Given $f$ an analytic function in open $D \subset \mathbb C$, $z_0$ is an isolated singularity defined as $B(z_0;r)\backslash\{z_0\} \subset D$, then know that $f$ can be written as an expansion of laurent series $$f(z)=\sum_{k=1}^{\infty}b_{-k}(z-z_0)^{-k}+\sum_{k=0}^{\infty}a_k(z-z_0)^k$$ in $B(z_0;r)\backslash\{z_0\}$, If there are infinite $b_{-k}\neq 0$, prove that $$\nexists\lim_{z\to z_0}(z-z_0)^mf(z)$$ for all nautural number $m$.
My attempt: Assume $\exists m: \lim (z-z_0)^m f(z)=\lim (z-z_0)^m (\sum_{k=1}^{\infty}b_{-k}(z-z_0)^{-k}+\sum_{k=0}^{\infty}a_k(z-z_0)^k)=c$, then $\lim(z-z_0)^m \sum_{k=0}^{\infty}a_k(z-z_0)^k=\lim(z-z_0)^m\lim \sum_{k=0}^{\infty}a_k(z-z_0)^k=0\cdot0=0$ as $\sum_{k=0}^{\infty}a_k(z-z_0)^k$ continuous in $B(z_0;r)$, then $\lim (z-z_0)^m \sum_{k=1}^{\infty}b_{-k}(z-z_0)^{-k}=$ $$\lim_{z\to z_0} \sum_{k=m}^{\infty}b_{-k}(z-z_0)^{-k+m}=c$$
What can I do next?
Edit: In my comment it’s $a_{-k}=0$
Edit2: I follow these definitions for classifications of isolated singularities:
Removable singularity: No singular part of laurant series.
Pole of order m: singular part with finite $a_{-k}\neq 0$, $a_{-m}\neq 0$ is the last term $\neq 0$ and $a_{-k}=0$ $\forall k>m$.
Essential singularity: singular part with infinite many $a_{-k}\neq 0$.
and I try to use these definitions to imply other than using other definitions.
Best Answer
If $\lim (z-z_0)^m f(z)$ exists then $(z-z_0)^m f(z)$ has a removable singularity at $z_0$, so there is a power series expansion $(z-z_0)^m f(z)=\sum\limits_{n=0}^{\infty} a_n(z-z_0)^{n}$. This gives a Laurent series expansion $f(z)=\sum\limits_{n=0}^{\infty} a_n(z-z_0)^{n-m}=\sum\limits_{n=-m}^{\infty} a_{n+m}(z-z_0)^{n}$. (Recall that Laurent series expansion is unique). Hence, there are only finitely many terms in negative powers of $z-z_0$.