# Epsilon delta proof.

## Homework Statement

if |x-3| < ε/7 and 0 < x ≤ 7 prove that |x^2 - 9| < ε

## The Attempt at a Solution

So ths is what I did so far.

|x+3|*|x-3| < ε (factored out the |x^2 - 9|)

|x+3|*|x-3| < |x+3|* ε/7 < ε (used the fact that |x-3| < ε/7)

|x+3|* ε/7 *7 < ε*7|x-3| < ε/7*7 (multiplied both sides of the inequality by 7)

I suck at epsilon delta proofs and have no idea where to go from here.

SammyS
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## Homework Statement

if |x-3| < ε/7 and 0 < x ≤ 7 prove that |x^2 - 9| < ε

## The Attempt at a Solution

So ths is what I did so far.

|x+3|*|x-3| < ε (factored out the |x^2 - 9|)

|x+3|*|x-3| < |x+3|* ε/7 < ε (used the fact that |x-3| < ε/7)

|x+3|* ε/7 *7 < ε*7|x-3| < ε/7*7 (multiplied both sides of the inequality by 7)

I suck at epsilon delta proofs and have no idea where to go from here.
It looks to me like the problem should read something like:
if |x-3| < ε/7 and 0 < x-3 ≤ 1, prove that |x2 - 9| < ε​
Otherwise you can only show that |x2 - 9| < (10/7) ε .

More importantly, you are to prove that |x2 - 9| < ε. The way you are trying to go about this is to assume the fact which you are trying to prove. That is not a valid way to go about this.

I had a typo originally. See the correction in Red.

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How do i get started on this proof then/?

SammyS
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Homework Helper
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How do i get started on this proof then/?

If x-3 ≤ 1, what does that say about x+3 ?

x+3<7

SammyS
Staff Emeritus
Homework Helper
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x+3<7
Correct.

I suppose I should have asked "What if -1 ≤ x-3 ≤ 1" or any suitable left hand value.

So, if 5 ≤ x+3 ≤ 7, then you can definitely say that |x+3| ≤ 7 . Correct?

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