What is the proof for nested limits?

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Discussion Overview

The discussion revolves around the proof of the limit of composite functions, specifically the expression lim x→a f(g(x)) = f(lim x→a g(x)). Participants explore the conditions under which this statement holds true, particularly focusing on continuity requirements for the functions involved.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant asserts that the limit of composite functions holds under certain conditions but seeks a general proof.
  • Another participant counters that the theorem is not universally true and requires continuity of f at the limit of g(x) as x approaches a.
  • Several participants emphasize that even if limits exist, continuity is necessary to avoid discrepancies, suggesting that simple examples can illustrate this point.
  • A participant introduces the idea that the definition of functions can vary, questioning the criteria for what constitutes a "contrived" function.
  • One participant attempts to clarify the reasoning behind the limit of composite functions by discussing continuity and the behavior of g(x) as x approaches a.

Areas of Agreement / Disagreement

Participants generally disagree on the universality of the limit of composite functions without continuity assumptions. Multiple competing views remain regarding the conditions necessary for the theorem to hold.

Contextual Notes

Limitations include the dependence on continuity assumptions and the potential for inequalities even when limits exist. The discussion does not resolve these mathematical nuances.

swampwiz
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AIUI, this is a law of proofs:

lim x→a f( g( x ) ) = f( lim x→a g( x ) )

I have searched for an explanation of this proof, but have been unable to find one, although I did find a page that was for certain types of functions of f( x ), just not a proof for a function in general.
 
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Not true in general. You need continuity assumptions. Even if all the limit exists, then still it is possible to have inequality, as playing with simple examples should indicate you.
 
lurflurf said:
of interest
https://teachingcalculus.com/2019/08/26/limit-of-composite-functions/

It is not true in general. That theorem is in any calculus book requiring lim x→a g( x ) exist and f continuous at that value.
In many examples of interest only one of the two conditions hold. We then need to find another method in those cases.

I was referring to functions that are the same expression, not some contrived function that is defined by different expressions for different sections of the domain.
 
swampwiz said:
I was referring to functions that are the same expression, not some contrived function that is defined by different expressions for different sections of the domain.
That is a rather arbitrary criterium. Take any function ##F## on ##\mathbb{R}## that is, in your opinion, contrived. Take another function ##G## on ##\mathbb{R}## that you don't find contrived. Define the function ##H## on ##\mathbb{R}## by ##H(x) = F(G(x))##. Then ##H## can be written using one and the same expression for all ##x \in \mathbb{R}##. Do you find ##H## to be contrived, or not?
 
Suppose that g(x) is continuous near x = a and f(u) is continuous near u = g(a). Then by definition the limit of f(g(x)) as x → a is what happens to f(g(x)) when x gets closer and closer to a. This means we are plugging into g(x) values of x close to a. By continuity of g(x) near x = a, these g(x)'s will get closer and closer to g(a). So whatever g(a) may be, the limit of f(g(x)) as x → a is the same as the limit of f(u) as u → g(a). (Are you with me so far?)

Now since f(u) is continuous at u = g(a) the limit of f(u) as u → g(a) is the same as plugging in g(a) for u, or in other words f(g(a)).
 

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