PG4.3 Limits of functions: ε and δ

Simulate

Check the claim with numbers

Lesson 2.1.1 found $\lim_{x \to 3} (x^2 + 1) = 10$ by substituting. Test it the way lesson PG4.1 tested sequences, with a tolerance $\varepsilon$ for the output. Then look for a window of half-width $\delta$ around 3 that keeps every output within $\varepsilon$ of 10.

The cell takes $\delta$ to be the smaller of 1 and $\frac{\varepsilon}{7}$ , a choice the proof will explain. It tries 1,998 inputs inside the window, all but 3 itself, and reports the worst distance from 10.

import math def f(x): return x**2 + 1 for eps in [1, 0.1, 0.01]: delta = min(1, eps / 7) xs = [3 + delta * (k / 1000) for k in range(-999, 1000) if k != 0] worst = max(abs(f(x) - 10) for x in xs) print(f"eps = {eps}: delta = {delta:.6f}, worst distance {worst:.6f}, widest delta {math.sqrt(9 + eps) - 3:.6f}")

⌘/Ctrl + Enter runs the cell

Each worst distance is below its $\varepsilon$ : 0.876653 for $\varepsilon = 1$ , and 0.008565 for $\varepsilon = 0.01$ . The last column is the widest window that would work. The proof's $\delta$ is a little smaller, and that is fine.

A function with a jump fails the test. Near 0, $\frac{|x|}{x}$ is $-1$ on the left and 1 on the right (lesson 2.1.2).

⌘/Ctrl + Enter runs the cell

However small the window, it holds inputs with outputs $-1$ and 1. With $\varepsilon = 0.5$ no $L$ is within 0.5 of both, since they are 2 apart.

Conjecture

\lim_{x \to 3} (x^2 + 1) = 10

Three tolerances worked with the same rule for $\delta$ . A proof must show that the rule works for every $\varepsilon > 0$ , and for every input in the window, not just 1,998 of them.

The definition

The definition has the same shape as for sequences, with the same tolerance $\varepsilon$ . The window $\delta$ (the Greek letter delta) takes the place of $N$ , because $x$ must be close to $a$ rather than far along.

Rule: Limit of a function

\begin{gathered} \lim_{x \to a} f(x) = L \text{ means:} \\ \text{for every } \varepsilon > 0 \text{ there is a } \delta > 0 \\ \text{such that } 0 < |x - a| < \delta \\ \text{implies } |f(x) - L| < \varepsilon \end{gathered}

Only inputs in the domain of $f$ count. The condition $0 < |x - a|$ leaves out $x = a$ itself: as in lesson 2.1.1, the value at $a$ plays no part.

In words: however narrow the band around $L$ , some window around $a$ sends every input in it, apart from $a$ , into the band. The window may depend on $\varepsilon$ , but not on $x$ . It is chosen first, and then every $x$ in it must work.

Bound the extra factor

For the conjecture, $|f(x) - 10| = |x^2 - 9| = |x - 3|\,|x + 3|$ . The window controls $|x - 3|$ , but the factor $|x + 3|$ needs a bound of its own.

So agree first that $\delta \le 1$ . Then $|x - 3| < 1$ means $2 < x < 4$ (lesson 0.3.3), so $5 < x + 3 < 7$ and $|x + 3| < 7$ .

Now $|f(x) - 10| < 7|x - 3| < 7\delta$ , and $7\delta \le \varepsilon$ when $\delta \le \frac{\varepsilon}{7}$ . Both demands hold for $\delta = \min\left(1, \frac{\varepsilon}{7}\right)$ , the smaller of the two. The cap at 1 is a free choice: another cap works too, with its own bound in place of 7.

Format 1

Reorder the steps

Here is the ε–δ proof of the conjecture, with its lines shuffled. Put them in order with the arrows, or drag a line by its handle.

Since $|x - 3| < \delta \le \frac{\varepsilon}{7}$ , this is less than $7 \cdot \frac{\varepsilon}{7} = \varepsilon$ .
Since $\delta \le 1$ , $2 < x < 4$ , so $5 < x + 3 < 7$ and $|x + 3| < 7$ .
So every such $x$ has $|f(x) - 10| < \varepsilon$ . As $\varepsilon > 0$ was arbitrary, the limit is 10.
Then $|f(x) - 10| = |x - 3|\,|x + 3| < 7|x - 3|$ , because $|x - 3| > 0$ .
Take any $x$ with $0 < |x - 3| < \delta$ .
Let $\varepsilon > 0$ , and set $\delta = \min\left(1, \frac{\varepsilon}{7}\right)$ , which is positive.

A bound that needs no cap

Lesson 2.1.1 handled $\sqrt{x}$ near 4 by multiplying by $\sqrt{x} + 2$ . The same trick gives a $\delta$ for $\lim_{x \to 4} \sqrt{x} = 2$ , and this time the bound holds on the whole domain, $x \ge 0$ .

|\sqrt{x} - 2| \le \frac{|x - 4|}{2}

import math for x in [3.9, 3.99, 4.01, 4.1]: print(x, round(abs(math.sqrt(x) - 2), 6), round(abs(x - 4) / 2, 6)) for eps in [0.1, 0.01]: delta = 2 * eps xs = [4 + delta * (k / 1000) for k in range(-999, 1000) if k != 0] print(eps, delta, round(max(abs(math.sqrt(x) - 2) for x in xs), 6))

⌘/Ctrl + Enter runs the cell

Format 2

Fill the gap

The proof that $\lim_{x \to 4} \sqrt{x} = 2$ , with its key steps left out. Type each missing piece in $x$ , or as a number, then check.

x \ge 0

How to type answers

Powers with ^, brackets round a power that is a sum: r^k, r^(k+1).
Fractions with / and brackets: (1-r)/(1+r). Multiply with * or side by side: 2k+1.
Press Enter in any box to check them all.

Format 3

Find the error

A proof that $\lim_{x \to -2} x^2 = 4$ . One line breaks the rules of the definition. Select it.

\lim_{x \to a} f(x) = L

Limits of functions: ε and δ

Check the claim with numbers

The definition

Rule: Limit of a function

Bound the extra factor

Reorder the steps

A bound that needs no cap

Fill the gap

Find the error

Summary card

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