Let $f(z) = \frac{ e^z - 1}{\sin z }$. Im trying to find $Res(f,z_0=0) $
I know that $(z-0)f(z)= \frac{ z(e^z-1)}{\sin z }$. And the limit as $z \to 0$ is of the form $0/0$ so applying lhopitals rule, we get that the limit is actually $0$. Can we conclude that the residue is then $0$ ?
In my notes, it says that the limit $\lim_{z \to z_0} (z-z_0) f(z) $ must be nonzero and then it equals the residue. Is there other ways to find the residue?