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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'.

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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?

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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 $z$.

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