Can $ f(z)$ be analytic in a deleted neighborgood of $z_0$ under this condition?

Question

It comes from complex analysis.

Let $f(z)$ be a complex function.

Can $ f(z)$ be analytic in a deleted neighborgood of $z_0$ when

$\lim (z-z_0)^nf(z)$ as $z \rightarrow z_0$ does not exist for any interger $n$?

I guess when $f(z)$ has an essential singularity at $z_0$,

it can be analytic in a deleted neighborhood of $z_0$ and satisfying above condition.

Is it right? And also I'd like to know whether there is another case satisfying the condition or not.

Answer 1

You are right and there are no other cases: isolated singularities are either removable, poles or essential singularities, and if you have a pole of order $m$, then the above limit exists for $n=m$.

Answer 2

Yes, you are right.

The function $f$ has a singularity at $z_0$. If it was a removable singularity, then $\lim_{z\to z_0}(z-z_0)^nf(z)$ would be $0$ for each natural $n$. If it was a pole of order $m$, then the limit would exist if and only if $n\geqslant m$. So, yes, it is an essential singularity.