PG4.1 What a limit means: ε and N

Simulate

Check the claim with numbers

Lesson 2.1.3 found that $\frac{3x + 1}{x + 2}$ heads for 3 as $x$ grows, missing it by $\frac{5}{x + 2}$ . Take it as a sequence, $a_n = \frac{3n + 1}{n + 2}$ for whole numbers $n \ge 1$ . Lesson 1.5.1 said what $a_n \to 3$ means: for every tolerance, from some term on, every term is that close to 3.

The cell tests exactly that. For three tolerances it finds the last term outside the band, checking 20,000 terms with exact fractions.

from fractions import Fraction def a(n): return Fraction(3 * n + 1, n + 2) for eps in [Fraction(1, 10), Fraction(1, 100), Fraction(1, 1000)]: outside = [n for n in range(1, 20001) if abs(a(n) - 3) >= eps] print(f"tolerance {float(eps)}: last term outside is n = {max(outside)}")

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The last term outside is $n = 48$ for 0.1, 498 for 0.01 and 4998 for 0.001. Each is the $n$ where the miss $\frac{5}{n + 2}$ equals the tolerance. A smaller tolerance needs a later start, but some start always seems to work.

A sequence with no limit fails the test. The terms of $(-1)^n$ jump between $-1$ and 1 for ever (lesson 1.5.1).

print([(-1) ** n for n in range(1, 11)]) for L in [-1, 0, 1]: far = [n for n in range(1000, 1011) if abs((-1) ** n - L) >= 0.5] print(f"L = {L}: n from 1000 to 1010 at least 0.5 away: {far}")

⌘/Ctrl + Enter runs the cell

Take the tolerance 0.5. For $L = 1$ every odd term stays outside the band, for $L = -1$ every even one, and for $L = 0$ all of them. No $L$ can work: the band is 1 wide, and the terms are 2 apart.

Conjecture

\frac{3n + 1}{n + 2} \to 3 \quad \text{as } n \to \infty

Three tolerances worked, each from some term on. A proof must handle every tolerance, however small, with a rule that names the starting term.

The definition

Lesson 1.5.1 described a limit in words. The precise version names the tolerance $\varepsilon$ (the Greek letter epsilon) and the starting term $N$ .

Rule: Limit of a sequence

\begin{gathered} a_n \to L \text{ means: for every } \varepsilon > 0 \\ \text{there is a whole number } N \text{ with} \\ |a_n - L| < \varepsilon \text{ for every } n > N \end{gathered}

$|a_n - L| < \varepsilon$ says that $a_n$ lies strictly between $L - \varepsilon$ and $L + \varepsilon$ (lesson 0.3.3). The number $N$ may depend on $\varepsilon$ , and usually does.

Read it as a game, in which a challenger names a tolerance $\varepsilon > 0$ . You must answer with a whole number $N$ such that every term after the $N$ -th is within $\varepsilon$ of $L$ . The limit is $L$ if you can always answer.

The order matters: $\varepsilon$ comes first, and $N$ is chosen after it. Stage PG1 practised reading statements of this shape.

The simulation found the smallest answers, 48, 498 and 4998. Any larger $N$ works too, since the terms after it are among those already checked. So a proof may pick a convenient $N$ , not the smallest one.

Writing an ε–N proof

First comes scratch work, which is not part of the proof. Work out the distance $|a_n - L|$ , and find which $n$ make it less than $\varepsilon$ . For our sequence, $3 - a_n = \frac{5}{n + 2}$ , and $\frac{5}{n + 2} < \varepsilon$ when $n + 2 > \frac{5}{\varepsilon}$ .

Then write the proof. Start from "let $\varepsilon > 0$ ", name $N$ in terms of $\varepsilon$ , and show that every $n > N$ works. Here any whole number $N \ge \frac{5}{\varepsilon}$ will do: it is a little larger than needed, and simpler.

Such an $N$ exists for every $\varepsilon$ , because some whole number is larger than any given real number. This is the Archimedean property; lesson PG4.4 proves it from the completeness of the real numbers.

N

import math from fractions import Fraction def a(n): return Fraction(3 * n + 1, n + 2) for eps in [Fraction(1, 10), Fraction(1, 100), Fraction(1, 1000)]: N = math.ceil(5 / eps) print(float(eps), N, all(abs(a(n) - 3) < eps for n in range(N + 1, N + 10001)))

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Format 1

Reorder the steps

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

For this $n$ , $|a_n - 3| = \left| \dfrac{3n + 1 - 3(n + 2)}{n + 2} \right| = \dfrac{5}{n + 2}$ .
By the Archimedean property, choose a whole number $N$ with $N \ge \dfrac{5}{\varepsilon}$ .
So $|a_n - 3| < \varepsilon$ for every $n > N$ . As $\varepsilon > 0$ was arbitrary, $a_n \to 3$ .
Now take any whole number $n > N$ .
Since $n + 2 > N \ge \dfrac{5}{\varepsilon}$ , this is less than $5 \div \dfrac{5}{\varepsilon} = \varepsilon$ .
Let $\varepsilon > 0$ be given.

Bound, do not solve

Solving $|a_n - L| < \varepsilon$ exactly can be hard. It is enough to find a simpler expression that is larger than the distance and still tends to 0.

Take $a_n = \frac{n^2 + 1}{n^2 + n}$ , which tends to 1. Making its distance from 1 less than $\varepsilon$ exactly means solving a quadratic inequality. Instead, compare the distance with $\frac{1}{n}$ .

\frac{1}{n}

from fractions import Fraction def a(n): return Fraction(n * n + 1, n * n + n) for n in [1, 2, 10, 100, 1000]: print(n, round(float(1 - a(n)), 6), 1 / n) eps = Fraction(1, 1000) print(max(n for n in range(1, 5001) if abs(a(n) - 1) >= eps))

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Format 2

Fill the gap

The proof that $a_n = \frac{n^2 + 1}{n^2 + n} \to 1$ , with its key steps left out. Type each missing piece in $n$ , or as a number, then check.

n \ge 1

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 of a false claim. One line claims more than the lines before it have shown. Select it.

a_n \to L

What a limit means: ε and N

Check the claim with numbers

The definition

Rule: Limit of a sequence

Writing an ε–N proof

Reorder the steps

Bound, do not solve

Fill the gap

Find the error

Summary card

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