Limit of a sequence of matrices  I'm preparing or my exam in linear algebra and I'm stuck with a question. I've tried to find some help in my textbook (Linear Algebra and its applications, 4th Edition, By David C. Lay). I can't find anything about it (maybe because the question is written in danish and I'm having trouble translating it right?).
I'm asked to find the limiting value, $\lim_{n \to \infty}A^nx$ where
$$
A =  \begin{bmatrix}
       0.25 & -0.75 & 0.75           \\[0.3em]
       -0.5 & 0 & 0.5 \\[0.3em]
       0.25 & -0.25 & 0.75
     \end{bmatrix}
$$
$$
x = \begin{bmatrix}
     2 \\[0.3em]
     3 \\[0.3em]
     3
\end{bmatrix}
$$
How am I supposed to solve this? I'm not asking you to calculate the answer for me, but I'm asking for the right way to solve this kind of problem.
 A: $A$ is clearly diagonalisable, as the eigenvalues of $A$ are $-1/2,1/2,1$. Therefore, $A = PDP^{-1}$, where
$$D = \begin{bmatrix}
       -0.5 & 0 & 0           \\[0.3em]
       0 & 0.5 & 0 \\[0.3em]
       0 & 0 & 1
     \end{bmatrix}$$
Now, $A^n = PD^nP^{-1}$, therefore, $\displaystyle \lim_{n \rightarrow \infty} A^n = \lim_{n \rightarrow \infty} PD^nP^{-1} = P X P^{-1}$, where,
$$X = \begin{bmatrix}
       0 & 0 & 0           \\[0.3em]
       0 & 0 & 0 \\[0.3em]
       0 & 0 & 1
     \end{bmatrix}$$
Now, all that remains is to find $P,P^{-1}$. I'll leave this easy calculation, as the method is clear.
A: You can see that the sup norm of matrix A, i.e., $\max\{a_{ij}\}$, is 3/4. Then $A:\mathbb{R}^n \rightarrow \mathbb{R}^n$ is a contraction. Since $R^n$ is a complete metric space there is a unique $y\in \mathbb{R}^n$ such that $Ay=y$ and we have $y=\lim_{n\to \infty}A^nz$, for every $z\in \mathbb{R}^n$, in particular for the given $x$. So, find y such that $Ay=y$. This limit do not depends of the norm because all norms in $\mathbb{R}^n$ are equivalents. 
