Epsilon-Delta Proofs: Linear Functions

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Textbook Reference

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Try This First: The Challenge Game

Think of the epsilon-delta definition as a challenge game. Your opponent names a tolerance $\varepsilon > 0$ (how close they want $f(x)$ to the limit $L$). You must respond with a distance $\delta > 0$ (how close $x$ must be to $a$) that guarantees the tolerance is met.

For $f(x) = 3x$ and the limit $\lim_{x \to 2} 3x = 6$:

Your opponent says: “I want $f(x)$ within 0.3 of 6.” ($\varepsilon = 0.3$.)

Predict: how close must $x$ be to 2 for $f(x)$ to be within 0.3 of 6? Try $\delta = 0.1$ and verify.

If $|x - 2| < 0.1$, then $|3x - 6| = 3|x - 2| < 3(0.1) = 0.3$. Works.

But the definition requires this for EVERY $\varepsilon > 0$, not just $\varepsilon = 0.3$. The proof finds $\delta$ in terms of $\varepsilon$.

Quantity-First Framing

The epsilon-delta definition makes “limit” mathematically precise. The intuition “as $x$ moves toward $a$, $f(x)$ moves toward $L$” is sharpened to: for every tolerance $\varepsilon > 0$ on the output, there exists a tolerance $\delta > 0$ on the input such that being within $\delta$ of $a$ (but not at $a$) guarantees being within $\varepsilon$ of $L$.

For linear functions $f(x) = mx + b$, finding $\delta$ is straightforward because $|f(x) - L|$ is directly proportional to $|x - a|$.

Prerequisite Check

• State the epsilon-delta definition: $\lim_{x \to a} f(x) = L$ means for every $\varepsilon > 0$, there exists $\delta > 0$ such that $0 < |x - a| < \delta$ implies $|f(x) - L| < \varepsilon$
• Solve $|3x - 6| < \varepsilon$ for $|x - 2|$
• Understand that $\delta$ depends on $\varepsilon$ but not on the specific $x$

Quick Reference

Proof structure for linear limits. To prove $\lim_{x \to a}(mx + b) = ma + b$:

1. Scratch work (not part of the proof): start from $|f(x) - L| < \varepsilon$ and work backwards to find what bound on $|x - a|$ is needed.
2. State $\delta$: choose $\delta = \varepsilon/|m|$ (derived from scratch work).
3. Verify: assume $0 < |x - a| < \delta$ and prove $|f(x) - L| < \varepsilon$.

Key Concepts

1. Scratch Work: Working Backwards

Before writing a proof, find what $\delta$ works. This is scratch work, not part of the formal proof.

Example 1. Prove $\lim_{x \to 2} 3x = 6$.

Scratch: Need $|3x - 6| < \varepsilon$. Factor: $|3(x - 2)| = 3|x - 2|$. So need $3|x-2| < \varepsilon$, i.e., $|x - 2| < \varepsilon/3$. Choose $\delta = \varepsilon/3$.

Formal proof:

Given $\varepsilon > 0$. Let $\delta = \varepsilon/3$.

Suppose $0 < |x - 2| < \delta$. Then: \[ |3x - 6| = |3(x-2)| = 3|x-2| < 3\delta = 3 \cdot \frac{\varepsilon}{3} = \varepsilon. \]

Since $\varepsilon > 0$ was arbitrary, $\lim_{x \to 2} 3x = 6$. $\square$

2. The General Linear Case

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

Scratch: $|f(x) - L| = |(mx + b) - (ma + b)| = |m(x-a)| = |m||x - a|$.

Need $|m||x-a| < \varepsilon$: choose $\delta = \varepsilon/|m|$.

Example 2. Prove $\lim_{x \to -1}(5x + 3) = -2$.

$L = 5(-1) + 3 = -2$.

Scratch: $|(5x+3) - (-2)| = |5x + 5| = 5|x + 1|$. Need $5|x+1| < \varepsilon$: choose $\delta = \varepsilon/5$.

Proof: Given $\varepsilon > 0$. Let $\delta = \varepsilon/5$. If $0 < |x + 1| < \delta$: \[ |(5x+3)+2| = 5|x+1| < 5\delta = \varepsilon. \] Done. $\square$

3. Two Representations: Number Line and Algebraic

Number line picture. The condition $0 < |x - a| < \delta$ means $x$ is in the interval $(a - \delta, a + \delta)$ but not equal to $a$. The conclusion $|f(x) - L| < \varepsilon$ means $f(x)$ is in the interval $(L - \varepsilon, L + \varepsilon)$.

For $\lim_{x \to 2} 3x = 6$ with $\varepsilon = 0.3$, $\delta = 0.1$:

• Input interval: $(1.9, 2.1)$ excluding $x = 2$.
• Output interval: $(5.7, 6.3)$.
• Every $x$ in $(1.9, 2.1) \setminus \{2\}$ gives $f(x) = 3x \in (5.7, 6.3)$.

Algebraic translation: the size of the output interval ($2\varepsilon$) is controlled by the size of the input interval ($2\delta$) through the slope: output-width $= |m| \times$ input-width. Since $|m| = 3$, the input interval must be $1/3$ as wide as the output interval.

4. Ask Why: Why Must $\delta$ Work for ALL $\varepsilon$?

A single $(\varepsilon, \delta)$ pair only establishes the limit for that specific tolerance. The definition requires it for every $\varepsilon > 0$ -- no matter how small the demanded tolerance, you can always find a sufficient input restriction.

If there were some $\varepsilon > 0$ for which no $\delta$ worked, the limit would fail to exist. For linear functions, $\delta = \varepsilon/|m|$ always works (for any $\varepsilon$), confirming the limit exists.

5. The Case $m = 0$: Constant Functions

For $f(x) = b$ (constant): $|f(x) - b| = 0 < \varepsilon$ for any $x$ and any $\varepsilon > 0$. So ANY $\delta > 0$ works. Choose $\delta = 1$ (or any positive number).

Named Misconception: limit-as-unreachable-barrier

The definition requires $0 < |x - a| < \delta$ (strictly: $x \neq a$). This is sometimes misread as “$x$ cannot equal $a$, so the limit is something the function never reaches at $a$.”

The correct reading: the limit asks what value $f(x)$ approaches as $x$ gets close to $a$ -- the behavior near $a$, not at $a$. The limit may equal $f(a)$ (continuity) or differ from $f(a)$ (removable discontinuity); the definition treats both cases the same way. The exclusion $x \neq a$ is a logical feature of “approaching,” not a barrier.

Common Errors

Leveled Practice

Level 1 -- Computing $\delta$

Problem 1. For $\lim_{x \to 3}(2x - 1) = 5$, find $\delta$ in terms of $\varepsilon$ and verify.

Problem 2. Prove $\lim_{x \to -2}(4 - 7x) = 18$.

Level 2 -- Writing the Proof

Problem 3. Prove $\lim_{x \to 0} cx = 0$ for any constant $c$.

Level 3 -- Low-Floor-High-Ceiling Extension

Problem 4 (Extension).

(a) (Floor) For $\lim_{x \to a}(mx + b) = ma + b$: does the choice of $\delta$ depend on $a$ and $b$? Explain.

(b) (Mid) Prove $\lim_{x \to a} x = a$ (the limit of the identity function). This is the simplest possible case.

(c) (Ceiling) A student claims: “Since $\delta = \varepsilon/3$ works for $\lim_{x \to 2} 3x = 6$, any smaller $\delta' < \varepsilon/3$ also works.” Is this correct? Explain why or why not, and what this means about uniqueness of $\delta$.

Common Misconceptions

Mastery Checklist

• State the epsilon-delta definition of the limit
• Use scratch work to find $\delta = \varepsilon/|m|$ for a linear limit
• Write a complete formal epsilon-delta proof for a linear limit
• Explain why smaller $\delta$ values also work
• Prove the limit of a constant function using any valid $\delta$

Mental Model

Epsilon-delta is the challenge game. Your opponent names any output tolerance $\varepsilon$; you name an input tolerance $\delta$ that guarantees the output is within $\varepsilon$ of $L$ whenever the input is within $\delta$ of $a$.

For linear functions, the game is easy: the slope $|m|$ converts input tolerance to output tolerance. If you restrict the input to within $\varepsilon/|m|$, the output is automatically within $\varepsilon$. The proof formalizes this: state $\delta$, show the algebra works.

Connections

Within MATH161

• Limit laws: The epsilon-delta definition provides the rigorous foundation that the limit laws rest on. The limit of a sum $= $ sum of limits is proved using epsilon-delta.
• Continuity: A function is continuous at $a$ if $\lim_{x\to a} f(x) = f(a)$; the epsilon-delta framework makes continuity precise.
• Quadratic proofs: The next lesson extends to quadratic and polynomial limits, which require one additional step (bounding $|x - a|$ to control the quadratic term).

Back to Calculus I Skills | Next: Epsilon-Delta Proofs: Quadratic Functions