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I have $2$ questions regarding the following limit law:

Suppose that $\lim_{x \to a}g(x)$ exists and is equal to $l$, and that $f$ is continuous at $l$. Then, $\lim_{x \to a}f\bigl(g(x)\bigr)$ exists, and $$ \DeclareMathOperator{\epsilon}{\varepsilon} \lim_{x \to a}f\bigl(g(x)\bigr)=f(l)=f\left(\lim_{x \to a}g(x)\right) \, . $$

My questions are:

  1. Have I stated this limit law correctly?
  2. How do you prove this limit law?

Here is my attempted proof:

Let $\epsilon>0$. We wish to find a $\delta>0$ such that, for all $x$,

If $0<|x-a|<\delta$, then $|f(g(x))-f(l)|<\epsilon$.

We are given that $f$ is continuous at $l$, i.e. that there exists a $\delta'>0$ such that, for all $y$,

If $|y-l|<\delta'$, then $|f(y)-f(l)|<\epsilon$.

We are also given that $\lim_{x \to a}g(x)=l$, meaning that there is a $\delta>0$ such that

If $0<|x-a|<\delta$, then $|g(x)-l|<\delta'$

Since $g(x)$ is a number $y$ satisfying $|y-l|<\delta'$, we get that $|f(g(x))-f(l)|<\epsilon$. Hence, if $0<|x-a|<\delta$, then $|f(g(x))-f(l)|$, and so $\lim_{x \to a}f(g(x))=f(l)=f\left(\lim_{x \to a}g(x)\right)$, completing the proof.

Joe
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    Yes what you have is right. Is there anything specifically about the statement/proof you're unsure about? – peek-a-boo Jun 18 '21 at 19:12
  • There's an issue with domains. What if $l$ is an isolated point in the domain of $f$? Then $f$ is automatically continuous there, but $f(g(x))$ may be undefined in a punctured neighbourhood of $a$ so that $\lim_{x \to a} f(g(x))$ becomes meaningless. – Hans Lundmark Jun 18 '21 at 20:17
  • A related question: https://math.stackexchange.com/questions/167926/formal-basis-for-variable-substitution-in-limits – Hans Lundmark Jun 18 '21 at 20:24
  • @HansLundmark: I suppose I should change the hypothesis "$f$ is continuous at $l$" to "$f$ is continuous at $l$ and $l$ is a limit point of the domain of $f$". Is that correct? – Joe Jun 18 '21 at 21:04
  • No, that's not enough, since $g$ may still take only values where $f$ is undefined. So I think you'll actually want to require $f$ to be defined in a punctured neighbourhood of $a$. Anyway, it all becomes much simpler if you phrase things in terms of continuity instead of limits: a composition of continuous functions is continuous, and that's it (no exceptional cases to worry about). – Hans Lundmark Jun 18 '21 at 21:10
  • @HansLundmark: When you said "$f$ has to be defined in a punctured neighbourhood of $a$", did you mean "$f$ has to be defined in a punctured neighbourhood of $l$"? – Joe Jun 18 '21 at 21:17
  • Yes, it should be $l$, of course. Sorry. Maybe some weaker assumption would work too, but that should be enough to be on the safe side, at least. – Hans Lundmark Jun 18 '21 at 21:20
  • Okay, thanks for clarifying. I don't quite understand your comment about "a composition of continuous functions is continuous". I don't see why that is relevant given that $g$ is not required to be continuous at $a$—we just need $\lim_{x \to a}g(x)$ to exist; it doesn't have to be equal to $g(a)$ for the limit law to be applicable. Am I misunderstanding you? – Joe Jun 18 '21 at 21:25
  • @HansLundmark: Okay, thanks for clarifying. I don’t quite understand your comment about “a composition of continuous functions has to be continuous”. I don’t see why that is relevant because $g$ is not required to be continuous at $a$—we just need $\lim_{x \to a}g(x)$ exist; it doesn’t need to be equal to $g(a)$ for the limit law to apply. Am I misunderstanding you? – Joe Jun 18 '21 at 22:34
  • If $\lim_{x \to a} g(x)$ exists, then you can extend or redefine $g$ to be continuous at $a$. – Hans Lundmark Jun 19 '21 at 05:35

2 Answers2

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If $g(x)$ is continuous at $a$ and $f(x)$ is continuous at $g(a)$

Then for any $\epsilon_1$ there exists a $\delta_1$ such that $|x-a|<\delta_1 \implies |g(x) - g(a)| < \epsilon_1$

and

for any $\epsilon_2$ there exists a $\delta_2$ such that $|y-g(a)|<\delta_2 \implies |f(y) - f(g(a))| < \epsilon_2$

Choose $\epsilon_1$ to be less less than $\delta_2$

Doug M
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Your proof is fine.

So my remarks would go towards minor issues for a better wording (I think).

There is no need for the $0<|x-a|$ an absolute value is always positive, and it is not mandatory that $x\neq a$.

Also I would not state the conclusion as the starting point, it is a bit confusing, start at "We are given...".

Finally I would use proper quantifiers rather than "if, then, let", and repeating it mathematically the line below.

The following formulation is less verbose, and just fine:

  • $g$ has limit $l$ at $a$ so $\quad\forall \delta'>0,\ \exists \delta>0\text{ s.t } |x-a|<\delta\implies |g(x)-l|<\delta'$
  • $f$ is continuous at $l$ so $\quad \forall \epsilon>0,\ \exists \delta'>0\text{ s.t } |y-l|<\delta'\implies |f(y)-f(l)|<\epsilon$

Consequently:

$$\forall \epsilon>0, \exists\delta>0\text{ s.t }|x-a|<\delta\implies |g(x)-l|<\delta'\implies |f(g(x))-f(l)|<\epsilon$$

And the theorem is proved.

zwim
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    Thanks for this answer. Regarding "there is no need...": since $g$ is not required to be continuous or even defined at $a$, I don't see why $f(g(a))$ should equal $f(l)$. – Joe Jun 18 '21 at 20:02
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    "absolute value is always positive" ummm surely you mean non-negative $\geq 0$. Several textbooks impose the strict inequality in the definition of limits, precisely to exclude the behavior at the point in question, so what the OP wrote is fine. – peek-a-boo Jun 18 '21 at 20:53
  • @peek-a-boo Yeah, this is the eternal debate on terms on MSE. Basically (I generalize, don't be offended) in France (Italy and probably other EU countries) we consider things loosely ($\mathbb N$ contains $0$, positive is $\ge 0$, increasing = $\nearrow$ or constant, etc...) while this is considered strict in the U.S. But anyway, $x\neq a$ is not necessary here, the exclusion you are referring to is for the definition of a limit point of a set where we consider neighbourhoods not containing the point. – zwim Jun 19 '21 at 02:12