How to replicate the result of symbolic limit?

Question

I am trying to check whether the limit mentioned in the equation (3.5) of this textbook (page 14 of the PDF) does really simplify into the claimed expression.

I am trying to simplify this limit in particular: $$ \lim_{k\to\infty} \left(\mathbf{1} + \alpha_a\frac{X_a}{k}\right)^k = \sum_{m=0}^{\infty} \frac{1}{m!}\left(\alpha_aX_a\right)^m \equiv e^{\alpha_aX_a}, $$ where $X_a$ is a matrix and $\alpha_a$ is a scalar.

To make this case as simple as possible, I tried to simplify the limit using the following concrete matrix $X = \begin{bmatrix}0 & -1\\1 & 0\end{bmatrix}$.

The result of the limit simplification should then be $\begin{bmatrix}\cos\alpha & -\sin\alpha\\\sin\alpha & \cos\alpha\end{bmatrix}$, as is mentioned in equation (3.6) of the textbook.

However, Instead of that result, MMA simplified the limit into the $2\times2$ identity matrix. This is the expression I entered into MMA:

A = {
  {0, -a},
  {a, 0}
}

Limit[(IdentityMatrix[2] + A/N)^N, N -> Infinity] // MatrixForm

How can I replicate the textbook results?

Answer 1

Clear["Global`*"]

Use MatrixPower and MatrixFunction

A = {{0, -a}, {a, 0}};

Limit[MatrixPower[IdentityMatrix[2] + A/n, n], n -> Infinity] // 
  FullSimplify

(* {{Cos[a], -Sin[a]}, {Sin[a], Cos[a]}} *)

MatrixFunction[Exp, A]

(* {{Cos[a], -Sin[a]}, {Sin[a], Cos[a]}} *)