# Prove that $f$ can have at most finitely many zeros.

Let $$f : \Bbb C \longrightarrow \Bbb C$$ be an entire function such that $$\lim\limits_{z \to 0} \left \lvert f \left (\dfrac {1} {z} \right ) \right \rvert = \infty.$$ Then show that $$f$$ can have at most finitely many zeros.

What I can observe is that if $$f$$ has infinitely many zeros and zero set of $$f$$ is bounded then by Bolzano-Weierstrass theorem it has a limit point. Then by identity theorem it follows that $$f \equiv 0$$ on $$\Bbb C.$$ But then the above limit becomes $$0,$$ a contradiction. Hence if the zero set of $$f$$ has infinite cardinality then the set has to be unbounded. Now how do I proceed? Any help in this regard will be appreciated.

There exists $$\delta >0$$ such that $$|z| <\delta$$ implies $$|f(\frac 1 z)| >1$$. In particular $$|z| >\frac 1 {\delta}$$ implies $$f(z) \neq 0$$. So all the zeros are in $$\{z: |z| \leq \frac1 {\delta}\}$$. If there are infinitely many zeros in this compact set there would be a limit point for the zeros which leads to a contradiction. .