Epsilon-Delta Proof for Limit of x^2sin(1/x) as x Approaches 0

  • Context: Graduate 
  • Thread starter Thread starter rman144
  • Start date Start date
  • Tags Tags
    Proof
Click For Summary

Discussion Overview

The discussion focuses on proving the limit of the function \( x^2 \sin(1/x) \) as \( x \) approaches 0 using the epsilon-delta method. Participants explore different approaches and conditions for establishing this limit.

Discussion Character

  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant proposes using the epsilon-delta definition with \( \delta = \sqrt{\epsilon} \) to show that the limit holds.
  • Another participant suggests that the limit can be shown by establishing that \( |x^2 \sin(1/x)| \leq x^2 < \epsilon \) for sufficiently small \( x \).
  • A third participant agrees with the use of \( \delta = \sqrt{\epsilon} \) and notes that \( |\sin(1/x)| \) is bounded by 1, which supports the argument.
  • A later reply reiterates the sufficiency of the approach using \( \delta = \sqrt{\epsilon} \) and emphasizes the bounding of the sine function.

Areas of Agreement / Disagreement

Participants generally agree on the use of \( \delta = \sqrt{\epsilon} \) and the bounding behavior of the sine function, but there is no consensus on the most rigorous approach or whether the proposed methods are sufficient without further clarification.

Contextual Notes

Some assumptions about the behavior of the sine function and the conditions under which the limit holds are not fully explored, leaving some mathematical steps unresolved.

rman144
Messages
34
Reaction score
0
I need to show that:

[itex]\lim_{x \to 0} x^{2}sin(1/x)=0[/itex]

Using the epsilon-delta method. I figured delta=sqrt[epsilon] would make the limit hold, but wanted to be sure. Thanks in advance.
 
Physics news on Phys.org
rman144 said:
[itex]\lim_{x \to 0} x^{2}sin(1/x)=0[/itex]

I believe it can be show that if [tex]\epsilon[/tex] = [tex]\delta[/tex][itex]abs(x^{2}sin(1/x))\leq x^{2}[/itex] < [tex]\epsilon[/tex]
 
[tex]\delta = \sqrt{\varepsilon}[/tex] is ok too, |sin(...)| will be always less or equal to 1, so [tex]\varepsilon \left|\sin(1/\sqrt{\varepsilon})\right| \leq \varepsilon[/tex]
 
Liwuinan said:
[tex]\delta = \sqrt{\varepsilon}[/tex] is ok too, |sin(...)| will be always less or equal to 1, so [tex]\varepsilon \left|\sin(1/\sqrt{\varepsilon})\right| \leq \varepsilon[/tex]

This is more sufficient.:)
 

Similar threads

  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 25 ·
Replies
25
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 31 ·
2
Replies
31
Views
3K