Definition: $f$ has an isolated singularity about $z=0$ if there exists a punctured disc about $z=0$ such that it is holomorphic in that punctured disc.
I read about the characterisation of a pole in terms of the following
$f$ has a pole order $n$ about $z=0$ whenever there exists positive constants $C_1, C_2$ and $\delta>0$ such that $C_1|z|^{-n}<|f(z)|<C_2|z|^{-n}$ for all $z\in B(0, \delta)-\{0\}$.
Now, it is easy to see that $f$ has a singularity at $z=0$ but how can we see that this singularity is actually isolated?
I wanted to know the singularity is isolated because there was a lemma in my textbook saying
Suppose $f$ has an isolated singularity at $z_0$ then $f$ has a pole there if and only if $f(z)\to\infty$ as $z\to z_0$.
Hence I wanted to relate two things together.
However, in order to show it is isolated, we need to show it is holomorphic on some punctured disc about $z=0$. How do we proceed with only knowing that it is bounded in the above fashion on $B(0, \delta)-\{0\}?$
Many thanks in advance!