Chapter A: Appendix A: Limits and Proofs (ε–δ)

Why this matters“Gets close” is intuitive; “gets arbitrarily close for any ε you name” is a proof. This appendix turns the intuition you have used all term into the precise ε–δ definition — the backbone of real analysis.

1 · The ε–δ definition

lim_{x→a} f(x) = L means: for every ε > 0 there exists δ > 0 such that whenever 0 < |x−a| < δ, we have |f(x) − L| < ε.

Read it as a game: an opponent picks ε (how close f(x) must be to L). You respond with a δ (how close x must stay to a) that guarantees the win.

2 · The limit game, decoded

“For every ε > 0” = opponent goes first. “There exists δ > 0” = your move, allowed to depend on ε. “Whenever 0<|x−a|<δ then |f(x)−L|<ε” = the payoff. The δ may be tiny; the point is it exists for every ε.

3 · Worked example — proving lim (2x+1) = 5 as x→2

Worked example 1

Given ε > 0, choose δ = ε/2. If 0<|x−2|<δ, then |(2x+1)−5| = 2|x−2| < 2δ = ε. Done. The opponent’s ε dictates your δ; nothing more is needed.

4 · A harder example — lim x² = 4 as x→2

Worked example 2

We need |x²−4| = |x−2|·|x+2| < ε. First confine x: if |x−2| < 1 then |x+2| < 5. So pick δ = min(1, ε/5). Then |x²−4| < (ε/5)·5 = ε. The trick: temporarily trap x near a so the extra factor is bounded.

5 · Proving limit laws

The ε–δ game proves the laws you used intuitively: if f and g both have limits, so do f±g, f·g, f/g (when defined), and sandwiching works. For example, |((f+g)−(L+M))| ≤ |f−L| + |g−M| < ε/2 + ε/2 by choosing δ for each. Proofs become bookkeeping, not magic — that’s their strength.

6 · Formal continuity

f is continuous at a if it is defined there and lim_{x→a} f(x) = f(a). In ε–δ terms: for every ε > 0 there is δ > 0 with 0<|x−a|<δ ⇒ |f(x)−f(a)|<ε. Same game, with L = f(a).

Watch out

Do not let δ depend on x. The moment you choose δ after seeing x, the proof collapses: the definition requires one δ that works for every x in the punctured neighborhood. Bound the stray factors first, then pick δ.

Chapter formula sheet lim_{x→a} f(x) = L ⟺ ∀ε>0 ∃δ>0: 0<|x−a|<δ ⇒ |f(x)−L|<ε
Trick: trap |x−a|≤1 to bound stray factors
Sum: |f+g −(L+M)| ≤ |f−L| + |g−M| · Product: add cross terms
Continuous at a ⟺ defined at a and limit = value
Left/right limits = one-directional games

Common mistakes

Letting δ depend on x — δ may depend on ε and a, never on the actual x you are testing.
Dropping the 0< part: x=a itself is excluded; f(a) may be anything.
Forgetting to prove the win — producing a δ without checking the implication.
Thinking the game is a single play; it must be won for every ε.