8. Let $a_{n}$ denote the number of $n$-digit decimals in the interval $[0,1)$ that do not contain the digit 9, and let $S_{n}$ denote the sum of these decimals. Then
$$
\lim _{n \rightarrow \infty} \frac{S_{n}}{a_{n}}=
$$
$\qquad$
8. $\frac{4}{9}$.

Obviously, $a_{n}=9^{n}$, where the numbers with $1,2,3, \cdots, 8$ appearing in the $i$-th digit $(i=1,2, \cdots, n)$ of the decimal are each $9^{n-1}$. Then
$$
\begin{aligned}
S_{n} & =9^{n-1}(1+2+3+\cdots+8)\left(\frac{1}{10}+\frac{1}{10^{2}}+\frac{1}{10^{3}}+\cdots+\frac{1}{10^{n}}\right) \\
& =4 \times 9^{n-1}\left(1-\frac{1}{10^{n}}\right) .
\end{aligned}
$$

Therefore,
$$
\lim _{n \rightarrow \infty} \frac{S_{n}}{a_{n}}=\lim _{n \rightarrow \infty}\left(\frac{4}{9}\left(1-\frac{1}{10^{n}}\right)\right)=\frac{4}{9} .
$$