Reading an epsilon-delta proof is one thing. Writing one is another. The structure is always the same: scratch work to find the right δ, then a formal proof that the chosen δ works. These examples show both phases explicitly — because skipping the scratch work is why students end up with proofs that look like they came from nowhere.
Recall the Definition
limx→a f(x) = L means:
For every ε > 0, ∃ δ > 0 such that:
0 < |x − a| < δ ⟹ |f(x) − L| < ε
The proof strategy: start with |f(x) − L|, manipulate it until you can relate it to |x − a|, then choose δ in terms of ε.
The Two-Phase Approach
Every epsilon-delta proof has two parts: scratch work (find what δ should be) and formal proof (verify the choice works). Always do the scratch work first.
Example 1 — lim(x→2) 3x = 6
📋 Prove lim(x→2) 3x = 6
Scratch|3x − 6| = 3|x − 2|. Want this < ε. Need |x−2| < ε/3. Choose δ = ε/3.
ProofGiven ε > 0, let δ = ε/3. If 0 < |x−2| < δ, then |3x−6| = 3|x−2| < 3δ = ε. ✓
Example 2 — lim(x→4) (2x+1) = 9
📋 Proof
Scratch|(2x+1)−9| = |2x−8| = 2|x−4|. Need 2|x−4| < ε → δ = ε/2
ProofLet δ = ε/2. If 0 < |x−4| < δ, then |2x+1−9| = 2|x−4| < 2δ = ε ✓
Example 3 — lim(x→3) x² = 9 (Quadratic)
📋 The quadratic case requires bounding |x+3|
Scratch|x²−9| = |x−3||x+3|. Need to bound |x+3|.
Restrict δ≤1If |x−3| < 1, then 2 < x < 4, so |x+3| < 7.
So|x²−9| < 7|x−3|. Need 7|x−3| < ε → |x−3| < ε/7.
Chooseδ = min(1, ε/7)
ProofLet δ = min(1, ε/7). If 0 < |x−3| < δ, then δ ≤ 1 so |x+3| < 7, and δ ≤ ε/7 so |x²−9| = |x−3||x+3| < (ε/7)(7) = ε ✓
Example 4 — lim(x→0) x·sin(1/x) = 0
📋 Uses the bound |sin(θ)| ≤ 1
Note|x·sin(1/x) − 0| = |x||sin(1/x)| ≤ |x|·1 = |x−0|
Chooseδ = ε
ProofIf 0 < |x| < δ = ε, then |x·sin(1/x)| ≤ |x| < ε ✓
Example 5 — lim(x→2) (x²+3x) = 10
📋 Polynomial of degree 2
Scratch|x²+3x−10| = |(x−2)(x+5)| = |x−2|·|x+5|
BoundIf δ ≤ 1: |x−2| < 1 → 1 < x < 3 → |x+5| < 8
Chooseδ = min(1, ε/8)
Proof|x²+3x−10| = |x−2||x+5| < (ε/8)·8 = ε ✓
Example 6 — lim(x→1) (1/x) = 1
📋 Rational function — must keep x away from 0
Scratch|1/x − 1| = |1−x|/|x| = |x−1|/|x|. Need to bound 1/|x| away from 0.
Restrict δ≤½|x−1| < ½ → ½ < x < 3/2 → 1/|x| < 2
Chooseδ = min(½, ε/2)
Proof|1/x−1| = |x−1|/|x| < 2|x−1| < 2·(ε/2) = ε ✓
Frequently Asked Questions
Why do we restrict δ ≤ 1 in polynomial proofs?
We need to bound the extra factor (like |x+3| in x²−9). Restricting δ ≤ 1 puts x in a bounded interval where we can compute a numerical bound for the extra factor. The δ = min(1, ε/M) pattern ensures both restrictions are satisfied simultaneously.
How do I know what to restrict δ to?
The standard choice is δ ≤ 1, which works for most polynomial limits. For limits at other values of a, you might use δ ≤ |a|/2 to keep x bounded away from zero (needed for rational functions). The key is picking a restriction that lets you bound the "extra" factor.