# Understanding Proposition $3.14$ in Ullrich's Complex Made Simple

In Ullrich's Complex Made Simple, Proposition $$3.14$$ part $$ii$$ states

If $$f \in H(D'(z,r))$$ (holomorphic in the punctured disk) has an essential singularity at $$z$$ if and only if $$f(D'(z,\rho))$$ is dense in $$\mathbb C$$ for all $$\rho \in (0,r)$$.

I understand the if part which refers as Casorati–Weierstrass theorem. But in the proof to only if part I am having trouble to fill a gap. The proof proceeds by contradiction.

If $$f$$ has a pole or removable singularity at $$z$$ then $$f(w)$$ has (finite or infinite) limit as $$w \to z$$. [This part is OK. The limit would be $$0$$ if $$z$$ is removable singularity and infinite if $$z$$ is a pole.] Then it claims the range $$f(D'(z, \rho))$$ would be obviously not dense. I am having trouble to understand the 'obvious part'.

Suppose that $$\lim_{w\to z}f(w)=l(\in\mathbb{C})$$. Take $$\varepsilon>0$$. There is the a $$\delta>0$$ such that $$\lvert w-z\rvert<\delta\wedge w\neq z\implies\bigl\lvert f(w)-l\bigr\rvert<1$$. That is,$$f\bigl(D'(z,\delta)\bigr)\subset D(l,1).$$But then$$\overline{f\bigl(D'(z,\delta)\bigr)}\subset\overline{D(l,1)}\varsubsetneq\mathbb C.$$Can we deal with case in which $$f$$ has a pole on $$z$$ now?
If the limit is $$L$$ (finite), then $$f(w)$$ is within distance $$1$$ of $$L$$ if $$w$$ is sufficiently close to $$z$$. The disk of radius $$1$$ around $$L$$ is not dense in $$\mathbb C$$. Similarly, if the limit is $$\infty$$, then $$|f(w)| > 1$$ if $$w$$ is sufficiently close to $$z$$.