (log (n))$$ and take logarithms
          $$log(y)=log (log (n))times log(n)-nlog left(frac32right)=nleft(log (log (n))times frac log(n)n-log left(frac32right) right)$$ When $n to infty$, since $frac log(n)n to0 $, you have
          $$log(y) sim -n log left(frac32right) to -inftyimplies y=e^log(n) to 0$$






          share|cite|improve this answer









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            2












            $begingroup$

            $(ln ln n) (ln n) - n ln (3/2)=n[frac (ln ln n) (ln n) n - ln (3/2)] to -infty$ 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









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            • $begingroup$
              Same time, same answer !
              $endgroup$
              – Claude Leibovici
              Nov 15 '18 at 6:06










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            2 Answers
            2






            active

            oldest

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            Consider
            $$y=fracF_1F_2=left(frac32right)^-n n^log (log (n))$$ and take logarithms
            $$log(y)=log (log (n))times log(n)-nlog left(frac32right)=nleft(log (log (n))times frac log(n)n-log left(frac32right) right)$$ When $n to infty$, since $frac log(n)n to0 $, you have
            $$log(y) sim -n log left(frac32right) to -inftyimplies y=e^log(n) to 0$$






            share|cite|improve this answer









            $endgroup$

















              3












              $begingroup$

              Consider
              $$y=fracF_1F_2=left(frac32right)^-n n^log (log (n))$$ and take logarithms
              $$log(y)=log (log (n))times log(n)-nlog left(frac32right)=nleft(log (log (n))times frac log(n)n-log left(frac32right) right)$$ When $n to infty$, since $frac log(n)n to0 $, you have
              $$log(y) sim -n log left(frac32right) to -inftyimplies y=e^log(n) to 0$$






              share|cite|improve this answer









              $endgroup$















                3












                3








                3





                $begingroup$

                Consider
                $$y=fracF_1F_2=left(frac32right)^-n n^log (log (n))$$ and take logarithms
                $$log(y)=log (log (n))times log(n)-nlog left(frac32right)=nleft(log (log (n))times frac log(n)n-log left(frac32right) right)$$ When $n to infty$, since $frac log(n)n to0 $, you have
                $$log(y) sim -n log left(frac32right) to -inftyimplies y=e^log(n) to 0$$






                share|cite|improve this answer









                $endgroup$



                Consider
                $$y=fracF_1F_2=left(frac32right)^-n n^log (log (n))$$ and take logarithms
                $$log(y)=log (log (n))times log(n)-nlog left(frac32right)=nleft(log (log (n))times frac log(n)n-log left(frac32right) right)$$ When $n to infty$, since $frac log(n)n to0 $, you have
                $$log(y) sim -n log left(frac32right) to -inftyimplies y=e^log(n) to 0$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 15 '18 at 6:06









                Claude LeiboviciClaude Leibovici

                122k1157134




                122k1157134





















                    2












                    $begingroup$

                    $(ln ln n) (ln n) - n ln (3/2)=n[frac (ln ln n) (ln n) n - ln (3/2)] to -infty$ 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









                    $endgroup$












                    • $begingroup$
                      Same time, same answer !
                      $endgroup$
                      – Claude Leibovici
                      Nov 15 '18 at 6:06















                    2












                    $begingroup$

                    $(ln ln n) (ln n) - n ln (3/2)=n[frac (ln ln n) (ln n) n - ln (3/2)] to -infty$ 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









                    $endgroup$












                    • $begingroup$
                      Same time, same answer !
                      $endgroup$
                      – Claude Leibovici
                      Nov 15 '18 at 6:06













                    2












                    2








                    2





                    $begingroup$

                    $(ln ln n) (ln n) - n ln (3/2)=n[frac (ln ln n) (ln n) n - ln (3/2)] to -infty$ 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









                    $endgroup$



                    $(ln ln n) (ln n) - n ln (3/2)=n[frac (ln ln n) (ln n) n - ln (3/2)] to -infty$ 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 Nov 15 '18 at 6:00









                    Kavi Rama MurthyKavi Rama Murthy

                    62.5k42262




                    62.5k42262











                    • $begingroup$
                      Same time, same answer !
                      $endgroup$
                      – Claude Leibovici
                      Nov 15 '18 at 6:06
















                    • $begingroup$
                      Same time, same answer !
                      $endgroup$
                      – Claude Leibovici
                      Nov 15 '18 at 6:06















                    $begingroup$
                    Same time, same answer !
                    $endgroup$
                    – Claude Leibovici
                    Nov 15 '18 at 6:06




                    $begingroup$
                    Same time, same answer !
                    $endgroup$
                    – Claude Leibovici
                    Nov 15 '18 at 6:06

















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