Recipe

Prove a limit of a linear function with epsilon-delta

Use when: you must prove $\lim_{x\to a}(mx+b)=L$ from the definition, and $f$ is a straight line

Goal: a complete written proof: a formula for delta in terms of epsilon, plus a forward chain from the assumption to the tolerance

Taught in Writing an ε-δ proof · a later section extends this recipe with more cases

Quick recipe

Name a, L, and f(x)
Scratch work, simplify the output distance
Scratch work, solve for the window and choose delta
Open the proof with the three setup lines
Run the chain forward to epsilon
Conclude

1
Read the three numbers off the limit statement before doing any algebra.

Example

In $\lim_{x\to2}(3x+1)=7$ : $a=2$ , $L=7$ , $f(x)=3x+1$ .

Why: every later step uses these three, and mixing up $a$ and $L$ is the fastest way to prove the wrong statement.
2
Write $|f(x)-L|$ , combine the constants, and factor until it reads (a positive constant) times $|x-a|$ .

Example

$|(3x+1)-7|=|3x-6|=|3(x-2)|=3|x-2|$ .
- Not yet: a factor that still depends on x is left over — covered in ε-δ proofs beyond straight lines
Why: the tolerance is about the output and the window is about the input, so this rewrite is the bridge between them.
3
Require (constant) $\cdot|x-a|<\varepsilon$ , divide by the constant, and choose $\delta$ to be the result.

Example

$3|x-2|<\varepsilon$ exactly when $|x-2|<\varepsilon/3$ , so choose $\delta=\varepsilon/3$ .

Why: dividing by a positive constant keeps the inequality's direction, and any smaller positive delta would also be acceptable.
4
Write "Let $\varepsilon>0$ be given. Choose $\delta=\ldots$ . Suppose $0<|x-a|<\delta$ ."

Why: each line answers one phrase of the definition, in the definition's own order: every epsilon, some delta, then if-then.
5
Starting from $|f(x)-L|$ , rewrite it as in the scratch work, then use $|x-a|<\delta$ and substitute your $\delta$ until the chain ends at $\varepsilon$ .

Example

$|(3x+1)-7|=3|x-2|<3\delta=3\cdot\dfrac{\varepsilon}{3}=\varepsilon$ .

Why: this chain is the actual proof; it shows the chosen delta forces the tolerance for every input in the window.
6
Write "Since $\varepsilon>0$ was arbitrary, $\therefore\ \lim_{x\to a}f(x)=L$ ."

Why: the conclusion is only earned once the line before it says $|f(x)-L|<\varepsilon$ .

Common mistakes

Presenting the backward scratch work as if it were the proof.
Writing "Choose $\delta=\varepsilon/3$ " and then concluding without the chain that shows it works.
Swapping the roles: "Let $\delta>0$ be given, choose $\varepsilon$ ."
Dropping the absolute value bars partway through the chain.
Using the slope $m$ instead of its size $|m|$ when the slope is negative.