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

[evry190]

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?

 

[tiny-tim]

Hi evry190!

(have a delta: δ and an epsilon: ε )

Where's your ε?

(an epsilon delta proof has to have an epsilon …

the clue's in the name! )​

Start again, and try to prove that |2x^2-2| < ε.

 

[evry190]

but in the problem the epsilon is 1

 

[tiny-tim]

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? ​