Finding $\text{Res}(f,0)$ where $f(z)=\frac{1}{z^2\sin(z)}$ I am trying to determine the residue of $z=0$ where $f(z)=\frac{1}{z^2\sin(z)}$.
I have determined that $z=0$ is a pole of order $3$. Hence to compute the residue, I use $$\text{Res}(f,0)=\frac{1}{2}\lim_{z\to 0}\frac{\partial^2}{\partial z^2}\left(\frac{z}{\sin(z)}\right).$$
I am convinced I am doing something incorrect, as the calculations become very convoluted. I have computed that $$\frac{\partial^2}{\partial z^2}=\frac{z\sin^3(z)-2\sin^2(z)\cos(z)+2z\sin(z)\cos^2(z)}{\sin^4(z)},$$ but am unsure of how to further compute the residue, as I'm unable to solve this limit.
edit 
I have solved the problem via the repeated use of L'Hopitals rule, but this took the work of computing some ugly limits. Is there another way?
 A: $$\sin z=z-\dfrac{1}{3!}z^3+\dfrac{1}{5!}z^5-\dfrac{1}{7!}z^7+\cdots$$
then
\begin{align}
\dfrac{1}{z^2\sin z}
&= \dfrac{1}{z^3\left(1-\dfrac{1}{3!}z^2+\dfrac{1}{5!}z^4-\dfrac{1}{7!}z^6+\cdots\right)} \\
&= \dfrac{1}{z^3}\left(1+\dfrac{1}{6}z^2+(\dfrac{1}{36}-\dfrac{1}{120})z^4+\cdots\right) \\
&= \dfrac{1}{z^3}+\dfrac{1}{6z}+\dfrac{7z}{120}+\cdots
\end{align}
s0 $a_{-1}=\dfrac16$.
A: $\frac{\partial^2}{\partial z^2}(\frac z{\sin z})=\frac{\partial}{\partial z}\frac{\sin z-z\cos z}{\sin^2 z}=\frac{(\sin z-z\cos z)2\sin z\cos z+z\sin^3 z}{\sin^4 z}=\frac{2\cos z(\sin z-z\cos z)}{\sin^3 z}+\frac z{\sin z}=\frac{2\cos z}{\sin^2 z}-\frac{2z\cos^2 z}{\sin^3 z}+\frac z{\sin z}$.
Now for the middle term (as the other two are easier) , we use L'Hôpital's:  $\frac{2\cos^2 z-4 z\cos z\sin z}{3\sin^2 z\cos z}=\frac{2\cos z}{3\sin^2 z}-\frac{4z}{3\sin z}$ 
Now for the first term (as the other is easy), L'Hôpital's again:  $-\frac1{3\cos z}\to-\frac13$.
The limit of each term is now easy:
$-1-(-\frac13-\frac43)+1=\frac13$.
I get a limit of $\frac13$.
