8. Let the set $A=\{1,2, \cdots, n\}$, and let $S_{n}$ denote the sum of the elements in all non-empty proper subsets of $A$, and $B_{n}$ denote the number of subsets of $A$, then $\lim _{n \rightarrow \infty} \frac{S_{n}}{n^{2} B_{n}}=$ $\qquad$ .
8. $\frac{1}{4}$.

It is easy to know that $B_{n}=2^{n}$.
First, we examine the number of times the number 1 appears in all non-empty proper subsets of $A$. It is easy to see that it appears 1 time in the one-element sets, i.e., $\mathrm{C}_{n-1}^{0}$ times, $\mathrm{C}_{n-1}^{1}$ times in the two-element sets, $\mathrm{C}_{n-1}^{2}$ times in the three-element sets, $\cdots \cdots$, and $\mathrm{C}_{n-1}^{n-2}$ times in the $(n-1)$-element sets. Therefore, the total number of times the number 1 appears in all non-empty proper subsets of $A$ is $\mathrm{C}_{n-1}^{0}+\mathrm{C}_{n-1}^{1}+\cdots+\mathrm{C}_{n-1}^{n-2}$. Thus,
$$
S_{n}=(1+2+\cdots+n)\left(\mathrm{C}_{n-1}^{0}+\mathrm{C}_{n-1}^{1}+\cdots+\mathrm{C}_{n-1}^{n-2}\right)=\frac{n(n+1)}{2}\left(2^{n-1}-1\right) .
$$

Therefore,
$$
\lim _{n \rightarrow \infty} \frac{S_{n}}{n^{2} B_{n}}=\frac{n(n+1)\left(2^{n-1}-1\right)}{2 n^{2} \times 2^{n}}=\frac{1}{4} .
$$