In the first case of the master theorem we have:
$f(n)=O(n^{\log_{b}a-\epsilon})$. Isn't this equivalent to $f(n)=o(n^{\log_{b}a})$?
I haven't tried to formalize this but it seems intuitive (With my current understanding of big O notation, which may be flawed). However i'm sure this is not the case, since if it was that's how it would be formulated. Can someone supply a counter example?