8. In the sequence $\left\{a_{n}\right\}$, $a_{1}=1, a_{n+1}>a_{n}$, and $a_{n+1}^{2}+a_{n}^{2}+1=2\left(a_{n+1} a_{n}+a_{n+1}+\right.$ $\left.a_{n}\right), \lim _{n \rightarrow+\infty} \frac{S_{n}}{n a_{n}}=$ $\qquad$ .
8. $\frac{1}{3}$.
From the condition, we get $\left(a_{n+1}+a_{n}\right)^{2}-2\left(a_{n+1}+a_{n}\right)+1=4 a_{n+1} a_{n}$, hence $\left(a_{n+1}+a_{n}-1\right)^{2}=4 a_{n+1} a_{n}$, so $a_{n+1}+a_{n}-1= \pm 2 \sqrt{a_{n+1} a_{n}}$. Also, $a_{1}>1, a_{n+1}>a_{n} \geqslant a_{1}>1$, thus $a_{n}+a_{n+1}-1>0$, then $a_{n+1}+a_{n}-1=2 \sqrt{a_{n+1} a_{n}}$, which means $\left(\sqrt{a_{n+1}}-\sqrt{a_{n}}\right)^{2}=1$, i.e., $\sqrt{a_{n+1}}-\sqrt{a_{n}}=1$. Therefore, $\sqrt{a_{n}}=\sqrt{a_{1}}+(n-1)=n, a_{n}=n^{2}, S_{n}=\frac{1}{6} n(n+1)(2 n+1)$, then $\frac{S_{n}}{n \cdot a_{n}}=\frac{(n+1)(2 n+1)}{6 n^{2}}$. Therefore, $\lim _{n \rightarrow+\infty} \frac{S_{n}}{n a_{n}}=\frac{1}{3}$.