8. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=3, a_{n+1}=$ $9 \sqrt[3]{a_{n}}(n \geqslant 1)$. Then $\lim _{n \rightarrow \infty} a_{n}=$ $\qquad$ .
8.27.

Given $\log _{3} a_{n+1}=2+\frac{1}{3} \log _{3} a_{n}$, let $b_{n}=\log _{3} a_{n}$, then
$$
\begin{array}{l}
b_{n+1}=\frac{1}{3} b_{n}+2, b_{1}=1 \\
\Rightarrow b_{n+1}-3=\frac{1}{3}\left(b_{n}-3\right) \\
\Rightarrow b_{n}-3=\left(b_{1}-3\right)\left(\frac{1}{3}\right)^{n-1}=-2 \times\left(\frac{1}{3}\right)^{n-1} \\
\Rightarrow b_{n}=-\frac{2}{3^{n-1}}+3 \Rightarrow a_{n}=3^{-\frac{2}{3^{n-1}}+3} . \\
\text { Also } \lim _{n \rightarrow \infty}\left(-\frac{2}{3^{n-1}}+3\right)=3, \text { so } \lim _{n \rightarrow \infty} a_{n}=3^{3}=27 .
\end{array}
$$