[Math] Removable singularity of $f(z)=\dfrac{\sin^2 z}{z}$

Question

I am having trouble understanding when a function might have a removable singularity over a pole.

For example:
$$f(z)=\frac{\sin^2 z}{z}$$

I believe the pole is at $z=0$. However, if we take the taylor expansion of $f(z)$ apparently the pole vanishes. I do not understand how and where does the pole vanish that it becomes a removable singularity.

Answer 1

$$\lim_{z \to 0}(z-0)\cdot f(z) =0$$ hence $0$ is removable singularity

this link may provide more clarity http://en.wikipedia.org/wiki/Removable_singularity