Calculus I · Limits and Continuity · lesson

Epsilon-Delta Proofs for Linear Functions

Learn Epsilon-Delta Proofs for Linear Functions with plain-language explanations, guided examples, worked homework methods, interactive checks, and exam-style p

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Where this chapter fits

Chapter 6: Formal limits

Translate the intuitive neighborhood picture into epsilon-delta language, constructive proofs, graph windows, and counterexamples.

Reading lens: How small must the input window be to force every allowed output into the requested tolerance band? Keep that question in view while reading Epsilon-Delta Proofs for Linear Functions; the worked mathematics is evidence for the idea, not a substitute for it.

This page connects Epsilon-Delta Definition: An Introduction to Epsilon-Delta Proofs for Quadratic Functions. Read the explanation first, predict each example’s next move, and only then compare the written solution.

Concept

Learning objectives

Construct a $\delta$ directly from a requested $\varepsilon$ for a linear function.

Linear Functions: The Cleanest Proofs

Guided walkthrough

$f(x)=2x$

Prove

\lim_{x\to3}2x=6.

Show worked solution

We want the output error to be less than $\varepsilon$ :

|2x-6|<\varepsilon.

Factor:

|2x-6|=2|x-3|.

So it is enough to require

2|x-3|<\varepsilon.

Divide by $2$ :

|x-3|<\frac\eps2.

Choose

\boxed{\delta=\frac\eps2}.

Then if $0<|x-3|<\delta$ ,

|2x-6|=2|x-3|<2\delta=2\left(\frac\eps2\right)=\varepsilon.

Therefore,

\lim_{x\to3}2x=6.

Interactive checkepsilon-flow-01

For $f(x)=2x$ at $x=3$ , choose $\delta$ in terms of $\varepsilon$ .

Your work stays on this device. No account or AI grader is used.

Show hint

$|2x-6|=2|x-3|$ .

Attempt once to unlock the solution

Submit an answer first. The hint is available now.

This proof has two phases.

• Discovery phase. Start from the desired output inequality and work backward to guess a useful $\delta$ . • Proof phase. State the chosen $\delta$ , assume the input inequality, and work forward to the output inequality.

Worked example

A general linear function

Prove

\lim_{x\to2}(3x+1)=7.

Show worked solution

Discovery. We need

|(3x+1)-7|<\varepsilon.

Simplify:

|3x-6|=3|x-2|.

Thus, requiring

|x-2|<\frac\eps3

is enough.

Proof. Let $\varepsilon>0$ . Choose

\delta=\frac\eps3.

0<|x-2|<\delta,

then

|(3x+1)-7|=3|x-2|<3\delta=\varepsilon.

Therefore,

\boxed{\lim_{x\to2}(3x+1)=7}.

Summary

For $f(x)=mx+b$ , a standard choice is

\delta=\frac\varepsilon{|m|}

when $m\ne0$ . The output changes $|m|$ times as much as the input, so the input tolerance must be $|m|$ times smaller.

Source & rights

Original instruction with traceable references.

The exposition is original. No Active Calculus exercise is reproduced verbatim. Public-domain examples were modernized and recomposed when used as inspiration.

The verified handoff declares original composition and requires owner provenance review. BetterGrades-original material remains separate from public-domain references; no source textbook PDF is published here.

Chapter 6: Formal limits

Translate the intuitive neighborhood picture into epsilon-delta language, constructive proofs, graph windows, and counterexamples.

Linear Functions: The Cleanest Proofs

Guided walkthrough

$f(x)=2x$

Prove

\lim_{x\to3}2x=6.

Show worked solution

We want the output error to be less than $\varepsilon$ :

|2x-6|<\varepsilon.

Factor:

|2x-6|=2|x-3|.

So it is enough to require

2|x-3|<\varepsilon.

Divide by $2$ :

|x-3|<\frac\eps2.

Choose

\boxed{\delta=\frac\eps2}.

Then if $0<|x-3|<\delta$ ,

|2x-6|=2|x-3|<2\delta=2\left(\frac\eps2\right)=\varepsilon.

Therefore,

\lim_{x\to3}2x=6.

Interactive checkepsilon-flow-01

For $f(x)=2x$ at $x=3$ , choose $\delta$ in terms of $\varepsilon$ .

Your work stays on this device. No account or AI grader is used.

Show hint

$|2x-6|=2|x-3|$ .

Attempt once to unlock the solution

Submit an answer first. The hint is available now.

This proof has two phases.

• Discovery phase. Start from the desired output inequality and work backward to guess a useful

\delta

. • Proof phase. State the chosen

\delta

, assume the input inequality, and work forward to the output inequality.

Worked example

A general linear function

Prove

\lim_{x\to2}(3x+1)=7.

Show worked solution

Discovery. We need

|(3x+1)-7|<\varepsilon.

Simplify:

|3x-6|=3|x-2|.

Thus, requiring

|x-2|<\frac\eps3

is enough.

Proof. Let $\varepsilon>0$ . Choose

\delta=\frac\eps3.

0<|x-2|<\delta,

then

|(3x+1)-7|=3|x-2|<3\delta=\varepsilon.

Therefore,

\boxed{\lim_{x\to2}(3x+1)=7}.

Summary

For $f(x)=mx+b$ , a standard choice is

\delta=\frac\varepsilon{|m|}

when $m\ne0$ . The output changes $|m|$ times as much as the input, so the input tolerance must be $|m|$ times smaller.

Source & rights

Original instruction with traceable references.

The exposition is original. No Active Calculus exercise is reproduced verbatim. Public-domain examples were modernized and recomposed when used as inspiration.

Epsilon-Delta Proofs for Linear Functions

Chapter 6: Formal limits

Learning objectives

Linear Functions: The Cleanest Proofs

\(f(x)=2x\)

A general linear function

Original instruction with traceable references.

Epsilon-Delta Proofs for Linear Functions

Chapter 6: Formal limits

Learning objectives

Linear Functions: The Cleanest Proofs

\(f(x)=2x\)

A general linear function

Original instruction with traceable references.

Chapter 6: Formal limits

Linear Functions: The Cleanest Proofs

Related resources

Original instruction with traceable references.

Chapter 6: Formal limits

Linear Functions: The Cleanest Proofs

Related resources

Original instruction with traceable references.