Both numerator and denominator have simple zeros at zero, and $1-e^{-z}=z+O(z^2)$. Their quotient tends to one and extends holomorphically with that value. Hence the singularity is \b[removable], and $\boxed{\operatorname{Res}_0 f=0}$ because its <Laurent series> has no negative powers.

Solution

Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2008 printed variant / 5 / i / Solution

 Solution (source code)