Conjecturing the Limit and Finding Delta for Sinusoidal Function

  • Context: Graduate 
  • Thread starter Thread starter John O' Meara
  • Start date Start date
  • Tags Tags
    Delta Limit
Click For Summary

Discussion Overview

The discussion revolves around the limit of the function f(x) = sin(2x)/x as x approaches 0, specifically conjecturing the limit value and determining a corresponding delta (δ) for a given epsilon (ε). The context includes mathematical reasoning and the use of graphing utilities to explore the behavior of the function near the limit.

Discussion Character

  • Mathematical reasoning, Exploratory

Main Points Raised

  • One participant conjectures that the limit L = 2 as x approaches 0 and seeks to find a delta such that |f(x) - L| < ε for ε = 0.1.
  • Another participant argues that it is possible to find a delta even if the function does not go above and below the limit, as long as f(x) remains within the specified bounds.
  • A third participant describes their attempts to use a graphing calculator to find a specific x value where f(x) approaches 1.9, noting that the interval they found does not directly correspond to delta.
  • Another participant clarifies that any positive delta that satisfies the condition regarding f(x) being between 0 and L ± ε is acceptable, regardless of whether it is the largest possible delta.

Areas of Agreement / Disagreement

Participants express differing views on the necessity of the function's behavior around the limit and the definition of delta. There is no consensus on the specific value of delta or the method to find it.

Contextual Notes

Participants have not resolved the specific value of delta, and there are assumptions about the behavior of the function that remain unexamined. The discussion also reflects varying interpretations of the conditions required for delta.

John O' Meara
Messages
325
Reaction score
0
Let [tex]f(X)=\frac{\sin(2x)}{x}[/tex] and use a graphing utility to conjecture the value of L = [tex]\lim_{x->0}f(x) \mbox{ then let } \epsilon =.1[/tex] and use the graphing utility and its trace feature to find a positive number [tex]\delta[/tex] such that [tex]|f(x)-l|< \epsilon \mbox{ if } 0 < |x| < \delta[/tex]. My conjecture of the limit L = 2, therefore if that is the case then [tex]1.9< f(x) < 2.1[/tex]. Since the maximum value of f(x) < 2, the graphing utility will not be able to find delta will it? What is the value of delta if L=2?Thanks.
 
Physics news on Phys.org
It will be able to. The function does not need to go both above and below. You just need to find a number [tex]\delta[/tex] so that f(x) is between those numbers whenever x is in the interval [tex][-\delta,\delta][/tex].
 
I set up my graphing calculator Xmin =.2758..., Xmax = .2775... Then f(x)=y=1.8999954 gives x_0=.27596197. The length 0 to x_0 is not = delta but this half interval has the property that for each x in the interval (except possibly for x=0) the values of f(x) is between either 0 and L+epsilon or L-epsilon. delta could be much smaller than x_0. I still have not found delta.
 
Any positive number delta that has that property ("for each x in the interval (except possibly for x=0) the values of f(x) is between either 0 and L+epsilon or L-epsilon") is a solution to the problem as long as it does not specifically ask for the largest possible number delta with that property.
 

Similar threads

  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 25 ·
Replies
25
Views
4K
  • · Replies 16 ·
Replies
16
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K