(12) The sequences $\left\{a_{n}\right\},\left\{b_{n}\right\}$ satisfy $a_{1}=1, b_{1}=7$, and $\left\{\begin{array}{l}a_{n+1}=b_{n}-2 a_{n}, \\ b_{n+1}=3 b_{n}-4 a_{n}\end{array}\right.$, then the value of $\lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}$ is . $\qquad$
(12) From the given conditions, we have $a_{2}=b_{1}-2 a_{1}=5$. Since $b_{n}=a_{n+1}+2 a_{n}$, it follows that
$$
a_{n+2}+2 a_{n+1}=b_{n+1}=3 b_{n}-4 a_{n}=3\left(a_{n+1}+2 a_{n}\right)-4 a_{n},
$$

which simplifies to $a_{n+2}=a_{n+1}+2 a_{n}$. Therefore,
$$
a_{n+2}+a_{n+1}=2\left(a_{n+1}+a_{n}\right),
$$

hence
$$
a_{n+1}+a_{n}=\left(a_{2}+a_{1}\right) \cdot 2^{n-1}=3 \cdot 2^{n},
$$

and thus
$$
a_{n+1}-2^{n+1}=-\left(a_{n}-2^{n}\right),
$$

which leads to
$$
a_{n}=2^{n}+(-1)^{n}.
$$

Furthermore,
$$
\begin{aligned}
b_{n}= & a_{n+1}+2 a_{n} \\
= & 2^{n+1}+(-1)^{n+1}+2\left[2^{n}+(-1)^{n}\right] \\
& =2^{n+2}+(-1)^{n}.
\end{aligned}
$$

Therefore,
$$
\lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}=\lim _{n \rightarrow \infty} \frac{2^{n}+(-1)^{n}}{2^{n+2}+(-1)^{n}}=\frac{1}{4}.
$$