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

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.

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an isolated singularity is removable if the limit exists at that point. We can then "fill in" that point and the singuarity is removed. –  anegligibleperson Nov 28 '12 at 16:38

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

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

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Directly by arithmetic of limits:

$$\frac{\sin^2z}{z}=\frac{\sin z}{z}\cdot\sin z\xrightarrow [z\to 0]{} 1\cdot 0 = 0$$

which means, just as with real analysis, that the discontinuity (singularity) is removable.

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