8. Let $a_{n}$ be the coefficient of the $x$ term in the expansion of $(3-\sqrt{x})^{n}$ $(n=2,3,4, \cdots)$. Then $\lim _{n \rightarrow \infty}\left(\frac{3^{2}}{a_{2}}+\frac{3^{3}}{a_{3}}+\cdots+\frac{3^{n}}{a_{n}}\right)=$ $\qquad$ .
8,18
By the binomial theorem, we know that $a_{n}=C_{n}^{2} \cdot 3^{n-2}$. Therefore,
$$
\begin{array}{l}
\frac{3^{n}}{a_{n}}=\frac{3^{2} \times 2}{n(n-1)}=18\left(\frac{1}{n-1}-\frac{1}{n}\right) . \\
\lim _{n \rightarrow \infty}\left(\frac{3^{2}}{a_{2}}+\frac{3^{3}}{a_{3}}+\cdots+\frac{3^{n}}{a_{n}}\right) \\
=\lim _{n \rightarrow \infty} 18\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdots+\frac{1}{n-1}-\frac{1}{n}\right) \\
=\lim _{n \rightarrow \infty} 18\left(1-\frac{1}{n}\right)=18 .
\end{array}
$$