1. The problem statement, all variables and given/known data 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 2. Relevant equations 3. 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?