if $A$ is $2 \times 2$ matrix find $\lim_{n \to \infty} A^n$ if $A$ is $2 \times 2$ given by
$$A= \begin{bmatrix}
 1& \frac{\theta}{n} \\ 
-\frac{\theta}{n} & 1
\end{bmatrix}$$
matrix find $$\lim_{n \to \infty} A^n$$ 
I tried like this:
$$A-I=\begin{bmatrix}
 0& \frac{\theta}{n} \\ 
-\frac{\theta}{n} & 0
\end{bmatrix}$$
So
$$(A-I)^2= -\frac{\theta^2}{n^2} I$$ Generalizing we get
$$(A-I)^n=-\frac{\theta^n}{n^n} I$$ 
any clue or hint here?

Let p(t) be a polynomial function with variable t.
Let $ p(t)=\det(tI-A) $.
$$p(t)=t^2 -2t+1+ \frac{θ^2 }{n^2}$$
By Cayley-Hamilton theorem, p(matrix A)= zero matrix
Set p(t)=0, t= 1 (+ or -) θi/n, and these are the eigenvalues.
