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

> The formal definition of a limit — the ε–δ game — and how it proves the limit laws you have been using all along.

**Canonical:** https://calculus.kitzo.work/chapter/epsilon-delta  
**Level:** Advanced  
**Estimated time:** 40 minutes  
**Exercises:** 8 (auto-marked; answers stay server-side)

## Topics
- The ε–δ definition
- The limit game
- Formal limit examples
- Proving limit laws
- Formal continuity

**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.

## Practice exercises

1. The ε–δ definition quantifies over:

- A. every ε > 0 and some δ > 0
- B. one fixed ε only
- C. all x equally spaced
- D. finite δ = 1

2. In the game, δ may depend on:

- A. ε (and a)
- B. x
- C. f(x)
- D. nothing

3. A standard starting move in ε–δ proofs of polynomial limits is to restrict:

4. To prove lim_{x→2} (2x+1) = 5, given ε choose δ =

- A. ε/2
- B. ε
- C. 2ε
- D. 1

5. In the proof of lim_{x→2} (2x+1)=5 with δ=ε/2, if ε=0.1, what is δ?

6. Continuity at a in ε–δ terms is:

- A. defined at a and lim_{x→a} f(x) = f(a)
- B. monotone near a
- C. derivative exists at a
- D. ε = δ

7. The full chain of words: lim f = L means _____ ε>0 there is δ>0 such that …

8. The function f(x)=x² has limit 4 at x=2; a typical δ choice is:

- A. min(1, ε/5)
- B. ε
- C. always 1
- D. 10

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