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Do you know any book that covers the "infinite limits and limits at infinity" notion for a real-valued function ?

I ask because I find the theorem $4.33$ from Rudin Principles of Mathematical Analysis on that topic quite unprecise.

Here are three points that make think the theorem as presented by Rudin is unprecise:

  1. He doesn't mention the case where $x$ is a limit point of $E$, not belonging to $E$.
  2. He wants $V\cap E$ to be nonempty. For me, it would be $V\cap E \backslash \{x\}$ instead.
  3. He doesn't say which set $x$ belongs to.

For all these reasons I am looking for an alternative text, which covers the topic.

PS: Here is the theorem from Rudin:

Let $f$ be a real function defined on $E \subset R$. We say that $f(t) \rightarrow A$ as $t \rightarrow x$ where $A$ and $x$ are in the extended real number system, if for every neighborhood $U$ of $A$ there is a neighborhood $V$ of $x$ such that $V \cap E$ is not empty, and such that $f(t) \in U$ for all $t \in V \cap E$, $t \neq x$.

niobium
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    I don't really see any problem with his definition. Regarding (3), he writes explicitly that $x$ belongs to the extended reals. But there's no assumption that $x \in E$, so (1) and (2) are also taken care of. – Hans Lundmark Apr 13 '23 at 20:10
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    @HansLundmark - Point (3) is indeed false, but it does not handle points (1) or (2). If $x$ is an isolated point in $E$, then for small enough $V, V\cap E = {x} \ne \emptyset$, and the condition on $f$ holds vacuously. But even if (2) is corrected to $V\cap E \setminus {x}$ is not empty, we would have for $$f : (0,1) \cup {2}\to\Bbb R : x\mapsto \begin{cases}0&0<x<1\1&x=2\end{cases}$$ that $\lim_{x\to 2}f(x) = 0$, which violates the concept that $x$ is approaching $2$ in the limit. $x$ needs to be restricted to accumulation points of $E$ in $\overline{\Bbb R}$. – Paul Sinclair Apr 14 '23 at 16:32
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    Just to make it a little clearer why that last limit is not desirable: We could instead consider the function $$f : (0,1) \cup {2}\cup(3,4)\to\Bbb R : x\mapsto \begin{cases}0&0<x<1\1&x=2\2&3<x<4\end{cases}$$ for which $$\lim_{x\to 2}f(x) = 0\\lim_{x\to2}f(x) = 2$$ are both true by this definition. However, both (1) and (2) are completely handled by just requiring $x$ be an accumulation point of $E$ (every neighborhood of $x$ contains points of $E$ other than $x$). – Paul Sinclair Apr 14 '23 at 16:48
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    Oh, I see. I think you are right! If $x$ is the only point in $V \cap E$, then the “for all” condition indeed becomes vacuously true. So contrary to what he says, Def. 4.33 does not agree with Def. 4.1 when $A$ and $x$ are real. But can't it all be fixed (without further changes) simply by adding the requirement that $x$ be a limit point of $E$, like in Def. 4.1? Then $x$ is never the only point in $V \cap E$. – Hans Lundmark Apr 14 '23 at 17:39
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    I found some previous question and/or answers, where others have noticed precisely this error in Def. 4.33: https://math.stackexchange.com/questions/1365208, https://math.stackexchange.com/questions/3581828, https://math.stackexchange.com/a/3873565/1242, https://math.stackexchange.com/questions/2102463. – Hans Lundmark Apr 14 '23 at 17:51
  • "But even if (2) is corrected to $V∩E∖ {x}$ is not empty, we would have for..." If $x$ is an isolated point, $V∩E∖ {x}$ is empty; hence, the limit of $f$ as $t$ goes to $x$ is not defined. – niobium Apr 15 '23 at 09:50
  • According, to this document (see «page 98, line 7») changing $V \cap E$ nonempty to $V \cap E \backslash {x}$ nonempty makes the definition valid – niobium Apr 15 '23 at 09:55

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