Compute $\lim_{N\to \infty} A_N$ where $\{A_N\}$ is a sequence of matrices 
Given matrix, $$A_N = \begin{bmatrix} 1 & \frac{\pi}{2N}\\ \frac{-\pi}{2N} & 1\end{bmatrix}^N$$ compute $$\lim_{N \rightarrow \infty} A_N$$

I took the logarithm of both sides but was not able to figure out the limit. Any suggestions on how to approach this problem?

The characteristic polynomial of $A$ is given by:
$$ \det \begin{bmatrix} x-1 & \frac{-\pi}{2N} \\  \frac{\pi}{2N} & x-1\end{bmatrix} =(x-1)^2+\frac{\pi^2}{4N^2}$$
To find the eigenvalues of $A$, set the characteristic polynomial equal to zero, and solve for $x$. This gives us the complex conjugate eigenvalues $x=\frac{i\pi}{2N}+1$ and $x=\frac{-i\pi}{2N}+1$.
Now, $2$ linearly independent eigenvectors of $A$ would then be: $x=\begin{bmatrix} i \\ -1 \end{bmatrix}$ and $x=\begin{bmatrix} i \\ 1 \end{bmatrix}$, corresponding to the eigenvalues  $x=\frac{i\pi}{2N}+1$ and $x=\frac{-i\pi}{2N}+1$ respectively ( This part should be relatively easy to obtain ).
Hence, $A$ is diagonalisable, and we have $A=PDP^{-1}$, where the diagonal matrix $D$ is such that $D = \begin{bmatrix} 1+\frac{i\pi}{2N} & 0 \\ 0 & 1-\frac{i\pi}{2N} \end{bmatrix}  $ and $P= \begin{bmatrix} i & i \\ -1 & 1  \end{bmatrix}$. In addition, we may easily evaluate $P^{-1}$ to be: $P^{-1}= \frac{1}{2i} \begin{bmatrix} 1 & -i \\ 1 & i  \end{bmatrix}$.
Since $A^{N}$=$(PDP^{-1})^N$=$PD^NP^{-1}$, and $D^N \rightarrow \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix}$ as $N \rightarrow \infty$ , we have that $A^N=P\begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix}P^{-1}$ as $N \rightarrow \infty$ . Furthermore, this is simply equal to: $\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$, giving us the desired limit of the matrix.
