Calculus I · Limits and Continuity · lesson

Epsilon-Delta Proofs for Quadratic Functions

Concept

Learning objectives

Use a preliminary restriction such as ∣x−a∣<1|x-a|<1 to bound a factor that depends on xx; choose δ\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 xx.

Worked example
Show worked solution

We need to make

∣x2−4∣|x^2-4|

small. Factor:

∣x2−4∣=∣x−2∣∣x+2∣.|x^2-4|=|x-2||x+2|.

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

∣x−2∣<1.|x-2|<1.

Then

1<x<3,1<x<3,

so

3<x+2<5,3<x+2<5,

and therefore

∣x+2∣<5.|x+2|<5.

Now

∣x2−4∣=∣x−2∣∣x+2∣<5∣x−2∣.|x^2-4|=|x-2||x+2|<5|x-2|.

To make this less than ε\varepsilon, require

∣x−2∣<ε5.|x-2|<\frac\varepsilon5.

We need both restrictions, so choose

δ=min⁡{1,ε5}.\boxed{\delta=\min\left\{1,\frac\varepsilon5\right\}}.

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

0<∣x−2∣<δ,0<|x-2|<\delta,

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

∣x2−4∣=∣x−2∣∣x+2∣<5(ε5)=ε.|x^2-4| =|x-2||x+2| <5\left(\frac\varepsilon5\right) =\varepsilon.

Therefore,

lim⁡x→2x2=4.\boxed{\lim_{x\to2}x^2=4}.
Concept

The minimum notation means, "Use whichever restriction is smaller." If ε/5\varepsilon/5 is tiny, use it. If ε/5\varepsilon/5 is larger than 11, keep the local bound 11. Both jobs must be done at once.

Worked example

A shifted square

Prove

lim⁡x→1x2=1.\lim_{x\to1}x^2=1.
Show worked solution

Factor the output error:

∣x2−1∣=∣x−1∣∣x+1∣.|x^2-1|=|x-1||x+1|.

If ∣x−1∣<1|x-1|<1, then 0<x<20<x<2, so ∣x+1∣<3|x+1|<3. Therefore,

∣x2−1∣<3∣x−1∣.|x^2-1|<3|x-1|.

Choose

δ=min⁡{1,ε3}.\boxed{\delta=\min\left\{1,\frac\varepsilon3\right\}}.

Then

∣x2−1∣<3δ≤ε.|x^2-1|<3\delta\le\varepsilon.

After the explanation

Use the section idea

Reading lens

How small must the input window be to force every allowed output into the requested tolerance band?

Mental model

Epsilon sets the demanded vertical accuracy; delta is the horizontal promise you choose so every permitted nearby input meets that demand.

Decision

Work backward from the desired output inequality, isolate an input-distance bound, then state a positive delta that is no larger than that bound.

Common trap

A proof must control every eligible input in the punctured window; checking examples or choosing delta after seeing the input is not enough.

Check yourself

Formal understanding means you can translate between bands, inequalities, and words, then verify the implication from delta to epsilon in forward order.

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.

Vocab
Limit
Math glossaryLimit
lim⁡x→af(x)=L\lim_{x\to a}f(x)=L

The value a function approaches as its input approaches a target.

Learn more
Function
Math glossaryFunction
y=f(x)y=f(x)

A rule or relation assigning exactly one output to each allowed input.

Learn more
Continuity
Math glossaryContinuity
lim⁡x→af(x)=f(a)\lim_{x\to a}f(x)=f(a)

At a point, the function value exists and equals the limit there.

Learn more
Domain
Math glossaryDomain
dom⁡f\operatorname{dom}f

The set of permitted input values.

Learn more
Math glossary