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  1. For the definition of $\lim_{(x,y) \to (x_0,y_0)}g(x,y)=L$ below, is the following correct?

    • 1.1.

    Let $g$ be a real function defined on an open deleted neighbourhood of some point $(x_0,y_0) \in \mathbb R^2$.

    • 1.2. I understand the above translates to

    Let $G \subseteq \mathbb R^2$. Let $(x_0,y_0) \in \mathbb R^2$. Let $g: G \to \mathbb R$. Let $U$ be an open neighbourhood of $(x_0,y_0)$ with $U \ \setminus \{(x_0,y_0)\} \subseteq G$.

    • 1.3. Then $\lim_{(x,y) \to (x_0,y_0)}g(x,y)=L$ is defined as

    For all $\varepsilon > 0$, there exists $\delta > 0$ s.t. $|g(x,y)-L| < \varepsilon$ whenever $0 < \sqrt{(x-x_0)^2+(y-y_0)^2} < \delta$ and $(x,y) \in $ (either $G$ or $U$ or $U \ \setminus \ \{(x_0,y_0)\}$ - not sure? see next)

  2. As for $G$, $U$ or $U \ \setminus \ \{(x_0,y_0)\}$, does it make a difference?

    • 2.1. I think there's no difference between $U$ or $U \ \setminus \ \{(x_0,y_0)\}$ because we already have '$0 <$'.

    • 2.2. As for $G$ and $U$, not sure.

BCLC
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1 Answers1

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Yes you are right there's no difference between $U$ or $U \ \setminus \ \{(x_0,y_0)\}$ because we already have the strict inequality.

For the second issue, keeping the restriction to $U$ we should refer to it but we can also not consider the restriction to $U$ using that $0 < \sqrt{(x-x_0)^2+(y-y_0)^2} < \delta$ and $(x,y) \in G$.

user
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  • @JohnSmithKyon Yes in this case we are considering a restriction of the function around the limit point $f:U\subseteq G \to \mathbb R^2$ and this is fine. – user Oct 06 '21 at 02:23
  • Note: user is responding to a comment which said something like 'thanks user! does it really have to be $U$ instead of $G$? please consider helping in the complex case'. maybe i accidentally deleted the comment without posting an edited version. – BCLC Oct 06 '21 at 02:27
  • user, what's an example of a problem if we use $G$ instead of $U$ please? – BCLC Oct 06 '21 at 02:28
  • user are you sure it's $U$ instead of $G$? in the complex version it is $G$ i believe – BCLC Oct 06 '21 at 02:45
  • @JohnSmithKyon For example some point near $(x_0,y_0)$ such that $f$ blows out. – user Oct 06 '21 at 02:46
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    @JohnSmithKyon I'll take a look also to this! What is the reference for the version you are using here? – user Oct 06 '21 at 02:47
  • thanks! for the real limit: it's a combination of my class notes and stewart calculus. but i'm trying to make the definition to be a little more precise with the domains, where $z$ is, etc. for the complex limit: combination of class notes and brown churchill – BCLC Oct 06 '21 at 02:49
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    @JohnSmithKyon In this form the definition is fine. But I understand your doubt and we can also eliminate $U$ by the definition using that $0 < \sqrt{(x-x_0)^2+(y-y_0)^2} < \delta$ and $(x,y) \in G$. – user Oct 06 '21 at 02:57
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    @JohnSmithKyon I've update answer accordingly. – user Oct 06 '21 at 03:00
  • thanks user. anyway bottom line the definition is the same/equivalent for $(x,y) \in G$ vs $(x,y) \in U$ vs $(x,y) \in U \ \setminus \ (x_0,y_0)$? – BCLC Oct 06 '21 at 03:18
  • For the definition we don’t need to introduce $U$ ( not wrong but unnecessary), therefore $(x,y)\in G$ is fine. – user Oct 06 '21 at 03:31
  • user, in what way is $U$ unnecessary? we want an open set involved right? do you mean to assume $z_0$ is a limit point of $G$? do you mean to assume $G$ is open in $\mathbb C$? or what? Note that $G$ isn't a region or domain or whatever. It's just subset of $\mathbb C$ that happens to be the domain of a complex function. – BCLC Oct 06 '21 at 17:47
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    @JohnSmithKyon Because by the condition$ 0 < \sqrt{(x-x_0)^2+(y-y_0)^2} < \delta$ with $(x,y) \in G$ we are selecting points in a open ball around $(x_0,y_0)$. – user Oct 06 '21 at 19:57
  • user, i (sort of) asked my instructor about this earlier. does this make a difference for real or complex re the $U$ is unnecessary given $G$? – BCLC Oct 07 '21 at 12:26