TITLE: Why $z=0$ is a removable singularity of $\frac{iz-e^{iz}+1-\frac{z^2}{2}}{z^3}$?
QUESTION [0 upvotes]: Why $z=0$ is a removable singularity of $$\frac{iz-e^{iz}+1-\frac{z^2}{2}}{z^3}$$
I tried to show $\lim_{z\to 0}z\frac{iz-e^{iz}+1-\frac{z^2}{2}}{z^3}=\lim_{z\to 0}\frac{iz-e^{iz}+1-\frac{z^2}{2}}{z^2}=0$, however, I cannot see why this limit is $0$? 
Could anyone kindly help? Thank you!

REPLY [0 votes]: $e^{iz} = 1+iz-{1 \over 2!} z^2 - i {1 \over 3!} z^3 + \sum_{k=4}^\infty {1 \over k!} (iz)^k$.
Let $g(z) = - i {1 \over 3!} + \sum_{k=4}^\infty {1 \over k!} (iz)^{k-3}$
and note that $g$ is entire.
Note that ${e^{iz} - 1-iz+{1 \over 2!} z^2 \over z^3} = g(z) $ for $z \neq 0$.
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