Recipe

Prove a limit of a nonlinear function with epsilon-delta

Use when: the output distance $|f(x)-L|$ factors as $|x-a|$ times a leftover factor that still depends on $x$ , as for polynomials, square roots, and fractions

Goal: a complete written proof with $\delta=\min(r,\varepsilon/M)$ , where $r$ keeps $x$ in a safe interval and $M$ bounds the leftover factor there

Taught in ε-δ proofs beyond straight lines · extends Prove a limit of a linear function with epsilon-delta

Quick recipe

Factor out the input distance
Restrict delta to a safe radius r
Bound the leftover factor by a constant M
Choose delta as the minimum
Write the proof, proving the bound before using it
Conclude

1
Rewrite $|f(x)-L|$ as $|x-a|$ times a leftover factor. Factor a polynomial; multiply a square-root difference by its conjugate; put a difference of fractions over a common denominator.

Example

$|x^2-4|=|x-2|\,|x+2|$ , with leftover factor $|x+2|$ .

Why: the window controls only $|x-a|$ , so it must appear as its own factor.
2
Decide that $\delta\le r$ . Use $r=1$ unless the window would then leave the domain or reach a zero of a denominator; in that case use a smaller $r$ .

Example

With $r=1$ , every $x$ in the window satisfies $|x-2|<1$ , so $1<x<3$ .
- f is a polynomial — $r=1$ always works.
- f has a square root or a denominator — choose $r$ no larger than the distance from $a$ to the edge of the domain, and strictly smaller than the distance from $a$ to the nearest zero of a denominator.
Why: the leftover factor can only be bounded on a known interval that avoids trouble spots.
3
On the interval from Step 2, find a number $M$ that the leftover factor never exceeds.

Example

$1<x<3$ gives $3<x+2<5$ , so $|x+2|<5$ and $M=5$ .

Why: a constant bound turns the problem back into the linear case, (constant) times $|x-a|$ .
4
Choose $\delta=\min(r,\ \varepsilon/M)$ .

Example

$\delta=\min(1,\ \varepsilon/5)$ .

Why: the first entry keeps the bound from Step 3 valid; the second makes $M|x-a|<\varepsilon$ .
5
Write "Let $\varepsilon>0$ be given. Choose $\delta=\min(r,\varepsilon/M)$ . Suppose $0<|x-a|<\delta$ ." Then use $\delta\le r$ to derive the bound, and finish the chain with $\delta\le\varepsilon/M$ .

Example

$|x^2-4|=|x-2|\,|x+2|\le 5|x-2|<5\delta\le5\cdot\dfrac{\varepsilon}{5}=\varepsilon$ .

Why: in the proof the bound is a consequence of the chosen delta, so it has to be derived inside the proof.
6
Write "Since $\varepsilon>0$ was arbitrary, $\therefore\ \lim_{x\to a}f(x)=L$ ."

Why: the conclusion is earned by the inequality $|f(x)-L|<\varepsilon$ on the line before it.

Common mistakes

Choosing $\delta=\varepsilon/|x+2|$ , a "delta" that depends on $x$ .
Choosing $\delta=\varepsilon/M$ without the minimum, which fails for large epsilon.
Using the bound $|x+2|<5$ in the proof without first deriving it from $\delta\le1$ .
Bounding the leftover factor wrongly, for example $|x+2|<3$ from $1<x<3$ .
For a fraction or root, restricting to a radius that lets $x$ reach a zero of the denominator or leave the domain.