$\frac{\sin(z)}{z}$ Bounded on $\mathbb{C}-\{0\}$? Is $\frac{\sin(z)}{z}$ bounded on $\mathbb{C}-\{0\}$?

Based on real analysis intuition it is, and even though complex $\sin$ is not bounded the $\frac{1}{z}$ makes me think it is.
Notice that it does have a limit at $z=0$. This means it has a removable singularity there. 

If it were bounded on $\mathbb{C}\setminus\{0\}$ it would be bounded everywhere. This is because it is also bounded near $z=0$.

Since it is holomorphic everywhere (notice at $z=0$, since it has a limit, it has a removable singularity), then by Liouville's theorem it would be constant. 

But it is not, because if it were then $\sin(z)=Cz$, for some constant $C$. But you know that $\sin$ vanishes in ($z=\pi k, k\in\mathbb{Z}$) more places than $Cz$ so they can't be equal.
No, it is not bounded. Since this function is analytic in its domain. if it is bounded then it would be constant by Liouvill's Theorem but function is no constant.