How do I choose delta in an epsilon delta proof without just guessing Concept

You find delta by working backwards, on scratch paper, and then you write the proof forwards. Those are two separate documents and your textbook only shows you the second one.

Scratch: start from what you want, |f(x) - L| < ε, and manipulate until you have |x - a| < something. For f(x) = 3x + 1 at a = 2, |3x + 1 - 7| = 3|x - 2|, and you want that under ε, so |x - 2| < ε/3. That is where delta came from.

Proof: let δ = ε/3, assume 0 < |x - 2| < δ, then |f(x) - 7| = 3|x - 2| < 3δ = ε. Done. Nothing mysterious, just the scratch work read right to left.

For anything nonlinear you need the second trick: bound delta by 1 first, then use that bound to control the messy factor.

Example, f(x) = x² at a = 3. |x² - 9| = |x - 3||x + 3|. The |x + 3| is the problem because it depends on x. So say up front you will take δ ≤ 1, which forces 2 < x < 4, which forces |x + 3| < 7. Now |x² - 9| < 7|x - 3|, and you finish with δ = min(1, ε/7).

That min(1, ...) pattern covers about ninety percent of the exercises you will be given.

Also worth saying out loud because nobody does: you are allowed to overshoot. Delta does not have to be the largest one that works. If ε/3 works then ε/10 works too. Proofs are not optimisation problems.

One habit that made these click for me: after you write down a delta, sanity check it with a number. Set ε = 0.03, compute your delta, pick an x that far from a, and see whether f(x) really lands inside the window. If it does not, your algebra dropped a factor somewhere.