5. 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 $\lim _{n \rightarrow \infty} \frac{S_{n}}{a_{n}}=$ $\qquad$
5. $\frac{4}{9}$.

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

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