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hgfhh123

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## Homework Statement

When constructing an Epsilon Delta proof, why do we need to make a stipulation? For example, in most proofs for limits of quadratic functions, it is stipulated, for example, that δ≤1. Why is this needed anyway?

This is my thought process for a quadratic:

Prove that lim(x -->3) x^2 = 9

## Homework Equations

## The Attempt at a Solution

For every ε > 0 there is δ > 0 such that |x^2 - 9| < ε whenever 0 < |x-3| < δ

|x+3||x-3| < ε

We will try to find a number K such that

|x+3||x-3| < K|x-3| < ε

Which will give us |x+3| < K and |x-3| < ε/K

This is the point at which it is generally stipulated that δ ≤ 1.

Instead, I used the original statement from above: |x-3| < δ

-δ + 3 < x < δ + 3

-δ + 6 < x + 3 < δ + 6

Now, if x+3 = |x+3|, then x+3 must be ≥0.

Therefore, -δ + 6 ≥ 0 --> δ≤6

-δ + 6 < |x + 3| < δ + 6 (with the condition that δ≤6)

Since |x+3| < K and |x + 3| < δ + 6 --> K ≥ δ + 6

We have found K for any δ > 0 we choose, so we have shown that |x+3| < K

for some value K, and so,

|x-3| < ε/K

Thus, we can say that δ ≤ ε/K to guarantee that this is true

Proof:

if 0 < |x-3| < ε/K then |x^2 - 9| < ε

We assumed that |x-3| < ε/K and that |x+3| < K

|x^2 - 9| = |x+3||x-3|

< K|x-3|

< K(ε/K)

< ε

My question is, did I do something wrong by not stipulating that δ is less than or equal to some value? Is my method logically correct?

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