A tough question on meromorphic functions- Conway This is a question I came across in JB Conways book. 



Let $f$ be meromorphic function in the punctured disk $D_r(z_0)$ \
  {$z_0$}. Suppose there is a sequence {$p_n$} of poles of $f$ in
  $D_r(z_0)$ \ {$z_0$} such that $\lim_{n\rightarrow \infty}p_n=z_0$.
  Show that for each $w\in \mathbb{C}$ there is a sequence {$z_n$} in
  $D_r(z_0)$ \ {$z_0$} such that $\lim_{n\rightarrow \infty}z_n=z_0$ and
  $\lim_{n\rightarrow \infty}f(z_n)=w$

I find this question very challenging and I am sorry for not showing any effort as I could not figure out anything. Hope someone can give me a hint so that I can work on it to get the answer. Thanks !
 A: I am not sure if this follows directly from the Casorati-Weierstraß theorem, but
I think that it can be proved in a similar way.
The statement

For each $w\in \mathbb{C}$ there is a sequence {$z_n$} in
  $D_r(z_0)$ \ {$z_0$} such that $\lim_{n\rightarrow \infty}z_n=z_0$ and
  $\lim_{n\rightarrow \infty}f(z_n)=w$

can be equivalently formulated as

For each $w\in \mathbb{C}$ and for each $\varepsilon > 0$ there is a 
  $z \in D_r(z_0) \setminus {z_0}$ with $\left| z-z_0 \right| < \varepsilon $
  and $\left| f(z) - w_0 \right| < \varepsilon $.

Let us assume that this is false. Then we have

There is a $w_0 \in \mathbb C$ and an $\varepsilon_0 > 0$ such 
  that $\left| f(z) - w_0 \right| \ge \varepsilon_0 $ for all
  $z \in D_r(z_0)  \setminus {z_0}$ with $\left | z-z_0 \right| < \varepsilon_0 $.

Then the function $g := 1/(f - w_0)$ is holomorphic and bounded in a punctured
disk with center $z_0$ and therefore has a removable singularity at $z_0$.
Therefore $f = w_0 + 1/g$ has a meromorphic extension to $U_{\varepsilon_0}(z_0)$. This contradicts
the assumption that the poles of $f$ accumulate at $z_0$.
