<s>[INST] What is $\lim_{n\to\infty} \frac{x_n}{y_n}$ provided $x_n=2y_{n}-y_{n-1}$ and $y_n=3y_{n-2}-y_{n-2}$ with $x_0=y_0=1$? Suppose $x_n$ and $y_n$ satisfy:
$$
        \begin{pmatrix}
        x_n \\
        y_n \\
        \end{pmatrix}=\begin{pmatrix}
        2&1 \\
        1&1 \\
        \end{pmatrix}\begin{pmatrix}
        x_{n-1} \\
        y_{n-1} \\
        \end{pmatrix} $$ 
with $x_0=y_0=1$. What is $\displaystyle\lim_{n\to\infty} \frac{x_n}{y_n}$ assuming this limit exists?
After doing some algebra, I've got $x_n=2y_{n}-y_{n-1}$ and $y_n=3y_{n-2}-y_{n-2}$, but cannot go further.
 [/INST] Hint: Let $z_{n}\equiv(x_{n},y_{n})^{\intercal}$. Then, we can rewrite
the recurrence as
$$
z_{n}=Mz_{n-1}\text{ (}n\geq1\text{)}
$$
where 
$$
M\equiv\begin{pmatrix}2 & 1\\
1 & 1
\end{pmatrix}.
$$
Note that $M$ is diagonalizable. That is, we can write $M=SJS^{-1}$
where
$$
S\equiv\begin{pmatrix}\frac{1}{2}\left(1-\sqrt{5}\right) & \frac{1}{2}\left(1+\sqrt{5}\right)\\
1 & 1
\end{pmatrix}\text{ and }J\equiv\begin{pmatrix}\frac{1}{2}\left(3-\sqrt{5}\right)\\
 & \frac{1}{2}\left(3+\sqrt{5}\right)
\end{pmatrix}
$$
By induction, we have $z_{n}=M^{n}z_{0}$. Note that
$$
M^{n}=\left(SJS^{-1}\right)^{n}=SJ^{n}S^{-1}.
$$
Can you figure out the rest?
</s>