Where is the value of the variable $\epsilon$ obtained in the following explanation my professor gave? My professor gave us this explanation from the textbook Introduction to Algorithms regarding the Master Method/Theorem:

As a first example, consider $$T(n)=9T(n/3)+n.$$ For this recurrence,
  we have $a=9$, $b=3$, $f(n)n$, and thus we have that
  $n^{log_b{a}}=n^{log_3{9}}=\Theta(n^2)$. Since
  $f(n)=O(n^{log_3{9-\epsilon}})$, where $\epsilon=1$, we can apply case
  1 of the master theorem and conclude that the possibility is
  $T(n)=\Theta(n^2)$.

I'm very confused where the "$\epsilon$ = 1" comes from. Above in another explanation it says "for some constant $\epsilon$ > 0" but that obviously gives us the possibility of any number from $1$ to $\infty$. How is $1$ obtained?

There is one function that is linear that solves it, $f(n)=-n/2$, but the general expression for $g(n)=f(n)+n/2$ is $g(n)=9g(n/3)$, so the obvious solution to that is $g(n)=An^2$.  I don't know what he means by $\epsilon=1$.
