Since $a_1, a_2, \dots, a_{n-1}$ all have exactly two possible values, so $T_n = 2^{n-1}$. Meanwhile, the frequency of $a_i = 1$ is the same as that of $a_i = 0$ for $1 \le i \le n-1$, and $a_n = 1$. Then
$$\begin{aligned} S_n &= \frac{1}{2} \times 2^{n-1} \times \left( \frac{1}{10} + \frac{1}{10^2} + \dots + \frac{1}{10^{n-1}} \right) + 2^{n-1} \times \frac{1}{10^n} \\ &= 2^{n-1} \times \frac{1}{18} \times \left( 1 - \frac{1}{10^{n-1}} \right) + 2^{n-1} \times \frac{1}{10^n}. \end{aligned}$$

So
$$\lim_{n \to \infty} \frac{S_n}{T_n} = \lim_{n \to \infty} \left[ \frac{1}{18} \left( 1 - \frac{1}{10^{n-1}} \right) + \frac{1}{10^n} \right] = \frac{1}{18}.$$