Maximum and Minimum Limits at Infinity Proof?

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Discussion Overview

The discussion revolves around proving that a continuous even function, which approaches the same limit \( l \) as \( x \) approaches both negative and positive infinity, must have either a maximum or a minimum. The participants explore the implications of the function's behavior at infinity and the conditions under which such extrema exist.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant expresses frustration with the proof and questions how to apply the definition of a limit at infinity to show the existence of a maximum or minimum without using the first derivative test.
  • Another participant points out that the conditions described could apply to a constant function, suggesting that the original claim may not hold universally.
  • A clarification is made regarding the limit notation, emphasizing that the limit as \( x \) approaches both positive and negative infinity equals \( a \), and that \( a \) is even.
  • Some participants suggest that unless the function is constant, there must be a point where the function changes from increasing to decreasing, indicating the presence of extrema.
  • One participant argues that if the function is not constant, there exist points where the function value differs from \( a \), leading to the conclusion that maximum and minimum values must exist within certain intervals due to continuity.
  • A later reply questions the selection of points \( x_1 \) and \( x_2 \) in the argument, suggesting that they should be chosen to ensure that the function values are less than \( f(x_0) \) for sufficiently large or small \( x \).

Areas of Agreement / Disagreement

Participants do not reach a consensus on the validity of the original claim. There are competing views regarding the necessity of the function being non-constant for the existence of extrema, and the discussion remains unresolved.

Contextual Notes

Participants highlight the limitations of the proof, particularly the implications of constant functions and the need for rigorous definitions and conditions when discussing limits at infinity.

Nebula
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I'm frustrated beyond belief with a proof.

Suppose we have an continuous even function with a domain of all real numbers. Now this function has limit as x goes to negative infinty equal to l and the limit as x goes to positive infinty is also equal to l.

I want to show that this function will either have a maximum or a minimum.

I'm not sure at all how to show this rigorously since I don't know how to apply the definition of a limit to limits at infinity. I think it has to do with bounds. And I need to do this without first derivative test.
 
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Aren't the specifications you give consistent with the constant function f(x) = 1?
 
Did you think my l was a 1. Maybe I should write it differently.

lim (x-> -oo) f(x) = lim (x-> oo) f(x) = a
and a is even.

Want to show f has either a minimum or a maximum.
 
intuitively this makes sense, but rigorously you could show that unless it is a consant function (for example y = 1) then there must be a point where it switches between a positive and negative slope. I am not entirely sure what level of "rigorousness" you want.
 
First, you still have the problem that was pointed out by both selfadjoint and T@p:
The constant function f(x)= a satisfies your conditions but does not have a maximum or minimum so the "theorem" as you stated it is not true.

If f(x) is NOT A CONSTANT FUNCTION, then there exist some x0 such that f(x0) is not equal to a and so is either larger than or less than a.

Assume f(x0)> a. Since limit as x-> infinity f(x)= a, there exist some x1> x0 such that f(x1)< f(x0).
Similarly, since limit as x-> -infinity f(x)= a, there exist some x2< x0 such that f(x2)< f(x0). Since f is continuous on the closed and bounded interval [x2, x1] it must have both maximum and minimum values there. Now show that f has a maximum on -infinity to infinity.
 
I need help. Find the minimum of y = Absolute value of (sinx + cosx + tanx + cotx +
secx + cscx) Thanks Ruth Jackson the_perfect_mom@hotmail.com
 
HallsofIvy said:
Assume f(x0)> a. Since limit as x-> infinity f(x)= a, there exist some x1> x0 such that f(x1)< f(x0).

I think you need to add something about x1 here. Choose it in such a way that if x>x1 then f(x)<f(x0). Similar change with how you select x2.
 

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