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I was about to ask the exact same question as presented here Change of Variables in Limits (Part 1) with some additional thoughts of mine.

I mean, we often do variable change in limits without thinking much (kind of mechanically we do it), but as said in the other question, I also noticed there's no formal theorem about this in any standard calculus / real analysis textbook. We were just taught to do it. Why?!

So... OK... since this question has already been asked here, this saves me a lot of typing, and I can just ask about confirmation of my additional thoughts.

I will just copy the claim here for easier reading:

Claim: If $\lim \limits_{x\to a}g(x)=b$, then $\lim \limits_{x\to a}f(g(x))=\lim \limits_{y\to b}f(y)$.

Basically we're changing/substituting $g(x)$ with $y$.

Now... I know as this stands it is not generally true.

Below are my thoughts. More or less, I just ask for their validation/confirmation, not for proofs. Of course if someone wants to present some formal proof, I don't mind. But I think I mainly need someone to validate these claims below.

  1. I think the following is true: if the limit on the RHS exists and is equal to a number $R$, then the one on the LHS exists and is also $R$. I cannot quite prove this formally (I will try again)... but is it true? So in other words, my thought is that (in some sense) the limit on the RHS is more general and the one on the LHS is more special.

  2. Is the statement in 1) also true if $R$ is $+\infty $ or $-\infty$ (and not necessarily a finite number)? I think it is true even then... but I cannot quite prove this formally, again... need to try out a bit harder. I just need confirmation if my claim is true, not a proof really.

  3. OK... If 1) and 2) are true, then I just realized that this change of variables has quite a deep meaning (in university we were more or less taught to do it mechanically which I didn't like at all btw). But in fact the idea is: you do the change of vars, and then you proceed and if you are able to calculate the RHS limit, then the LHS limit also exists and is the same value $R$. This is the logical/deep idea here. Is this realization of mine correct? Could anyone just confirm it. Or... is there anything even more subtle here?

  4. Another claim... If $g$ is an invertible function (in some neighborhood of $a$), then this equality of limits is really an equivalence, I mean... we can then ALSO claim that: if the LHS limit exists and is equal to the number $L$, then the RHS limit also exists and is equal to the same number $L$. Is this correct? I think I almost proved this formally. Is this claim true? Also, not quite sure if the same claim holds true even when $L$ is +/- infinity. I guess it does. Can anyone confirm?

  5. Finally I want to present a nice example/demonstration which I came up with myself. I just want someone to validate it please.

OK... we know $\lim \limits_{y\to 0}\frac{|y|}{y}$ does not exist.

This is my RHS limit. So $b=0$ and $f(y) = \frac{|y|}{y}$

Let's take $g(x) = (x-3)^2$ and $a = 3$.

We know that $\lim \limits_{x\to a}g(x) = \lim \limits_{x\to 3}g(x) = 0 = b$.

Also, obviously we have that:

$\lim \limits_{x\to 3}f(g(x)) = \lim \limits_{x\to 3}\frac{|g(x)|}{g(x)} = \lim \limits_{x\to 3}\frac{|(x-3)^2|}{(x-3)^2} = \lim \limits_{x\to 3} 1 = 1$

So in this case the LHS limit exists and is 1. The RHS limit though does not exist. I think this is because the function g in this case is not invertible (in any neighborhood of the point 3). Is that correct?

Is this example/observation correct? And in general... am I on the right track in all these thoughts.

  1. What led me too all these thoughts? The proof which I read of the Theorem about the derivative of the inverse function. They did one change of variables there which I didn't quite understand why it is justified. And it make me think further... And I think there it is justified because the word is about an invertible function.

  2. Are these proofs/theorems correct and formally sound?

https://math.stackexchange.com/q/456038
https://math.stackexchange.com/q/167948

Maybe they solve all my worries here, right?

peter.petrov
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    Wikipedia has a nice summary of conditions required to make this claim true. The $2$ answers you linked to are also worth looking at. https://en.wikipedia.org/wiki/Limit_of_a_function#Limits_of_compositions_of_functions – bjorn93 Jan 28 '20 at 19:27
  • @bjorn93 Oh, thanks very much. I wouldn't have found this section. By the way, I took the first answer and worked on it, thinking it over in details. And I can say it works fine in my case (for that proof of the formula for the derivative of the inverse function). – peter.petrov Jan 29 '20 at 11:41
  • Your point $4$ is the one which really is important and true. This is what we use in practice while doing substitution because we don't know in advance whether the limit will exist or not when making a substitution. I have discussed this in one of my answers. See https://math.stackexchange.com/a/1073047/72031 – Paramanand Singh Jun 27 '20 at 01:34

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Ok, this is all over the place, so I cannot answer all your questions because I don't really know what you are asking. Now,

  • the claim holds when $f$ is continuous at $b$. Continuity of $f$ at $b$ means precisely that if $b_n\to b$ then $f(b_n)\to f(b)$. So if $g(x)\to b$ as $x\to a$, you have $$\lim_{x\to a}f(g(x))=\lim_{y\to b}f(y).$$

  • More precisely, what one really needs is that $\lim_{y\to b}f(y)$ exists, not really continuity. When you see it like this, you'll see that the claim still holds if the limit of $f$ is infinite.

  • In your point 5, the reason why things fall apart is that $\frac{|y|}y$ is not continuous at zero. And $(x-3)^3$ approaches zero at $3$. The reason it fails with exponent $3$ and not $2$, say, is that you need a $g$ that changes sign after it takes the value zero, to exploit the discontinuity of $f(y)=|y|/y$ (because the lateral limits of $f(y)$ at zero exist and are different).

  • The answers you link look fine to me.

Martin Argerami
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