= Solution
Since $A^2=-\epsilon^2I$, its even powers are $A^{2k}=(-1)^k\epsilon^{2k}I$ and its odd powers are $A^{2k+1}=(-1)^k\epsilon^{2k}A$. Splitting the <matrix exponential> into these two absolutely convergent subseries gives the <matrix exponential for a scalar square>:
$$
\boxed{S=e^A=I\cos\epsilon+\frac{\sin\epsilon}{\epsilon}A=\begin{pmatrix}\cos\epsilon&\sin\epsilon\\-\sin\epsilon&\cos\epsilon\end{pmatrix}.}
$$
At $\epsilon=0$, the coefficient $\sin\epsilon/\epsilon$ has limiting value one and $A=0$, so $S=I$. The result is a planar rotation through angle $-\epsilon$.
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