# Singularity classification of $f(z)={e^{1\over z^2}\over z-1}$

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

• Another classification for a pole is that the negative power part of the Laurent series has finitely many terms. – Cameron Williams May 23 at 15:26

It is not true that $$\lim_{z\to0}f(z)=\infty$$. Take $$z_n=\frac in$$. Then $$\lim_{n\to\infty}z_n=0=\lim_{n\to\infty}f(z_n)$$.