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

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SUMMARY

The discussion focuses on proving that the expression n^(logc)/c^(logn) equals 1 as n approaches infinity, where c is a constant greater than 1. Participants explored various methods, including L'Hôpital's rule, but found it complicated due to logarithmic terms. The key insight provided by RGV simplifies the proof by transforming the expression into log(n^logc/c^logn), ultimately demonstrating that it converges to zero. This confirms that the original expression approaches 1 as n increases indefinitely.

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

RGV
 
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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