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in Edward B. Saff, Arthur David Snider Fundamentals of complex analysis, with applications 2003, it has the following claim

Suppose we are given a function $f$ that is analytic and nonzero at each point of a simple closed cantour $C$ and is meromorphic inside $C$. Under these conditions it can be shown that $f$ has at most a finite number of poles inside $C$. The proof of this depends on two facts: first. the only singularities of $f$ are isolated singularities(poles), and, second, that every infinite sequence of points inside $C$ has a subsequence that converges to some point on or inside $C$. Hence if $f$ had an infinite numbers of poles inside $C$, some subsequence of them would converge to a point that must be a singularity, but not an isolated singularity of $f$.

I am struggling to understand why the bolded part is correct, why would a sequence of poles converge to an essential singularity?

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If $f$ had infinitely many poles inside a bounded set, the sequence of poles would have a convergent subsequence which would converge to a non-isolated singularity of $f$.

It would not be an essential singularity which are isolated singularities ( $\sin(1/z)$ has an essential singularity at $z=0$ but $1/\sin(1/z)$ doesn't)

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If the point (call it $\ p\ $) were a pole of $\ f\ $, or $\ f\ $ were analytic at the point, then $\ f\ $ would have either a Laurent or power series expansion in a disc $\ D\ $ of positive radius centred on $\ p\ $, and would therefore be analytic in the punctured disc $\ D\setminus \{p\}\ $. This would contradict the existence of a sequence of poles of $\ f\ $ converging to $\ p\ $.

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