Epsilon-Delta Proofs for Polynomial Functions

5 min read

The New Challenge

For linear functions, $\vert f(x) - L\vert $ was simply a constant times $\vert x - a\vert $. Quadratics are trickier: when you factor $x^2 - a^2 = (x-a)(x+a)$, you get two factors that both change as $x$ moves.

The $(x - a)$ factor is small when $x$ is near $a$, which is good. But what about $(x + a)$? If we do not control it, we cannot bound the product.

The solution: first restrict $\vert x - a\vert $ to keep $(x + a)$ bounded, then choose $\delta$ to make the whole expression small. This leads to the famous $\delta = \min\{1, \varepsilon/K\}$ pattern.

Prerequisite Map

Quick Reference

Key Concepts

The Min Technique: Overview

The problem: For $\lim_{x \to a} x^2 = a^2$, we have: $$\vert x^2 - a^2\vert = \vert x - a\vert \cdot \vert x + a\vert $$

The factor $\vert x - a\vert $ is what we control. But $\vert x + a\vert $ could be arbitrarily large if we don’t restrict $x$.

The solution: A two-phase approach:

1. Phase 1: Assume $\vert x - a\vert < 1$ (an initial restriction). This bounds $\vert x + a\vert $.
2. Phase 2: With $\vert x + a\vert $ bounded by some constant $K$, choose $\delta$ so that $K\vert x - a\vert < \varepsilon$.

Final choice: $\delta = \min\{1, \varepsilon/K\}$ ensures both conditions are satisfied.

Why We Need the Minimum

The choice $\delta = \min\{1, \varepsilon/K\}$ ensures:

1. $\delta \leq 1$: So our bound on $\vert x + a\vert $ remains valid
2. $\delta \leq \varepsilon/K$: So $\vert x^2 - a^2\vert < K \cdot \delta \leq \varepsilon$

If we only used $\delta = \varepsilon/K$, that could be larger than 1 when $\varepsilon$ is large, invalidating our bound. Taking the minimum covers all cases.

Worked Example: $\lim_{x \to 3} x^2 = 9$

Goal: Prove that for every $\varepsilon > 0$, there exists $\delta > 0$ such that $0 < \vert x - 3\vert < \delta$ implies $\vert x^2 - 9\vert < \varepsilon$.

Phase 1: Bound the wandering factor

Factor: $\vert x^2 - 9\vert = \vert x - 3\vert \cdot \vert x + 3\vert $

Assume $\vert x - 3\vert < 1$. Then:

• $-1 < x - 3 < 1$
• $2 < x < 4$
• $5 < x + 3 < 7$
• So $\vert x + 3\vert < 7$

Phase 2: Make the product small

With $\vert x + 3\vert < 7$: $$\vert x^2 - 9\vert = \vert x - 3\vert \cdot \vert x + 3\vert < 7\vert x - 3\vert $$

We need $7\vert x - 3\vert < \varepsilon$, so $\vert x - 3\vert < \varepsilon/7$.

Final choice: $\delta = \min\{1, \varepsilon/7\}$

The formal proof:

Proof: Let $\varepsilon > 0$ be given. Choose $\delta = \min\{1, \varepsilon/7\}$.

Suppose $0 < \vert x - 3\vert < \delta$.

Since $\delta \leq 1$, we have $\vert x - 3\vert < 1$, which gives $2 < x < 4$, so $\vert x + 3\vert < 7$.

Then: $$\vert x^2 - 9\vert = \vert x - 3\vert \cdot \vert x + 3\vert < \vert x - 3\vert \cdot 7 < \delta \cdot 7 \leq \frac{\varepsilon}{7} \cdot 7 = \varepsilon$$

Therefore $\lim_{x \to 3} x^2 = 9$. $\square$

General Pattern for $\lim_{x \to a} x^2 = a^2$

Assume $\vert x - a\vert < 1$:

• $a - 1 < x < a + 1$
• $\vert x + a\vert < \vert a + 1\vert + \vert a\vert = \vert 2a + 1\vert $ or more simply, $\vert x + a\vert < 2\vert a\vert + 1$

Actually, a cleaner bound: if $\vert x - a\vert < 1$, then $\vert x\vert < \vert a\vert + 1$, so: $$\vert x + a\vert \leq \vert x\vert + \vert a\vert < (\vert a\vert + 1) + \vert a\vert = 2\vert a\vert + 1$$

Choose $\delta = \min\{1, \varepsilon/(2\vert a\vert + 1)\}$.

Extension: Cubic and Higher Polynomials

For $\lim_{x \to a} x^3 = a^3$, we have: $$\vert x^3 - a^3\vert = \vert x - a\vert \cdot \vert x^2 + xa + a^2\vert $$

The same technique applies:

1. Assume $\vert x - a\vert < 1$ to bound $\vert x^2 + xa + a^2\vert $ by some constant $K$
2. Choose $\delta = \min\{1, \varepsilon/K\}$

Practice Problems

The Min Technique: Summary

Common Mistakes

Common Misconceptions

Mastery Checklist

• I understand why linear proof techniques don’t directly work for $x^2$
• I can explain the two-phase approach (bound first, then choose $\delta$)
• I can bound $\vert x + a\vert $ by assuming $\vert x - a\vert < 1$
• I know why we use $\min\{1, \varepsilon/K\}$ instead of just $\varepsilon/K$
• I can write complete proofs for quadratic limits
• I can extend the technique to higher-degree polynomials

Mental Model

The Two-Step Dance:

Think of epsilon-delta proofs for polynomials as a negotiation:

1. First constraint: “I’ll only consider $x$ within 1 of $a$.” This keeps the non-linear parts under control.

2. Second constraint: “Within that region, I need $x$ even closer, within $\varepsilon/K$ of $a$.” This makes the product small.

3. Taking the minimum: You need both constraints satisfied simultaneously. The min ensures neither is violated.

Connections

Looking back:

• Linear Proofs had $\delta = \varepsilon/\vert m\vert $ with no minimum needed because there’s only one factor.
• The factoring $(x^2 - a^2) = (x-a)(x+a)$ is key. Without it, we cannot isolate $\vert x - a\vert $.

Looking ahead:

• This technique proves the Product Limit Law: $\lim (fg) = (\lim f)(\lim g)$.
• The same idea bounds more complex expressions in continuity proofs.

Last updated: 2026-01-22