# Questions on limits of functions

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?

• Last row , do you mean $z\to p$ , instead? – dmtri Jan 11 at 18:07
• Yes you are Right - to point out that it is an accumulationpoint – RM777 Jan 11 at 18:10
• By definition, in the limit of a function in $p$, we do not care about the value of the $f(p)$. – dmtri Jan 11 at 18:26
• 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: $\lim_{z\rightarrow p} f(z)= +\infty \iff \forall_{c\in\mathbb{R}}\exists_{\delta\in\mathbb{R}} \forall_{z\in\mathbb{R}}:|z-p|<\delta \Rightarrow f(z) > c$ – RM777 Jan 11 at 18:28
• For the Special case $lim_{z\rightarrow \infty}f(z)=+\infty \iff\forall_{ r \in \mathbb{R}}\exists_{a\in\mathbb{R}}\forall_{z\in D}:z\in(a,+\infty)\Rightarrow f(z)>r$ – 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 $$f(z)\to\infty$$.