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Classify the singularity points of $f(z)={e^{1\over z^2}\over z-1}$.

Obviously $z=1$ is a simple pole. Considering $z=0$, I know that $0$ is a pole iff $\lim_{z\to 0}f(z)=\infty$, and this is exactly the situation. But the solution is $z=0$ is essential singulaiar point (iff $\nexists\lim_{z\to0}f(z)$ and it's not $\infty$ either). What do I miss? Thanks

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  • $\begingroup$ Another classification for a pole is that the negative power part of the Laurent series has finitely many terms. $\endgroup$ – Cameron Williams May 23 at 15:26
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It is not true that $\lim_{z\to0}f(z)=\infty$. Take $z_n=\frac in$. Then $\lim_{n\to\infty}z_n=0=\lim_{n\to\infty}f(z_n)$.

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