4

A pole or removable or even essential singularity must be isolated a priori. But still we can try to talk about the limit of the function at the point even on a disk removing some (countable amount of) points.

A well-know example of non-isolated singularity will be $z=0$ for $$f(z)=\frac{1}{\sin(\frac{1}{z})}.$$

But $\lim_{z\to0}f(z)$ does not exist. My question is can there be a function with non-isolated singularity at $0$ with $\lim_{z\to0}|f(z)|=\infty$ or even $\lim_{z\to0}f(z)=c$?

CO2
  • 1,373
  • 1
    It seems you are calling "singularity" a point $a$ such that $f$ is meromorphic on $0 < |z-a| < \epsilon$. In that case or $f$ is analytic on $0 < |z-a| < \epsilon$ and one of the limit exists iff $f $ is meromorphic at $a$, or $f$ has infinitely many poles around $z=a$ and the limits don't exist (assume $\lim_{z\to a}|f(z)|=\infty$ and look at $1/f$) – reuns Nov 25 '18 at 22:33

1 Answers1

1

It's not entirely clear to me what you mean by "singularity". But the way in which you are using the term seems to indicate that you mean a point which is not in the domain of $f$ but which is in the closure of that domain. So, take $f$ to be the restriction of the identity function $z \mapsto z$, where the restricted domain is $D - \left(\{0\} \cup \{2^{-n} \mid n = 1,2,3,...\}\right)$. The singularity at $0$ is not isolated, because it is the limit of the singularities at $2^{-n}$. But $\lim_{z \to 0} f(z)=0$.

Lee Mosher
  • 120,280