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This question is a continuation of of my pains with

Change of Variables in Limits (Part 1)

Change of Variables in Limits (Part 2)

Limits of composite functions

I was re-reading the conditions (for this "chain rule" to hold true) in Wikipedia.

Chain rule:

If $$\lim\limits_{x \to a} g(x) = b$$ and
$$\lim\limits_{y \to b} f(y) = c$$
then $\lim\limits_{x \to a} f(g(x)) = \lim\limits_{y \to b} f(y) \ \ \ \ \ (\ \ \ = c \ \ \ \ \ ) \tag{*}$

What do I know so far?

For $a$ and $b$ and $c$ - all finite numbers, it is enough to assume one of these:
(1) f is continuous at b
or
(2) g does not take the value b in some set which looks like a
"punctured" neighborhood of $a$ i.e. in $(a-\delta, a)\cup(a, a + \delta)$
(for some $\delta \gt 0$)

and of course additionally (to (1) OR (2)) we need to have that

(3) $a$ is a limit point of $g$'s domain, $b$ is limit point of $f$'s domain

Then the chain rule holds! OK...

... and that means more precisely that:

a) if the RHS limit in $(*)$ exists, then the LHS limit exists and they are equal
b) if the LHS limit in $(*)$ exists, then the RHS limit exists and they are equal
And here I want to note that by "limit exists" I mean it exists and is finite.

So all this I know. I think I was able to prove it formally and to understand it well.

But what happens when we start allowing $a$ or $b$ or $c$ to be infinities (positive or negative)?

It is not clear to me what happens if $a$ or $b$ or $c$ or some combination of these is $+\infty$ or $-\infty$. I mean what does really happen if we want to be formal and rigorous? My head starts spinning because... one can just form too many statements and theorems when one allows for infinities.

I think the Wikipedia article covers only the cases when $a$ and $b$ and $c$ are finite numbers and not infinities.

So what is the final/rigorous truth here? When are we allowed to apply this chain rule (I mean in real single-variable analysis)?

Is there some easy way to remember when we can apply it and when not
(when we throw in infinities into the mix)?

I don't want to go into examples but I can... I will just mention one simple case. If $g$ is a bijection in some "punctured" neighborhood of $a$ (and here I allow for $a$ being +/- infinity), can we always use the chain rule i.e. does it always hold true in this case?

peter.petrov
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  • The chain rule concerns derivatives. – zhw. Jun 26 '20 at 20:46
  • I think I understood where my confusion (or part of it) comes from. In the Wikipedia article with equality (*) they don't mean b). They mean just a). Is this understanding correct? – peter.petrov Jun 27 '20 at 00:20
  • And then if so... my question is: if a, b, c can be infinities do we still need these conditions 1) and 2) for the equality (*) to hold true? And if so... how do we rephrase them for the cases where a,b,c (or some of them) are infinities?! – peter.petrov Jun 27 '20 at 00:23

1 Answers1

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Specifying the theorem for all cases in one place is clumsy. It is best the reader formulates the rules in his/her mind for individual cases. They need not be expressed explicitly in a textbook but can be given as exercises.

This is easily done provided one really understands the case when $a, b, c$ are finite. As an example let's make $b=-\infty, c=\infty$. And we can state

Theorem: If $\lim_{x\to a} g(x) =-\infty$ and $\lim_{x\to-\infty} f(x) =\infty$ then $\lim_{x\to a} f(g(x)) =\infty$.

Here the condition $(2)$ holds automatically as $g(x)$ can not equal $-\infty$ anyway.

  • 1
    Thanks a lot! So based on that... is it right to say that: when $b$ is $\infty$ or $-\infty$ then the chain rule always holds true since (2) is satisfied automatically? And also: when $b$ is finite then the chain rule holds true if (1) or (2) is true (regardless of if $a$ or $c$ are finite or not)? – peter.petrov Jun 27 '20 at 09:22
  • @peter.petrov: yes I think so. – Paramanand Singh Jun 27 '20 at 10:20
  • Yeah, I think so too. I was able to formulate a few particular statements/theorems just as you suggested and to prove them. So... I think what I mentioned in my previous comment is indeed so. Thanks a lot again. – peter.petrov Jun 27 '20 at 10:42