Example 1 Suppose
$$
(1+\sqrt{2})^{n}=x_{n}+y_{n} \sqrt{2}\left(x_{n}, y_{n} \in \mathbf{Z}\right) \text {. }
$$

Find $\lim _{n \rightarrow+\infty} \frac{x_{n}}{y_{n}}$.
(2000, Fudan University Admissions Exam for Recommended Students)
Solution 1 Construct the dual form.
Consider the dual form $(1-\sqrt{2})^{n}$, by the binomial theorem we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
(1+\sqrt{2})^{n}=x_{n}+y_{n} \sqrt{2}, \\
(1-\sqrt{2})^{n}=x_{n}-y_{n} \sqrt{2}
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
x_{n}=\frac{1}{2}\left[(1+\sqrt{2})^{n}+(1-\sqrt{2})^{n}\right], \\
y_{n}=\frac{1}{2 \sqrt{2}}\left[(1+\sqrt{2})^{n}-(1-\sqrt{2})^{n}\right] .
\end{array}\right.
\end{array}
$$

Therefore, $\lim _{n \rightarrow+\infty} \frac{x_{n}}{y_{n}}=\sqrt{2} \lim _{n \rightarrow+\infty} \frac{(1+\sqrt{2})^{n}+(1-\sqrt{2})^{n}}{(1+\sqrt{2})^{n}-(1-\sqrt{2})^{n}}$
$$
=\sqrt{2} \lim _{n \rightarrow+\infty} \frac{1+\left(\frac{1-\sqrt{2}}{1+\sqrt{2}}\right)^{n}}{1-\left(\frac{1-\sqrt{2}}{1+\sqrt{2}}\right)^{n}}=\sqrt{2} \text {. }
$$

Solution 2 Construct a recursive sequence.
Since $(1+\sqrt{2})^{n+1}=x_{n+1}+y_{n+1} \sqrt{2}$, and
$$
\begin{array}{l}
\left(x_{n}+y_{n} \sqrt{2}\right)(1+\sqrt{2}) \\
=\left(x_{n}+2 y_{n}\right)+\left(x_{n}+y_{n}\right) \sqrt{2},
\end{array}
$$

Therefore, $\left\{\begin{array}{l}x_{n+1}=x_{n}+2 y_{n} \\ y_{n+1}=x_{n}+y_{n} .\end{array}\right.$
Thus, $\lim _{n \rightarrow+\infty} \frac{x_{n}}{y_{n}}=\lim _{n \rightarrow+\infty} \frac{x_{n+1}}{y_{n+1}}=\lim _{n \rightarrow+\infty} \frac{x_{n}+2 y_{n}}{x_{n}+y_{n}}$. Let $\lim _{n \rightarrow+\infty} \frac{x_{n}}{y_{n}}=k>0$.
Then by $k=\frac{k+2}{k+1}$, solving gives $k=\sqrt{2}$ (negative value is discarded).