Let $f$ be meromorphic in $D=B(0,1)\setminus{0}$, and 0 is a limit point of poles of $f$. Show that for arbitrary $A\in\mathbb C$ + {$\infty\}$, there exists a sequence $\{z_n\}\in$ $D$, s.t. $\lim \limits_{n \to\infty}f(z_n)=A.$
Obviously there are 2 cases. Case 1: $ A=\infty.$ f(z) unbounded in arbitrarily small neighbourhood of 0. Then we can choose $f(z_1)>1$, $f(z_2)>2$, ... where $z_n\in B(0, |z_n-1|/2)\setminus{0}$. So OK. Case 2: $ A\in \mathbb C$. If $f(z)-A$ has zeros in arbitrarily small neighborhood of 0, then also Ok. Else, let $g = 1/(f-A)$. Then I get stuck.
Any hints? Thanks.
Otherwise $\frac{1}{f(z)-A}$ is meromorphic on $0 < |z| < 1$ and has a removable singularity at $z=0$ which means it is holomorphic on $|z| < \epsilon$ and meromorphic on $|z| < 1$, so that $f(z)$ is meromorphic too, contradicting it has infinitely many poles on $|z| \le 1/2$