what is the residue of $\sin(z^2)/z^7$ what is the residue of $$\frac{\sin(z^2)}{z^7} \text{ at } z=0.$$ I'm finding trouble at the limit part at the end.
I am using  the formula:
$$\operatorname{res}(f,z_0)=\lim_{z\to z_0}\frac{1}{(n-1)!} \frac{d^{n-1}}{dz^{n-1}} ((z-z_0)^n f(z))$$

Recall that
$$\sin(z)= \sum_{n=0}^\infty \frac{(-1)^{n} z^{2n+1}}{(2n+1)!}=z-\frac{z^3}{3!}+\frac{z^5}{5!}+\cdots$$
Which also means that
$$\sin(z^2)=z^2-\frac{z^6}{3!}+\frac{z^{10}}{5!}+\cdots$$
And thus
\begin{align}
\frac{1}{z^7}\sin{(z^2)}&=\frac{1}{z^7}\left( z^2-\frac{z^6}{3!}+\frac{z^{10}}{5!}+\cdots\right)\\
&=\frac{1}{z^5}-\frac{1}{3!}\frac{1}{z}+\cdots
\end{align}
And so in the Laurent series expansion, the coefficient of the $1/z$ term is the residue at $z=0$. And thus
$$\mathrm{Res}_{z=0}\left( \frac{\sin(z^2)}{z^7}\right)=-\frac{1}{3!}=-\frac{1}{6}$$
