 Solution

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