first off realize that
$\lim \limits_{n\to \infty} \dfrac{n^k}{3^n} = 0,~\forall k$
next divide top and bottom by $3^n$ to obtain
$\lim \limits_{n \to \infty}~\dfrac{2\left(1-\frac{n^3}{3^n}\right)}{3 - \frac{n^3}{3^n} - \frac{3n^2}{3^n} - \frac{3n}{3^n} - \frac{1}{3^n}}= \dfrac 2 3$