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Show that n^(logc)/c^(logn) =1 as n->inf

  1. Jan 31, 2012 #1
    1. The problem statement, all variables and given/known data
    Show that n^(logc)/c^(logn) =1 as n->inf where c is a constant greater than 1


    2. Relevant equations



    3. The attempt at a solution

    Tried L'hospitals. But the logs mess it up. Even if you assume that logc>1 then the top does eventually become a constant (second derivative). However the bottom gets too messy. Is there another method to start it?
     
  2. jcsd
  3. Jan 31, 2012 #2

    Ray Vickson

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    If f(n) = n^log(c) / c^log(n), look carefully at log (f(n)).

    RGV
     
  4. Jan 31, 2012 #3
    Thats log(n^logc/c^logn)= log(n^logc)-log(c^logn)=log(c)log(n)-log(n)log(c)=0.

    Ok i got it

    Thanks
     
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