8.129 Let the sequence $\left\{x_{n}\right\}$ satisfy $x_{1}=5$, and
$$x_{n+1}=x_{n}^{2}-2, n=1,2, \cdots$$

Find: $\lim _{n \rightarrow \infty} \frac{x_{n+1}}{x_{1} x_{2} \cdots x_{n}}$.
［Solution］ It is easy to prove by induction

Since
$$\begin{aligned}
x_{n}>2, n & =1,2, \cdots \\
x_{n+1}^{2}-4 & =\left(x_{n}^{2}-2\right)^{2}-4=x_{n}^{2}\left(x_{n}^{2}-4\right) \\
& =x_{n}^{2} x_{n-1}^{2}\left(x_{n-1}^{2}-4\right) \\
& =\cdots=x_{n}^{2} x_{n-1}^{2} \cdots x_{1}^{2}\left(x_{1}^{2}-4\right)=21\left(x_{1} x_{2} \cdots x_{n}\right)^{2},
\end{aligned}$$

Therefore, $\left(\frac{x_{n+1}}{x_{1} \cdots x_{n}}\right)^{2}=21+\frac{4}{\left(x_{1} \cdots x_{n}\right)^{2}}$.
From (1), we get
$$\frac{4}{\left(x_{1} \cdots x_{n}\right)^{2}}<\frac{4}{2^{2 n}}$$

Thus, $\lim _{n \rightarrow \infty} \frac{4}{\left(x_{1} \cdots x_{n}\right)^{2}}=0$. Therefore,
$$\lim _{n \rightarrow \infty} \frac{x_{n+1}}{x_{1} x_{2} \cdots x_{n}}=\sqrt{21}$$