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Under what conditions do we have the following, when $g$ is not defined at $q$:

$\lim_{x \to p} f(x) = q \land \lim_{y \to q} g(y) = Q \implies \lim_{x \to p} g(f(x)) = Q$

I know it holds when $g$ is continuous at $q$, but I wonder if that could be relaxed.

S11n
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    the weakest I would say is that $f$ is not locally constant in $p$, i.e. that for all $\varepsilon >0$, there is $x\in (p-\varepsilon ,p+\varepsilon )$ s.t. $f(x)\neq q$. – Surb Jun 01 '22 at 09:26
  • @Surb Thanks! I'll just clarify the question so it's clear it asks for the case when $g$ is not continuous at $q$. – S11n Jun 01 '22 at 09:30
  • The answer is the same... it always hold as far as $f$ is not locally constant (and $g$ is defined in a neigborhood of $g$ in the sense that there is $\delta >0$ s.t. $g$ is well defined on $(q-\delta ,q+\delta )\setminus {q}$). – Surb Jun 01 '22 at 09:39
  • @Surb Yes, you understood what I meant, and that's why I decided to clarify the question (because otherwise your answer would not be correct, and I wanted to prevent that from being pointed out). – S11n Jun 01 '22 at 09:42
  • @Surb Yes, if $g$ were defined but not continuous at $q$, your claim may not exactly hold without tweaking. Take for example $f(x) = \max{0,x}$ - then $\lim_{x\to0} g(f(x))$ may not exist. But I see what you're getting at. – Milten Jun 01 '22 at 09:45
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    @S11n But overall, what's happening is that $f$ and $g$ are basically continuous at the points, since they have the limits. Or precisely, they may only have removable discontinuities, which means we most often might just as well look at their "continuified" versions. – Milten Jun 01 '22 at 09:48
  • @Milten That's exactly what I was looking for! Could you please provide a full answer (with a proof)? – S11n Jun 01 '22 at 09:50
  • Related: https://math.stackexchange.com/questions/167926/formal-basis-for-variable-substitution-in-limits – Hans Lundmark Jun 01 '22 at 10:40
  • Another related answer: https://math.stackexchange.com/a/4463826/942378 – S11n Jun 04 '22 at 15:54

1 Answers1

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The fact that $f$ and $g$ have limits at the points means that we can define two functions $\tilde f$ and $\tilde g$ that are continuous at $p$ and $q$ respectively, and otherwise equal to $f$ and $g$. Now assume that

  • $g$ is defined at some value in the image $f(\omega'_p)$ of all punctured neighborhoods $\omega'_p$ of $p$,
  • if $g$ is defined and discontinuous at $q$, then $f(x)\ne q$ for all $x$ close enough to $p$ (i.e. in some punctured neighborhood around $p$).

The first condition means that $g\circ f$ takes values around $p$ (and I included it since you allowed for partially defined functions). The second condition ensures that $g\circ f$ is equal to $\tilde g\circ\tilde f$ close to $p$. But then $$ \lim_{x\to p}g(f(x)) = \lim_{x\to p}\tilde g(\tilde f(x)) = \tilde g(q) = Q. $$

Conversely, note that the first condition above is necessary to even define the limit of $g\circ f$ at $p$.

For the second condition, assume that the implication holds, that condition 1 holds, that $g$ is defined at $q$, and that $(x_n)_n$ is a sequence converging to $p$ with $x_n\ne p$ and $f(x_n)=q$ for all $n$. Then $$ g(q) = \lim_{n\to\infty} g(f(x_n)) = \lim_{x\to p}g(f(x)) = Q,$$ showing that $g$ is continuous at $q$. This shows that the second condition is necessary.

In conclusion, your implication is equivalent to the two conditions.

Milten
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    See also https://en.wikipedia.org/wiki/Limit_of_a_function#Limits_of_compositions_of_functions – Milten Jun 01 '22 at 10:54
  • Thanks! I suppose you are trying to show that the implication in the question title holds if and only if a) $g$ is continuous at $q$ or b) $f(x) \ne q$ for all $x$ close enough to $p$. The if part is clear, and you did it in the first part of your answer, by reducing the problem to continuous functions $\tilde{f}$ and $\tilde{g}$. However, I suppose in the second part of our answer (starting with "Conversely") was aiming to prove the only if direction, but it assumes that $g$ is defined at $q$, which is against the hypothesis in my question. Is it easy for you to fix the proof? – S11n Jun 04 '22 at 15:13
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    @S11n Condition 2 is equivalent to: $g$ is undefined at $q$ or $g$ is continuous at $q$ or $f\ne q$ around $p$. If you assume that $g$ is undefined at $q$, then it’s meaningless to ask about continuity (it can’t be continuous if it isn’t even defined). Furthermore, the implication is almost automatic, and is equivalent to just condition 1 in that case. If $g$ is only undefined at $q$, then it’s equivalent to $f$ not being identically equal to $q$ around $p$. – Milten Jun 04 '22 at 16:13
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    @S11n As for the proof, I don’t think anything needs fixing. We want to show $p\vee q\vee r$, and we do it by showing $\lnot p \land \lnot r\implies q$. Basically a contrapositive. – Milten Jun 04 '22 at 16:18
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    @S11n Let me just reemphasise: Your main question as it stands is odd. You assume $g$ is undefined at $q$, but then also talk about continuity at $q$ - those things are mutually exclusive. In a more usual setting, $g$ is defined everywhere (at least around $q$), and the conditions become simpler and less technical. You can see the precise conditions in that case on the wikipedia link. – Milten Jun 04 '22 at 21:25
  • Yes, I added that clarification so that it's clear that I knew about the case when $g$ is continuous, and I was looking for something else. For me your answer is enough, but for posterity it would definitely be more readable to explicitly name condition 1 and condition 2 (and also it is not clear which implication you refer to in your answer), because I thought you wanted to show the only if direction, but it seems like I was wrong, and you just wanted to show that either of the two conditions stated on wikipedia are enough for the implication in the question's title to hold. – S11n Jun 06 '22 at 08:08
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    @S11n No, you were right the first time. I prove that the implication you give in your question holds if and only if both of my two conditions hold. First I prove “if”, and then “only if” starting at “conversely”. – Milten Jun 06 '22 at 09:43