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I know that, if $f$ is continuous on $\lim\limits_{x \to a} g(x)=L$ then $\lim\limits_{x \to a} f(g(x))= \lim\limits_{y\to L} f(y) =f \left( \lim\limits_{x \to a } g(x) \right)$ $\ldots (*)$.

I was wondering if can you replace continuity of $f$ with some other condition? So I found one condition, which I'll write down below.

Asuming that $\lim\limits_{x \to a} g(x)=L$ and for every $\delta\gt0$ there exist an $\eta\gt 0$ such that $g$ takes all values on $(L-\eta,L+\eta)$, except perhaps $L$ itself, on $(a-\delta,a+\delta)-\{a\}$ then I want to show that $(*)$ holds. EDIT: I'm also assuming all the limits in $(*)$ exist.

I can show that $\lim\limits_{x \to a} f(g(x))= \lim\limits_{y \to L}f(y)$ but I'm not sure how to show the last part that the two limits also equal $f \left( \lim\limits_{x \to a}g(x) \right)$ or $f(L)$ which seems to assume continuity of $f$ on $L$ but I have to show that from only those conditions I have mentioned. Any help? And if it doesn't follow, can we produce a counterexample?

I have taken the condition from an answer here on stackexchange, you may also see my (embarrassing and clueless) comments below it. See the second to last paragraph of this answer, here.

Arturo Magidin
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William
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    The last part, that $\lim\limits_{y\to L}f(y) = f(L)$ is simply not true in general if $f$ is not continuous at $y=L$. – Jürgen Sukumaran Jan 19 '22 at 11:57
  • @TSF could you tell me what could the author have meant in the comment section? – William Jan 19 '22 at 12:02
  • @TSF I suspect there was some kind of miscommunication. (Though I'm still thankful to the author for engaging me despite my repeated cluelessness.) – William Jan 19 '22 at 12:04
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    They've made a mistake in their reasoning. I have added a comment explaining why their condition does not say anything about $f(L)$ (using the notation of that thread). – Jürgen Sukumaran Jan 19 '22 at 12:13
  • I wonder why the downvote? – William Jan 19 '22 at 16:16
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    What happened was that I was looking at the original query, phrased as $\lim_{x\to a}(f\circ g)(x) = \lim_{y\to g(a)}f(y)$, in the case where $g$ is continuous at $a$ and $f$ at $g(a)$. I had already discussed the case where $g$ is not assumed continuous, so that if $\lim_{x\to a}g(x) = L$, we would have $\lim_{x\to a}(f\circ g)(x) = \lim_{y\to L}f(y)$, provided $f$ was continuous at $L$. And then said we could replace "provided $f$ is continuous at $L$" with some other condition, including the one mentioned. There was miscommunication (cont) – Arturo Magidin Jan 19 '22 at 20:11
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    (cont) in replacing $\lim_{y\to L}f(y)$ with $f(L)$ (as happens in the original question if instead you phrase it as $\lim_{x\to a}(f\circ g)(x) = f\left(\lim_{x\to a}g(x)\right)$), which is your second equality. I was actually only talking about the first equality; and then we were talking a bit past each other. – Arturo Magidin Jan 19 '22 at 20:13

2 Answers2

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For the limit of a composite function you can write $$ \lim\limits_{x \to a} f(g(x))= \lim\limits_{y\to b} f(y) $$ under these assumptions:

  1. $\lim\limits_{x \to a} g(x)=b$;
  2. $b \notin g(U_\delta^*(a))$ for some deleted neighbourhood of $a$, $U_\delta^*(a) := (a-\delta,a)\cup(a,a+\delta)$.

The second condition just says that $g$ send (small) deleted neighbourhoods of $a$ to deleted neighbourhoods of $b$. To prove the claim above, just apply twice the definition of the limit of a function. (Of course, we interpret the equality above as saying either both limits exist and the equality holds, or they both do not exist.)

You cannot omit the second condition: just consider $g(x):= x \sin \frac1x$ and $f(0):=1$ and $f(x):=0$ otherwise. The limit on the left does not exist but the limit on the right exists and equals $0$.

To obtain $\lim\limits_{x \to a} f(g(x))=f \left( \lim\limits_{x \to a } g(x) \right)$, you need continuity of $f$. I mean, you could omit it in some special cases but you would not get anything reasonably general.

Roman Hric
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  • Right, I suppose "roughly" speaking. Those conditions say, as $x\to a$, $g(x)$ approaches $b$ but isn't equal to $b$. Or formally, there exists a deleted nbd of $a$ such that $g(x)≠b$ in that nbd. – William Jan 19 '22 at 16:20
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Counterexample: Let $g$ be the identity function and the limit of it as $x\to 0$ is $0$ and it trivially fulfills your condition. Now take your favorite function $f$ which is discontinuous at $0$ and $\lim\limits_{x\to 0}f(g(x)) $ doesn't exist.

Edit: Actually $f$ has to be a little more than discontinuous, it has to have a jump discontinuity or something like that, such that the limit at $0$ doesn't exist.

Manatee Pink
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    Are you sure? Take $f(x)=x$ for $x≠0$ then $f(g(x))= g(x)=x$ when $g(x)=x ≠0$. Now $\lim_{x\to 0}f(g(x))$ exists and is $0$. – William Jan 19 '22 at 12:00
  • Oh did you mean, $f(\lim\limits_{x\to 0}g(x))$ does not exist? – William Jan 19 '22 at 12:01
  • The limit $\lim\limits_{x\to 0}f(g(x))$ exists in your example @William but it is not equal to $f(0)$, even though $g$ is taking all the values in the deleted neighborhood $]\delta,\delta[$ of $0$. – Jürgen Sukumaran Jan 19 '22 at 12:04
  • @TSF Yes, see my next comment right below it. – William Jan 19 '22 at 12:05
  • @William, your edit is incorrect. If $g$ is the identity function then $f(g(x)) =f(x)$ and if $f$ is discontinuous at $0$ the limit cannot exist. Take for example the heaviside function. – Manatee Pink Jan 19 '22 at 12:36
  • Like, is this normal that someone else can just wildly change the content of an answer like that? That seems weird to say the least... – Manatee Pink Jan 19 '22 at 12:38
  • @ManateePink Sorry! But your answer before my edit wasn't correct either. See my comment. I, perhaps made it less incorrect but I was only going off on what you said. I never "made up" anything new in the answer or wildly changed anything. – William Jan 19 '22 at 12:50
  • Also, only the continuity of $f$ was replaced by the condition. Rest everything was same. For instance, $\lim\limits_{y \to L} f(y)$ needs to exist. – William Jan 19 '22 at 12:54
  • @William, check out my edit, the statement is correct now. However, even if it wasn't totally correct before, it is more important to preserve the intention of an answer rather than making it correct. Corrections by someone else should be made in the comments, not in edits. – Manatee Pink Jan 19 '22 at 12:58
  • @William, the condition that the limit of $f$ exists, is not part in the condition that is in your post. You only have a condition on $g$. – Manatee Pink Jan 19 '22 at 13:01
  • @ManateePink the limit can exist if $g$ is identity function and $f$ is discontinuous at $0$ - this was already demonstrated by using William's comment with $f(x)=x$ for $x\neq 0$ and $f(0)=100$. You answer was wrong and they tried to correct it to what they assumed you meant - this is normal on this site and there is always text + a log saying who edited an answer. Nothing was wildly changed nor is it weird for this website. – Jürgen Sukumaran Jan 19 '22 at 23:38