Continuity Proof: f(x) = x^3 [cos(pi/x^2) + sin(pi/x^2)]

  • Thread starter Thread starter JPanthon
  • Start date Start date
  • Tags Tags
    Continuity Proof
Click For Summary

Homework Help Overview

The discussion revolves around the continuity of the function f(x) = x^3 [cos(pi/x^2) + sin(pi/x^2)] for x≠0. Participants are exploring the conditions under which this function may be continuous, particularly at the point x=0.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • One participant attempts to apply the squeeze theorem to analyze the behavior of the function as x approaches 0 but expresses difficulty in determining the range of the trigonometric components. Another participant questions the exact nature of the continuity proof being sought, while a third notes that the function appears continuous for x≠0 but raises concerns about its definition at x=0.

Discussion Status

The discussion is ongoing, with participants providing various insights and questioning the assumptions underlying the continuity proof. Some guidance has been offered regarding the continuity of the function for x≠0, but there is no explicit consensus on the approach to take for x=0.

Contextual Notes

Participants are grappling with the implications of the function's definition at x=0 and the continuity criteria that apply in this context. There is an indication that the problem may not be as straightforward as initially perceived.

JPanthon
Messages
20
Reaction score
0

Homework Statement



f(x) = x^3 [cos(pi/x^2) + sin(pi/x^2)] for x≠0

Homework Equations





The Attempt at a Solution



I really am stuck.

I've tried squeeze theorem on [cos(pi/x^2) + sin(pi/x^2)], but I can't compute the range.

So, I tried doing it individually, squeezing -1 ≤ cos(pi/x^2) ≤ 1, but that doesn't work.

Any hints please?
 
Physics news on Phys.org
cos(θ)+sin(θ) = (√2)sin(θ + π/4)
 
What's the question exactly? You want to prove that that function is continuous?
 
The given function is obviously continuous for [itex]x\ne 0[/itex] because it is a composition of continuous functions. It obviously not continuous for x= 0 because it is not defined at x= 0.

Because I suspect the problem was not supposed to be that "obvious", please check again and tell us what the problem really is!
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 40 ·
2
Replies
40
Views
5K
  • · Replies 14 ·
Replies
14
Views
2K
Replies
3
Views
2K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 11 ·
Replies
11
Views
1K
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K