8. In the sequence $\left\{a_{n}\right\}$, it is known that $a_{1}=1, a_{n+1}>a_{n}$, and $a_{n+1}^{2}+a_{n}^{2}+1=2\left(a_{n+1}+\right.$ $\left.a_{n}+2 a_{n+1} a_{n}\right)$, find $\lim _{n \rightarrow \infty} \frac{S_{n}}{n a_{n}}$.
8. First, find the general term $a_{n}$ and the sum of the first $n$ terms $S_{n}$.

From the given, $\left(a_{n+1}+a_{n}-1\right)^{2}=4 a_{n} a_{n+1}$. Also, $a_{1}=1, a_{n+1}>a_{n}$, so $a_{n+1}+a_{n}-1>0$, and $\left(\sqrt{a_{n+1}}-\sqrt{a_{n}}\right)^{2}=1$. Therefore, $\sqrt{a_{n+1}}-\sqrt{a_{n}}=1$, which means the sequence $\left\{\sqrt{a_{n}}\right\}$ is an arithmetic sequence with the first term 1 and common difference 1. Hence, $\sqrt{a_{n}}=n, a_{n}=n^{2}$, and thus $S_{n}=1^{2}+2^{2}+\cdots+n^{2}=$ $\frac{n(n+1)}{6}(2 n+1)$. Therefore, $\lim _{n \rightarrow \infty} \frac{S_{n}}{n a_{n}}=\frac{1}{3}$.