12. Let $M_{n}=\left\{(\right.$ decimal $) n$-digit pure decimal $0 . \overline{a_{1} a_{2} \cdots a_{n}} \mid a_{i}$ takes only 0 or $1(i=1,2, \cdots, n-1)$, $\left.a_{n}=1\right\}, T_{n}$ is the number of elements in $M_{n}$, $S_{n}$ is the sum of all elements in $M_{n}$, then $\lim _{n \rightarrow \infty} \frac{S_{n}}{T_{n}}=$ $\qquad$ .
$$
\begin{array}{l}
T_{n}=2^{n-1}, S_{n}=2^{n-2}\left(\frac{1}{10}+\frac{1}{10^{2}}+\cdots+\frac{1}{10^{n-1}}\right)+2^{n-1} \cdot \frac{1}{10^{n}} \\
=2^{n-2}\left(\frac{1}{10}+\frac{1}{10^{2}}+\cdots+\frac{1}{10^{n-1}}+\frac{2}{10^{n}}\right)=2^{n-2}\left[\frac{\frac{1}{10}\left(1-\frac{1}{10^{n}}\right)}{1-\frac{1}{10}}+\frac{1}{10^{n}}\right] \\
=2^{n-2}\left[\frac{1}{9}\left(1-\frac{1}{10^{n}}\right)+\frac{1}{10^{n}}\right]=\frac{2^{n-2}}{9}\left(1+\frac{8}{10^{n}}\right) \Rightarrow \lim _{n \rightarrow \infty} \frac{S_{n}}{T_{n}}=\frac{1}{18} .
\end{array}
$$