# Thread: Epsilon - Delta Proof with Squares

1. ## Epsilon - Delta Proof with Squares

Hey i've been working on this problem for a few days now, just seeing if anyone else has some insight.

∀x ∈ R, ∀ε ∈ R+, ∃δ ∈ R+, ∀y ∈ R, |x – y| < δ ==> |x^2 – y^2| < ε

prove using epsilon-delta structured proof

ive got to the point where

|x^2 – y^2| < δ|x + y|

but am not completely stuck, I think I have to replace y with a constant somehow? any help would be great! thanks in advance!

2. I cannot read that very well. It also would help if you would add the entire question.

You mentioned that
$|x-y| < \delta$ and $|x^{2}-y^{2}| < \epsilon$

From there I would rewrite
$|x^{2}-y^{2}| < \epsilon$
$|(x-y) (x+y)| < \epsilon$
$|(x-y) * C| < \epsilon$

Then
$|(x-y)| < \frac{\epsilon}{C} = \delta$

To find C, I would look at:
if $0< |x-a| < \delta$ then $|f(x) - L| < \epsilon$

$a- \delta < x< a+ \delta$
In this case y would be a

But to be sure I would need the entire question. If you still need this problem, maybe you can submit the question asked.

3. Originally Posted by kale3193
Hey i've been working on this problem for a few days now, just seeing if anyone else has some insight.

∀x ∈ R, ∀ε ∈ R+, ∃δ ∈ R+, ∀y ∈ R, |x – y| < δ ==> |x^2 – y^2| < ε

prove using epsilon-delta structured proof

ive got to the point where

|x^2 – y^2| < δ|x + y|

but am not completely stuck, I think I have to replace y with a constant somehow? any help would be great! thanks in advance!
if i am not wrong this is just the prove that f(x) = x^2 is continuous for all x on R.
if we take delta = min{ |x|/2, 2epsilon/5|x| } for all x from R \ {0} the formula will work

and for x = 0 we can take delta = sqrt( epsilon )

you choose x and epsilon first and than you choose delta, so delta must depend just on x and epsilon, not on y

4. ## Okay but

How can you write < epsilon first?
I thought we had to assume |x-y| < delta first and show that it leads to |x^2 - y^2 | < Epsilon

5. Originally Posted by kale3193
How can you write < epsilon first?
I thought we had to assume |x-y| < delta first and show that it leads to |x^2 - y^2 | < Epsilon
i didnt wrote < epsilon first...
you need to find delta ( based on epsilon and x ) such that the statment in first post is true! Dont know if you learnd about continuosity of functions ( maybe you did, but didnt realise that very good, because this is just an example of that )