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

[bfpri]

Homework Statement

Show that n^(logc)/c^(logn) =1 as n->inf where c is a constant greater than 1

Homework Equations

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?

 

[Ray Vickson]

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

RGV

 

[bfpri]

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