Compute the limit of this expression of norms: Compute the limit, as n goes to infinity, of the quotient:
$$\frac{||A^{n+2}(x)||}{||A^n(x)||} $$, given the matrix $$
        \begin{bmatrix}
        0 & 3  \\
        -2 & 5  \\
        \end{bmatrix}$$ 
and the vector x = (1,0).
I diagonalized A, found its eigenvalues and eigenvectors, and arrive at the expression 
$A^{n+2}$ = $SD^{n+2}S^{-1}$ = $$\begin{bmatrix}
        3 & 1  \\
        2 & 1  \\ 
        \end{bmatrix}  \begin{bmatrix}
        2^{n+2} & 0  \\
        0 & 3^{n+2}  \\
        \end{bmatrix}\begin{bmatrix}
        1 & -1  \\
        -2 & 3  \\
        \end{bmatrix}$$, 
where the columns of S form a basis of eigenvectors of A for the 2-dimensional space.
Now, how do I actually compute: $${||A^{n+2}(x)||}?$$
Can I do this (from following my intuition):  $$\begin{bmatrix}
        3 & 1  \\
        2 & 1  \\ 
        \end{bmatrix}  \begin{bmatrix}
        2^{n+2} & 0  \\
        0 & 3^{n+2}  \\
        \end{bmatrix}\begin{bmatrix}
        1 & -1  \\
        -2 & 3  \\
        \end{bmatrix}* (1,0)^T$$,  and then just multiply everything through and arrive at some 2x1 column vector, with entries that depend on n, and then take the usual p-norm for vectors, say, the 2-norm, then take the limit as n goes to infinity?  
I tried this method of "brute force", which seems valid, but I am not getting the desired answer.
What's going wrong?  
Thanks,

Hint: note that
$$
\|A^n(x)\|
=
\left\| 
\pmatrix{
 3\cdot 2^n-2\cdot 3^n \\
 2 \cdot 2^n-2\cdot 3^n
}
\right\| = 
3^{n}\left\| 
\pmatrix{
 3\cdot (2/3)^n-2 \\
2\cdot (2/3)^n-2
}
\right\|
$$
and that as $n \to \infty$, we have
$$
\left\| 
\pmatrix{
 3\cdot (2/3)^n-2 \\
2\cdot (2/3)^n-2
}
\right\| \to 
\left\| 
\pmatrix{
-2 \\
-2
}
\right\|
$$
