Tips on proving this convergence. We have an inductively defined sequence $x_n=x_{n-1}+2y_{n-1}$ and $y_n=x_{n-1}+y_{n-1}$ where $x_n^2-2y_n^2=\pm 1$, where $x_0=1$ and $y_0=0$.
I need to prove that the sequence $\left(\frac{x_n}{y_n}\right)_{n=1}^\infty$ converges to $\sqrt2$.
Now I can see that $\left(\frac{x_n^2}{y_n^2}\right)_{n=1}^\infty$ converges to $2$, so that may be easier to prove.
It follows now that in order to prove this, $\forall\epsilon >0, \exists n>N$ s.t. $|\frac{x_n^2}{y_n^2}-2|<\epsilon$.
Can anybody point me in the right direction for solving this? 
 A: That is not an equivalent statement. For example, $\{ (-1)^n \}$ doesn't converge, but the sequence of squares of each term converges to 1. 
You will need to make an argument about the positive or negative nature of each term too. Once you've done that
Hint: $\frac {x_n ^2}{y_n^2} - 2 = \pm \frac {1}{y_n^2}$, and $y_n \rightarrow \pm \infty$, depending on your initial conditions.
A: The recursion can be written as
$$
\begin{bmatrix}x_n\\y_n\end{bmatrix}=\begin{bmatrix}1&2\\1&1\end{bmatrix}\begin{bmatrix}x_{n-1}\\y_{n-1}\end{bmatrix}
$$
The matrix has two eigenvalues: the roots of $\lambda^2-2\lambda-1=0$; that is, $\lambda=1\pm\sqrt{2}$.
The eigenvector with the eigenvalue of $1-\sqrt{2}$ is
$$
\begin{bmatrix}1&2\\1&1\end{bmatrix}\color{#C00000}{\begin{bmatrix}\sqrt{2}\\-1\end{bmatrix}}=\begin{bmatrix}-2+\sqrt{2}\\-1+\sqrt{2}\end{bmatrix}=(1-\sqrt{2})\color{#C00000}{\begin{bmatrix}\sqrt{2}\\-1\end{bmatrix}}
$$
The eigenvector with the eigenvalue of $1+\sqrt{2}$ is
$$
\begin{bmatrix}1&2\\1&1\end{bmatrix}\color{#C00000}{\begin{bmatrix}\sqrt{2}\\1\end{bmatrix}}=\begin{bmatrix}2+\sqrt{2}\\1+\sqrt{2}\end{bmatrix}=(1+\sqrt{2})\color{#C00000}{\begin{bmatrix}\sqrt{2}\\1\end{bmatrix}}
$$
Therefore,
$$
\begin{align}
\begin{bmatrix}x_n\\y_n\end{bmatrix}
&=\begin{bmatrix}x_0\\y_0\end{bmatrix}\begin{bmatrix}1&2\\1&1\end{bmatrix}^n\\
&=\left(a\begin{bmatrix}\sqrt{2}\\1\end{bmatrix}+b\begin{bmatrix}\sqrt{2}\\-1\end{bmatrix}\right)\begin{bmatrix}1&2\\1&1\end{bmatrix}^n\\
&=a(1+\sqrt2)^n\begin{bmatrix}\sqrt{2}\\1\end{bmatrix}+b(1-\sqrt2)^n\begin{bmatrix}\sqrt{2}\\-1\end{bmatrix}\\
&=a(1+\sqrt2)^n\left(\color{#C00000}{\begin{bmatrix}\sqrt{2}\\1\end{bmatrix}+\frac{b}{a}(2\sqrt{2}-3)^n\begin{bmatrix}\sqrt{2}\\-1\end{bmatrix}}\right)
\end{align}
$$
Assuming $a\ne0$, what is the limit of $x_n/y_n$ in red?
