3. Let $S_{n}$ and $a_{n}$ represent the sum and the number of $n$-digit decimals in the interval $[0,1)$ that do not contain the digit 9, respectively. Then the value of $\lim _{n \rightarrow \infty} \frac{a_{n}}{S_{n}}$ is $(\quad)$.
(A) $\frac{3}{4}$
(B) $\frac{5}{4}$
(C) $\frac{7}{4}$
(D) $\frac{9}{4}$
3. (D).

It is easy to know that $a_{n}=9^{n}$, where in the $i$-th digit $(i=1,2, \cdots, n)$ of the decimal, the numbers with $0,1,2, \cdots, 8$ each appear $9^{n-1}$ times. Then
$$
\begin{aligned}
S_{n}= & 9^{n-1}(1+2+\cdots+8) \times\left(\frac{1}{10}+\frac{1}{10^{2}}+\cdots+\frac{1}{10^{n}}\right) \\
= & 4 \times 9^{n-1}\left(1-\frac{1}{10^{n}}\right) .
\end{aligned}
$$

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