8・133 Let $x_{n}=(1+\sqrt{2}+\sqrt{3})^{n}, n=1,2,3, \cdots$ If $x_{n}$ is expressed as $x_{n}=q_{n}+r_{n} \sqrt{2}+s_{n} \sqrt{3}+t_{n} \sqrt{6}$, where $q_{n}, r_{n}, s_{n}$, and $t_{n}$ are all integers, find the following limits:
$$\lim _{n \rightarrow \infty} \frac{r_{n}}{q_{n}}, \lim _{n \rightarrow \infty} \frac{s_{n}}{q_{n}}, \lim _{n \rightarrow \infty} \frac{t_{n}}{q_{n}} .$$
[Solution] Let $u_{1}=1+\sqrt{2}+\sqrt{3}, u_{2}=1-\sqrt{2}+\sqrt{3}, u_{3}=1+\sqrt{2}-\sqrt{3}$, $u_{4}=1-\sqrt{2}-\sqrt{3}$, it is easy to see that
$$\begin{array}{l}
u_{1}^{n}=q_{n}+r_{n} \sqrt{2}+s_{n} \sqrt{3}+t_{n} \sqrt{6}, \\
u_{2}^{n}=q_{n}-r_{n} \sqrt{2}+s_{n} \sqrt{3}-t_{n} \sqrt{6}, \\
u_{3}^{n}=q_{n}+r_{n} \sqrt{2}-s_{n} \sqrt{3}-t_{n} \sqrt{6}, \\
u_{4}^{n}=q_{n}-r_{n} \sqrt{2}-s_{n} \sqrt{3}+t_{n} \sqrt{6} .
\end{array}$$

Therefore,
$$\begin{array}{l}
4 q_{n}=u_{1}^{n}+u_{2}^{n}+u_{3}^{n}+u_{4}^{n} \\
4 \sqrt{2} r_{n}=u_{1}^{n}-u_{2}^{n}+u_{3}^{n}-u_{4}^{n} \\
4 \sqrt{3} s_{n}=u_{1}^{n}+u_{2}^{n}-u_{3}^{n}-u_{4}^{n} \\
4 \sqrt{6} t_{n}=u_{1}^{n}-u_{2}^{n}-u_{3}^{n}+u_{4}^{n}
\end{array}$$

Since $\left|u_{k}\right|<\left|u_{1}\right|, k=2,3,4$, it follows that $\lim _{n \rightarrow \infty} \frac{u_{k}^{n}}{u_{1}^{n}}=0, k=2,3,4$. Therefore, $\lim _{n \rightarrow \infty} \frac{q_{n}}{u_{1}^{n}}=\frac{1}{4}$, which means $\lim _{n \rightarrow \infty} \frac{u_{1}^{n}}{q_{n}}=4$.
Thus,
$$\begin{array}{l}
\lim _{n \rightarrow \infty} \frac{r_{n}}{q_{n}}=\lim _{n \rightarrow \infty} \frac{r_{n}}{u_{1}^{n}} \cdot \frac{u_{1}^{n}}{q_{n}}=\frac{4}{4 \sqrt{2}}=\frac{1}{\sqrt{2}}, \\
\lim _{n \rightarrow \infty} \frac{s_{n}}{q_{n}}=\lim _{n \rightarrow \infty} \frac{s_{n}}{u_{1}^{n}} \cdot \frac{u_{1}^{n}}{q_{n}}=\frac{1}{\sqrt{3}}, \\
\lim _{n \rightarrow \infty} \frac{t_{n}}{q_{n}}=\lim _{n \rightarrow \infty} \frac{t_{n}}{u_{1}^{n}} \cdot \frac{u_{1}^{n}}{q_{n}}=\frac{1}{\sqrt{6}} .
\end{array}$$