Epsilon-Delta Proofs for Linear Functions

MATH161

Reference: Stewart 2.4 • Chapter: 1 • Section: 4

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Epsilon-Delta Proofs for Linear Functions

From Definition to Proof

You know what the epsilon-delta definition says—but how do you prove a specific limit using it? Linear functions provide the simplest starting point because the relationship between $\delta$ and $\varepsilon$ is direct and transparent.

The strategy is called "guess and verify":

Work backwards from what you need ($\vert f(x) - L\vert < \varepsilon$) to figure out what $\delta$ should be
Write the formal proof going forwards, showing your $\delta$ works

This two-step dance is the foundation of all epsilon-delta proofs.

Prerequisite Map

Quick Reference

Property	Value
Course	MATH161
Chapter.Section	1.7
Difficulty	Intermediate
Time	~15 minutes

Key Concepts

The Guess-and-Verify Strategy

Goal: Prove $\lim_{x \to a} f(x) = L$ using epsilon-delta.

Step 1 (The Guess): Start with $\vert f(x) - L\vert < \varepsilon$ and solve for $\vert x - a\vert $.

For linear $f(x) = mx + b$: $$\vert f(x) - L\vert = \vert (mx + b) - L\vert = \vert m\vert \vert x - a\vert $$

(where $L = ma + b$)

We need $\vert m\vert \vert x - a\vert < \varepsilon$, so $\vert x - a\vert < \frac{\varepsilon}{\vert m\vert }$.

Guess: $\delta = \frac{\varepsilon}{\vert m\vert }$

Step 2 (The Verification): Write the formal proof showing this $\delta$ works.

Proof Template for Linear Functions

Claim: $\lim_{x \to a} (mx + b) = ma + b$ for $m \neq 0$.

Proof:

Let $\varepsilon > 0$ be given. Choose $\delta = \frac{\varepsilon}{\vert m\vert }$.

Suppose $0 < \vert x - a\vert < \delta$. Then:

$$\begin{aligned} \vert (mx + b) - (ma + b)\vert &= \vert mx - ma\vert \\ &= \vert m(x - a)\vert \\ &= \vert m\vert \cdot \vert x - a\vert \\ &< \vert m\vert \cdot \delta \\ &= \vert m\vert \cdot \frac{\varepsilon}{\vert m\vert } \\ &= \varepsilon \end{aligned}$$

Therefore $\lim_{x \to a} (mx + b) = ma + b$. $\square$

Worked Example: $\lim_{x \to 3} (4x - 5) = 7$

The Guess (scratch work):

We need $\vert (4x - 5) - 7\vert < \varepsilon$. $$\vert (4x - 5) - 7\vert = \vert 4x - 12\vert = 4\vert x - 3\vert $$

So we need $4\vert x - 3\vert < \varepsilon$, i.e., $\vert x - 3\vert < \frac{\varepsilon}{4}$.

Guess: $\delta = \frac{\varepsilon}{4}$

The Verification (formal proof):

Proof: Let $\varepsilon > 0$ be given. Choose $\delta = \frac{\varepsilon}{4}$.

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

$$\vert (4x - 5) - 7\vert = \vert 4x - 12\vert = 4\vert x - 3\vert < 4 \cdot \frac{\varepsilon}{4} = \varepsilon$$

Therefore $\lim_{x \to 3} (4x - 5) = 7$. $\square$

The Special Case: Constant Functions

For $f(x) = c$ (constant), we have $\vert f(x) - c\vert = 0 < \varepsilon$ for any $x$.

So any $\delta > 0$ works. We can simply choose $\delta = 1$ (or any positive number).

The Special Case: Identity Function

For $f(x) = x$ with $\lim_{x \to a} x = a$: $$\vert f(x) - a\vert = \vert x - a\vert < \varepsilon$$

Choose $\delta = \varepsilon$. The "slope" is 1, so $\delta = \varepsilon/1 = \varepsilon$.

Practice Problems

Level 1 Finding Delta for a Given Epsilon

For the limit $\lim_{x \to 2} (3x + 1) = 7$, find a $\delta$ that works when $\varepsilon = 0.06$.

Thought Process

This is a linear function with slope $m = 3$. We have: $$\vert (3x + 1) - 7\vert = \vert 3x - 6\vert = 3\vert x - 2\vert $$

We need $3\vert x - 2\vert < 0.06$, so $\vert x - 2\vert < 0.02$.

Show Answer

$\delta = \frac{\varepsilon}{3} = \frac{0.06}{3} = 0.02$

Verification: If $\vert x - 2\vert < 0.02$, then $\vert (3x + 1) - 7\vert = 3\vert x - 2\vert < 3(0.02) = 0.06 = \varepsilon$. ✓

Level 2 Complete Proof: Positive Slope

Prove that $\lim_{x \to 5} (2x + 3) = 13$ using the epsilon-delta definition.

Thought Process

Scratch work:

$f(x) = 2x + 3$, $a = 5$, $L = 13$
$\vert f(x) - L\vert = \vert (2x + 3) - 13\vert = \vert 2x - 10\vert = 2\vert x - 5\vert $
Need $2\vert x - 5\vert < \varepsilon$, so $\vert x - 5\vert < \varepsilon/2$
Choose $\delta = \varepsilon/2$

Now write the formal proof.

Show Answer

Proof: Let $\varepsilon > 0$ be given. Choose $\delta = \frac{\varepsilon}{2}$.

Suppose $0 < \vert x - 5\vert < \delta$. Then:

$$\begin{aligned} \vert (2x + 3) - 13\vert &= \vert 2x - 10\vert \\ &= 2\vert x - 5\vert \\ &< 2 \cdot \frac{\varepsilon}{2} \\ &= \varepsilon \end{aligned}$$

Therefore $\lim_{x \to 5} (2x + 3) = 13$. $\square$

Level 3 Complete Proof: Negative Slope

Prove that $\lim_{x \to -1} (7 - 3x) = 10$ using the epsilon-delta definition.

Thought Process

Scratch work:

$f(x) = 7 - 3x = -3x + 7$, $a = -1$, $L = 10$
Check: $f(-1) = 7 - 3(-1) = 10$ ✓
$\vert f(x) - L\vert = \vert (7 - 3x) - 10\vert = \vert -3x - 3\vert = \vert -3\vert \vert x - (-1)\vert = 3\vert x + 1\vert $
Need $3\vert x + 1\vert < \varepsilon$, so $\vert x + 1\vert < \varepsilon/3$
Choose $\delta = \varepsilon/3$

The slope is $m = -3$, so $\vert m\vert = 3$.

Show Answer

Proof: Let $\varepsilon > 0$ be given. Choose $\delta = \frac{\varepsilon}{3}$.

Suppose $0 < \vert x - (-1)\vert < \delta$, i.e., $0 < \vert x + 1\vert < \delta$. Then:

$$\begin{aligned} \vert (7 - 3x) - 10\vert &= \vert -3x - 3\vert \\ &= \vert -3\vert \cdot \vert x + 1\vert \\ &= 3\vert x + 1\vert \\ &< 3 \cdot \frac{\varepsilon}{3} \\ &= \varepsilon \end{aligned}$$

Therefore $\lim_{x \to -1} (7 - 3x) = 10$. $\square$

Level 4 General Linear Proof

Let $f(x) = mx + b$ where $m \neq 0$. Prove that $\lim_{x \to a} f(x) = ma + b$ for any $a \in \mathbb{R}$.

(This is the general template for all linear functions.)

Thought Process

Work with general $m$, $b$, and $a$:

$L = ma + b$
$\vert f(x) - L\vert = \vert (mx + b) - (ma + b)\vert = \vert m(x - a)\vert = \vert m\vert \vert x - a\vert $
Need $\vert m\vert \vert x - a\vert < \varepsilon$
Since $m \neq 0$, we have $\vert m\vert > 0$, so $\vert x - a\vert < \varepsilon/\vert m\vert $
Choose $\delta = \varepsilon/\vert m\vert $

Show Answer

Proof: Let $\varepsilon > 0$ be given. Choose $\delta = \frac{\varepsilon}{\vert m\vert }$.

Suppose $0 < \vert x - a\vert < \delta$. Then:

$$\begin{aligned} \vert (mx + b) - (ma + b)\vert &= \vert mx - ma\vert \\ &= \vert m\vert \vert x - a\vert \\ &< \vert m\vert \cdot \frac{\varepsilon}{\vert m\vert } \\ &= \varepsilon \end{aligned}$$

Therefore $\lim_{x \to a} (mx + b) = ma + b$. $\square$

Note: The case $m = 0$ gives a constant function where any $\delta$ works.

Level 5 When the Naive Delta Fails

Consider $f(x) = \frac{x^2 - 4}{x - 2}$ for $x \neq 2$.

Simplify $f(x)$ for $x \neq 2$.
What is $\lim_{x \to 2} f(x)$?
A student writes: "$f(x) = \frac{x^2 - 4}{x - 2}$ is not linear, so the $\delta = \varepsilon/\vert m\vert $ formula doesn't apply." Is this correct? Prove the limit using epsilon-delta.
Explain why this function behaves like a linear function near $x = 2$.

Thought Process

For (a): Factor the numerator: $x^2 - 4 = (x-2)(x+2)$.

For (b): After canceling, $f(x) = x + 2$ for $x \neq 2$, so the limit should be $4$.

For (c): Since we're looking at $x \neq 2$ (the limit condition excludes $x = 2$), we can work with the simplified form $x + 2$.

For (d): The "hole" at $x = 2$ doesn't affect the limit because we exclude that point anyway.

Show Answer

For $x \neq 2$: $$f(x) = \frac{x^2 - 4}{x - 2} = \frac{(x-2)(x+2)}{x-2} = x + 2$$
$\lim_{x \to 2} f(x) = \lim_{x \to 2} (x + 2) = 4$
The student's concern is valid in form but not in substance. Here's the proof:
Proof: Let $\varepsilon > 0$ be given. Choose $\delta = \varepsilon$.

Suppose $0 < \vert x - 2\vert < \delta$. Since $x \neq 2$, we have:

$$\left\vert \frac{x^2 - 4}{x - 2} - 4\right\vert = \vert x + 2 - 4\vert = \vert x - 2\vert < \delta = \varepsilon$$

Therefore $\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = 4$. $\square$</li>
Near $x = 2$ (but not at $x = 2$), the function $\frac{x^2-4}{x-2}$ equals $x + 2$, which is linear with slope 1. The epsilon-delta definition only considers behavior near the point, not at it. So for limit purposes, this function is indistinguishable from the linear function $g(x) = x + 2$.

Common Mistakes

Mistake	Why It's Wrong	Correct Approach
Forgetting absolute value on $m$	$\delta$ must be positive	Use $\\|m\\|$ in denominator
Writing $\delta = \varepsilon/m$ when $m < 0$	This gives negative $\delta$	Use $\delta = \varepsilon/\\|m\\|$
Proving in the wrong direction	Starting with conclusion	Start with "Let $\varepsilon > 0$..."
Not showing $\delta > 0$	$\delta$ must be positive by definition	Since $\varepsilon > 0$ and $\\|m\\| > 0$, $\delta = \varepsilon/\\|m\\| > 0$

Mastery Checklist

[ ] I can perform the scratch work to find $\delta$ in terms of $\varepsilon$
[ ] I can write a complete formal epsilon-delta proof for a linear function
[ ] I know that $\delta = \varepsilon/\vert m\vert $ for $f(x) = mx + b$ with $m \neq 0$
[ ] I can handle the special case of constant functions (any $\delta$ works)
[ ] I can handle functions that simplify to linear functions near the limit point

Mental Model

The Scaling Factor:

For a linear function $f(x) = mx + b$:

The slope $m$ acts as a magnifier (or reducer) between input and output changes
If $\vert m\vert = 3$, then moving $x$ by 0.01 moves $f(x)$ by 0.03
To keep $f(x)$ within $\varepsilon$ of $L$, you need $x$ within $\varepsilon/\vert m\vert $ of $a$
Steeper slopes require tighter $\delta$ (smaller input window) for the same $\varepsilon$

Connections

Looking back:

Epsilon-Delta Definition gave us the framework; now we apply it.
The factoring $\vert mx + b - (ma + b)\vert = \vert m\vert \vert x - a\vert $ is why linear functions are the simplest case.

Looking ahead:

Polynomial Proofs require the "min" technique because the bound depends on more than just $\vert x - a\vert $.
These proofs form the basis for proving continuity of linear functions.

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Epsilon-Delta Definition	Skills Index	Epsilon-Delta Proofs: Polynomials

Last updated: 2026-01-22