(Consider the binomial $(3 x-1)^{2 n}=a_{2 n} x^{2 n}+a_{2 n-1} x^{2 n-1}+\cdots+a_{2} x^{2}+$ $a_{1} x+a_{0}$, let $T_{n}=a_{0}+a_{2}+\cdots+a_{2 n}, R_{n}=a_{1}+a_{3}+\cdots+a_{2 n-1}$, then $\lim _{n \rightarrow+\infty} \frac{T_{n}}{R_{n}}=$ $\qquad$ .
(8) -1 Hint: Taking $x=1$, we get $2^{2 n}=\sum_{i=0}^{2 n} a_{i}$; taking $x=-1$, we get $4^{2 n}=$ $\sum_{i=0}^{2 n}(-1)^{i} a_{i}$, thus $T_{n}=\frac{2^{2 n}+4^{2 n}}{2}, R_{n}=\frac{2^{2 n}-4^{2 n}}{2}$. Therefore,
$$
\lim _{n \rightarrow+\infty} \frac{T_{n}}{R_{n}}=\lim _{n \rightarrow+\infty} \frac{2^{2 n}+4^{2 n}}{2^{2 n}-4^{2 n}}=-1 .
$$