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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.:

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

Questions:

  • 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?

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  • $\begingroup$ 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
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(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 $f(z)\to\infty$.

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