\[
\left[\begin{array}{l}
\text { Square roots (continued). }
\end{array}\right]
\]

Let \((1+\sqrt{2}+\sqrt{3})^{\mathrm{n}}=p_{\mathrm{n}}+q_{\mathrm{n}} \sqrt{2}+r_{\mathrm{n}} \sqrt{3}+s_{\mathrm{n}} \sqrt{6}(n \geqslant 0)\). Find:
a) \(\lim _{n \rightarrow \infty} \frac{p_{n}}{q_{n}}\); b) \(\lim _{n \rightarrow \infty} \frac{p_{n}}{r_{n}}\); c) \(\lim _{n \rightarrow \infty} \frac{p_{n}}{s_{n}}\).
Let's use the method from problem $\underline{11.36}$. By choosing all possible combinations of signs for the numbers $\sqrt{2}$ and $\sqrt{3}$, we obtain the equalities

$$
\begin{aligned}
& \lambda_{1}^{n}=(1+\sqrt{2}+\sqrt{3})^{\mathrm{n}}=p_{\mathrm{n}}+q_{\mathrm{n}} \sqrt{2}+r_{\mathrm{n}} \sqrt{3}+s_{\mathrm{n}} \sqrt{6} \\
& \lambda_{2}^{n}=(1-\sqrt{2}+\sqrt{3})^{\mathrm{n}}=p_{\mathrm{n}}-q_{\mathrm{n}} \sqrt{2}+r_{\mathrm{n}} \sqrt{3}-s_{\mathrm{n}} \sqrt{6} \\
& \lambda_{3}^{n}=(1+\sqrt{2}-\sqrt{3})^{\mathrm{n}}=p_{\mathrm{n}}+q_{\mathrm{n}} \sqrt{2}-r_{\mathrm{n}} \sqrt{3}-s_{\mathrm{n}} \sqrt{6} \\
& \lambda_{4}^{n}=(1-\sqrt{2}-\sqrt{3})^{\mathrm{n}}=p_{\mathrm{n}}-q_{\mathrm{n}} \sqrt{2}-r_{\mathrm{n}} \sqrt{3}+s_{\mathrm{n}} \sqrt{6}
\end{aligned}
$$

By adding these equalities with coefficients $(1,1,1,1),(1,-1,1,-1),(1,1,-1,-1),(1,-1,-1,1)$, we find

$$
\begin{aligned}
& p_{\mathrm{n}}=\frac{1}{4}\left(\lambda_{1}^{n}+\lambda_{2}^{n}+\lambda_{3}^{n}+\lambda_{4}^{n}\right) \\
& q_{\mathrm{n}}=\frac{1}{4 \sqrt{2}}\left(\lambda_{1}^{n}-\lambda_{2}^{n}+\lambda_{3}^{n}-\lambda_{4}^{n}\right) \\
& r_{\mathrm{n}}=\frac{1}{4 \sqrt{3}}\left(\lambda_{1}^{n}+\lambda_{2}^{n}-\lambda_{3}^{n}-\lambda_{4}^{n}\right) \\
& s_{\mathrm{n}}=\frac{1}{4 \sqrt{6}}\left(\lambda_{1}^{n}-\lambda_{2}^{n}-\lambda_{3}^{n}+\lambda_{4}^{n}\right)
\end{aligned}
$$

From this, $\lim _{n \rightarrow \infty} \frac{p_{n}}{\boldsymbol{q}_{n}}=\sqrt{2}, \lim _{n \rightarrow \infty} \frac{p_{n}}{r_{n}}=\sqrt{3}, \lim _{n \rightarrow \infty} \frac{p_{n}}{s_{n}}=\sqrt{6}$.