Calculus I · Limits and Continuity · lesson

Epsilon-Delta Proofs for Quadratic Functions

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

Course progress44 of 47 core pages44 of 47

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 Quadratic Functions; the worked mathematics is evidence for the idea, not a substitute for it.

This page connects Epsilon-Delta Proofs for Linear Functions to Reading Epsilon and Delta From a Graph. Read the explanation first, predict each example’s next move, and only then compare the written solution.

Concept

Learning objectives

Use a preliminary restriction such as $|x-a|<1$ to bound a factor that depends on $x$ ; choose $\delta$ as the minimum of two requirements.

Nonlinear Functions: Controlling an Extra Factor

For a quadratic, factoring the output error creates an extra factor involving $x$ .

Worked example

Show worked solution

We need to make

|x^2-4|

small. Factor:

|x^2-4|=|x-2||x+2|.

The first factor is controlled by $\delta$ , but $|x+2|$ still depends on $x$ . First impose the simple restriction

|x-2|<1.

Then

1<x<3,

3<x+2<5,

and therefore

|x+2|<5.

Now

|x^2-4|=|x-2||x+2|<5|x-2|.

To make this less than $\varepsilon$ , require

|x-2|<\frac\eps5.

We need both restrictions, so choose

\boxed{\delta=\min\left\{1,\frac\eps5\right\}}.

Now write the forward proof. Let $\varepsilon>0$ , and choose the $\delta$ above. If

0<|x-2|<\delta,

then $|x-2|<1$ , so $|x+2|<5$ . Also, $|x-2|<\varepsilon/5$ . Hence

|x^2-4| =|x-2||x+2| <5\left(\frac\eps5\right) =\varepsilon.

Therefore,

\boxed{\lim_{x\to2}x^2=4}.

Concept

The minimum notation means, "Use whichever restriction is smaller." If $\varepsilon/5$ is tiny, use it. If $\varepsilon/5$ is larger than $1$ , keep the local bound $1$ . Both jobs must be done at once.

Worked example

A shifted square

Prove

\lim_{x\to1}x^2=1.

Show worked solution

Factor the output error:

|x^2-1|=|x-1||x+1|.

If $|x-1|<1$ , then $0<x<2$ , so $|x+1|<3$ . Therefore,

|x^2-1|<3|x-1|.

Choose

\boxed{\delta=\min\left\{1,\frac\eps3\right\}}.

Then

|x^2-1|<3\delta\le\varepsilon.

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.

Nonlinear Functions: Controlling an Extra Factor

For a quadratic, factoring the output error creates an extra factor involving

x

Worked example

Show worked solution

We need to make

|x^2-4|

small. Factor:

|x^2-4|=|x-2||x+2|.

The first factor is controlled by $\delta$ , but $|x+2|$ still depends on $x$ . First impose the simple restriction

|x-2|<1.

Then

1<x<3,

3<x+2<5,

and therefore

|x+2|<5.

Now

|x^2-4|=|x-2||x+2|<5|x-2|.

To make this less than $\varepsilon$ , require

|x-2|<\frac\eps5.

We need both restrictions, so choose

\boxed{\delta=\min\left\{1,\frac\eps5\right\}}.

Now write the forward proof. Let $\varepsilon>0$ , and choose the $\delta$ above. If

0<|x-2|<\delta,

then $|x-2|<1$ , so $|x+2|<5$ . Also, $|x-2|<\varepsilon/5$ . Hence

|x^2-4| =|x-2||x+2| <5\left(\frac\eps5\right) =\varepsilon.

Therefore,

\boxed{\lim_{x\to2}x^2=4}.

Concept

Worked example

A shifted square

Prove

\lim_{x\to1}x^2=1.

Show worked solution

Factor the output error:

|x^2-1|=|x-1||x+1|.

If $|x-1|<1$ , then $0<x<2$ , so $|x+1|<3$ . Therefore,

|x^2-1|<3|x-1|.

Choose

\boxed{\delta=\min\left\{1,\frac\eps3\right\}}.

Then

|x^2-1|<3\delta\le\varepsilon.

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.

Chapter 6: Formal limits

Nonlinear Functions: Controlling an Extra Factor

Related resources

Original instruction with traceable references.

Chapter 6: Formal limits

Nonlinear Functions: Controlling an Extra Factor

Related resources

Original instruction with traceable references.