Compute the curve integral $\int_{|z|=3}\dfrac{dz}{\sin^2 z}$ [trouble with limit] My attempt is as follows;
The region $\Omega$ which is enclosed by the curve $\gamma = \{z \in \mathbb{C}:|z|=3\}$ contains one singularity at $z=0$, a pole of order 2. Therefore, we may use the residue theorem which yields
$$\int_{\gamma}\dfrac{dz}{\sin^2z}= 2\pi i\, \text{Res}_f(0)\,.$$
When computing the residue according to definition for pole of order $k=2$, I arrive at
\begin{align*}
\text{Res}_f(0) &= \dfrac{1}{(2-1)!}\lim_{z \rightarrow 0} \dfrac{d}{dz}\left[\dfrac{z^2}{\sin^2(z)}\right]\\[5pt]
&=\lim_{z \rightarrow 0} \dfrac{2z[\sin z-z\cos z]}{\sin^3z}.\\[4pt]
\end{align*}
According to the solution manual, this limit is 0. I am unable to conclude this myself, so I was wondering if anyone here could provide a hint?
Many thanks in advance.
 A: Hint.
$$
\begin{align}
\frac{1}{\sin^2(z)}
&=\frac{1}{z^2}\left(1-\frac{z^2}{6}+\dots\right)^{-2}\\
&=\frac{1}{z^2}\left(1+\frac{z^2}{3}+\dots\right)\\
&=0\cdot \frac1z+\frac{1}{z^2}+\frac13+\dots
\end{align}
$$
where you use the fact that $\sin z=z-\frac{z^3}{6}+\cdots$.
Alternatively, observe that $\sin^2 (z)=\sin^2 (-z)$. So if you take the opposite direction of the contour $\gamma$, the integral remains the same:
\begin{align}
\int_\gamma f(z)dz
=&-\int_{\gamma}f(-w)dw&\textrm{(change of variable $w=-z$)}\\
=&\int_{-\gamma}f(-w)dw&\textrm{(definition of "negative" contour)}\\
=&\int_{-\gamma}f(w)dw&\textrm{($f(z)=f(-z)$)}\\
=&-\int_{\gamma}f(z)dz
\end{align}
A: Set $z=3e^{it};\;t\in[0,2\pi]$ and $dz=3 i e^{i t}dt$
The integral becomes
$$\int_0^{2\pi}\frac{3 i e^{i t}\,dt}{\sin^2\left(3e^{it}\right) }=\left[-\cot \left(3 e^{i t}\right)\right]_0^{2\pi}=-\cot 3+\cot 3=0$$
A: $$\lim_{z \to 0} \dfrac{2z[\sin z-z\cos z]}{\sin^3z}
=\lim_{z \to 0} \dfrac{2z\left[\left(z-\dfrac{z^3}{3!}+O(z^5)\right)-z\left(1-\dfrac{z^2}{2!}+O(z^4)\right)\right]}{z^3-O(z^5)}=\lim_{z \to 0}\frac z3.$$
