$f$ holomorphic on $D\setminus \{0\}$ and takes no values in $(-\infty,0],$ then $0$ removable > If $f$ is holomorphic on $D\setminus \\{0\\}$ and takes no values in $(-\infty,0]$ then $0$ is a removable singularity. I thought to prove this by elimination, but I can't really tell anything about the behavior of $f$ around $0$. How can one translate the information about the definition of $f$ in semi-open interval $(-\infty,0]$.

$G := \Bbb C \setminus (-\infty, 0]$ can be mapped conformally onto the unit disk. That is generally true for all _simply-connected_ domains due to the Riemann mapping theorem. In this particular case the mapping can be described explicitly as $$ \varphi(z) = \frac{\sqrt z - 1}{\sqrt z + 1} $$ where $\sqrt z$ is the holomorphic branch mapping $G$ onto the right halfplane.

Then $g := \varphi \circ f$ is holomorphic in $D\setminus \\{0\\}$ with values in the unit disk, i.e. $g$ is _bounded_. It follows that $g$ has a removable singularity at $z= 0$, and then the same holds for $f$.

One could also use the "Great Picard Theorem" which states that a holomorphic function takes on all possible complex values, with at most a single exception, infinitely often in a punctured neighbourhood of an essential singularity. But that is an "advanced" result in complex analysis.