Applicabilty of master theorem (case 1) for T(n)=9T(n/3)+nlogn
Citing CLRS 3ed pp. 94-95:


  In the first case, not only must $f(n)$ be smaller than $n^{\log_b a}$, it must be polynomially smaller. That is, $f(n)$ must be asymptotically smaller than $n^{\log_b a}$ by a factor of $n^\epsilon$ for some constant $\epsilon &gt; 0$.
  
  Note that the three cases do not cover all the possibilities for $f(n)$. There is a gap between cases 1 and 2 when $f(n)$ is smaller than $n^{\log_b a}$ but not polynomially smaller.


For your case we have:
$$\begin{align}
f(n) &amp; = n \log n\\
n^{\log_b a} &amp; = n^2\\
\end{align}$$

We can then show:
$$\begin{align}
f(n) &amp; = n \log n\\
&amp; &lt; n^{2-\epsilon} &amp; ^{*}\text{for } \epsilon &lt; 1\\
&amp; &lt; n^2 &amp; \text{for } \epsilon &gt; 0\\
\end{align}$$

Then you see we can pick a $\epsilon$ value in the range $(0, 1)$ to get a polynomial difference of at least $n^\epsilon$ because:
$$ f(n) &lt; n^{2-\epsilon} &lt; n^2$$

So yes, it is admissible, but your caution is noteworthy, because there are cases where the Master Theorem can not be applied in scenarios similar to this. One example would be the following:

$$T(n) = 2T\left(\frac{n}{2}\right) + \frac{n}{\log n}$$

You see $\frac{n}{\log n}$, while smaller than $n$, is not polynomially smaller, thus we cannot use the Master Theorem.

$^*$: A polylogarithmic function grows more slowly than any positive exponent