Isolated singularity is removable iff $\lim\limits_{z\to z_0} (z-z_0)f(z)=0$ Could someone explain a step in the following proof?

Theorem
An isolated singularity $z_0$ of $f$ is removable if and only if $\lim\limits_{z\to z_0} (z-z_0)f(z)=0$.
Proof ($\Leftarrow$)
Take $z_0=0$ which implies $\lim_{z\to 0} zf(z) = 0$. Since:
$$f(z) = \color{red}{\sum_{n=0}^\infty a_nz^n} + \color{blue}{\sum_{n=1}^\infty \frac{b_n}{z^n}}$$
then $z\color{red}{A(z)}\to 0$ (the holomorhic part), which implies $z\color{blue}{B(z)}\to 0$ as $z\to 0$.
This means that $\displaystyle g(z) := B\left(\frac{1}{z}\right) = \sum_{n=1}^\infty b_nz^n$ is a entire function for which $|g(z)| \leqslant |z|$ $\color{green}{(1)}$ if $|z|$ is big enough. By Liouville's theorem $g(z)$ is itself a polynomial of degree at most 1 which implies $g(z) = b_1z$ etc

Question regarding step (1)
Why is $|g(z)|$ bounded by $|z|$? Couldn't $g(z)$ be something like $g(z) = z+z^2$?

Since $zB(z)\to 0$ as $z \to 0$, we have that $g(z)/z \to 0$ as $|z|\to \infty$. Hence, letting $\varepsilon  = 1$, there corresponds a positive number $k$ such that $|g(z)/z| \le 1$ for all $|z| > k$, i.e., $|g(z)| \le |z|$ for all $|z| > k$.
