Existence of a limit implies that a function can be harmonic extended

[MathLearner123]

$\textbf{Theorem}$
If $u: \mathbb{D'} = \mathbb{D} \setminus \{0\} \to \mathbb{R}$ is harmonic and bounded, then $u$ extends to a function harmonic in $\mathbb{D}$.

In the next proof $\Pi^+$ is the upper half-plane.


$\textbf{Proof}$: Define a function $U: \Pi^{+} \rightarrow \mathbb{R}$ by $U(z)=u\left(e^{2 \pi i z}\right)$. Then $U$ is harmonic, being the composition of a harmonic function and a holomorphic function. Since $\Pi^{+}$ is simply connected, there exists $F \in H\left(\Pi^{+}\right)$ such that $U=\operatorname{R e}(F)$. Now, although $U(z+1)=U(z)$ we need not have $F(z+1)=F(z)$. But $\operatorname{R e}(F(z+1)-F(z))=U(z+1)-U(z)=0$ and so $F(z+1)-F(z)$ must be an imaginary constant: There exists $c \in \mathbb{R}$ such that $F(z+1)-F(z)= ic$ for all $z \in \Pi^{+}$. Choose a real number $\alpha \neq 0$ so that $c \alpha$ is a multiple of $2 \pi$ (you can take $\alpha=2 \pi / c$ unless $c=0$ ). Let
$$
E(z)=e^{\alpha F(z)}
$$
It follows that $E(z+1)=E(z)$, and so there exists $f \in H\left(\mathbb{D}^{\prime}\right)$ with
$$
E(z)=f\left(e^{2 \pi i z}\right) .
$$
Now the fact that $u$ is bounded shows that $f$ is bounded, so that $f$ has a removable singularity at the origin. The fact that $u$ is bounded also shows that $f$ is bounded away from $0$ , so in particular $f(0) \neq 0$. Hence $u$ has a limit at $0$ , since $u=\log (|f|) / \alpha$. (Since $f\left(e^{2 \pi i z}\right)=e^{\alpha F(z)}$ it follows that
$$
\left.\log \left(\left|f\left(e^{2 \pi i z}\right)\right|\right)=\alpha \operatorname{R e}(F(z))=\alpha U(z)=\alpha u\left(e^{2 \pi i z}\right) .\right)
$$

$\textbf{Question}$
Why the fact that $u$ has limit at $0$ implies that $u$ can be extended to a harmonic function $u : \mathbb{D} \to \mathbb{R}$?

 

[Office_Shredder]

It's an annoying quirk that you need double # instead of single $ to make in line latex, if you could edit your post to find and replace all instances it would be easier to read