Limits - Epsilon Delta Proof for 2x^2-2<1

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Homework Help Overview

The discussion revolves around an epsilon-delta proof related to the limit of the function 2x^2 as x approaches -1, specifically aiming to show that this limit equals 2. Participants are exploring the conditions under which the expression |2x^2 - 2| is less than 1.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to manipulate the inequality |2x^2 - 2| < 1 and expresses it in terms of |x + 1| and |x - 1|. They question the correctness of their derived delta value and how to determine the appropriate solution from two possible answers. Other participants raise the need for an epsilon value in the proof and seek clarification on its role in the problem.

Discussion Status

The discussion is active, with participants questioning the setup of the epsilon-delta proof and the original poster's understanding of the epsilon value. Some guidance has been offered regarding the necessity of including epsilon in the proof, but there is no explicit consensus on the approach to take or the correct interpretation of the problem.

Contextual Notes

Participants are working under the assumption that the epsilon value is set to 1, but there is confusion regarding its application in the context of the limit being discussed. The original poster expresses uncertainty about the relationship between their calculations and the epsilon value.

evry190
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Okay so here's the limit lim(as x --> -1) 2x^2=2

I have |2x^2-2| < 1
2|x + 1||x - 1| < 1
|x + 1||x - 1| < 1/2

and 0 < |x + 1 | < delta
-1 - delta < x < -1 + delta
|x - 1| < -2 + delta

|x + 1| < 1/2(-2 + delta)

so is delta = 1/2(-2 + delta) correct? if so, how do i know which answer (since there will be 2) is right?
 
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Hi evry190! :smile:

(have a delta: δ and an epsilon: ε :wink:)

Where's your ε?

(an epsilon delta proof has to have an epsilon …

the clue's in the name! :biggrin:)​

Start again, and try to prove that |2x^2-2| < ε. :smile:
 
but in the problem the epsilon is 1
 
evry190 said:
but in the problem the epsilon is 1

I don't understand … the problem was "the limit lim(as x --> -1) 2x^2=2" …

where's the ε = 1 in that? :confused:
 

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