Let $f:D\rightarrow\mathbb{C}, D\subseteq \mathbb{C}$ and $p$ an accumulationpoint of $D$($\iff$ for every neighbourhood of $p$ there exists at least one point different from $p$). Whether or not $p$ is in $D$, we can define the function, $\hat{f}_{p,a}:D\cup\{p\},a\in\mathbb{C}$ with $\hat{f}_{p,a}(z) = f(z)$, if $p\neq z \in D$ and $\hat{f}_{p,a}(z)=a$, if $z =p$.

One says that $f$ in $p$ converges against $a$ if $\hat{f}_{p,a}$ is continuous in $p$. i.e.:

$$\exists_{a\in\mathbb{C}}\forall_{\epsilon>0}\exists_{\delta>0}\forall_{p\neq z \in D}|z-p|<\delta\Rightarrow |f(z)-a|<\epsilon\tag{*}$$


  • In $(*)$, why $\forall_{p\neq z \in D}$?

  • What if $p=\pm\infty$. What would be considered a neighbourhood of $\pm\infty$? The definition for a neighbourhood of a point in $\mathbb{C}$ is $D_\epsilon(p):=\{z\in\mathbb{C}:|z-p|<\epsilon\}$

  • There are situations where $\lim_{z\rightarrow\pm \infty}f(z)=\pm\infty$ or more generally $\lim_{z\rightarrow p} f(z)=\pm\infty$. What would that mean in this context?

  • $\begingroup$ Last row , do you mean $z\to p$ , instead? $\endgroup$ – dmtri Jan 11 at 18:07
  • $\begingroup$ Yes you are Right - to point out that it is an accumulationpoint $\endgroup$ – RM777 Jan 11 at 18:10
  • $\begingroup$ By definition, in the limit of a function in $p$, we do not care about the value of the $f(p) $. $\endgroup$ – dmtri Jan 11 at 18:26
  • $\begingroup$ For the last two points I have an explaination now, the idea is that infinity is a special case.: $\infty$ accumulationpoint of D $\iff \forall z\in \mathbb{C}\exists r \in D : z\in D_r(0)$, if $D\subseteq \mathbb{R}, +\infty$ is accpoint.$\iff \forall r\in \mathbb{R} \exists_{a\in D}: a\in (r,+\infty) $analogously if -$\infty$ is a accumulationpoint. So for the General case: $\lim_{z\rightarrow p} f(z)= +\infty \iff \forall_{c\in\mathbb{R}}\exists_{\delta\in\mathbb{R}} \forall_{z\in\mathbb{R}}:|z-p|<\delta \Rightarrow f(z) > c$ $\endgroup$ – RM777 Jan 11 at 18:28
  • $\begingroup$ For the Special case $lim_{z\rightarrow \infty}f(z)=+\infty \iff\forall_{ r \in \mathbb{R}}\exists_{a\in\mathbb{R}}\forall_{z\in D}:z\in(a,+\infty)\Rightarrow f(z)>r$ $\endgroup$ – RM777 Jan 11 at 18:37

(1) The definition of continuity usually says $0<|z-p|<\delta$ but one could instead say $|z-p|<\delta$ with $z\neq p$.

(2) In the complex setting $-\infty$ is usually undefined. A neighborhood of $+\infty$ is an open set containing all $|z|>c$ for some $c$.

(3) If for all c there is a neighborhood U of p with $|f(z)|>c$ in U then $f(z)\to\infty$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.