(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Suppose that δ = min{1,e/7}. Show that

if |x - 3| < δ, then |x^2 - 9| < e

2. Relevant equations

Don't know.

3. The attempt at a solution

|x - 3| |x + 3| < e

Given : |x - 3| < δ => |x - 3| < e/7

|x + 3| |x-3| < e/7

|x + 3| 7|x - 3| < e

Assume δ < 1, => |x - 3| < 1 => -4 < x < -2

Sub (-4)

| (-4) + 3| 7|x-3| < e

|x-3| < e/7

Is this circular? Please give some feedback!

Thank you!

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# Delta/Epsilon (Need opinions on my solution, I am new to these)

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