5.1 Sequences and epsilon-N limits

This note starts chapter 5 by slowing down and making the first limit definition fully explicit. Before talking about limits of functions, first first study limits of sequences. That is the right order: a sequence moves along the number line in discrete steps, so it is the cleanest place to learn how a formal limit definition works.

A sequence is a function, not just a pattern

Many students first meet sequences as lists such as

1,2,3,4,\ldots

1,\frac12,\frac13,\frac14,\ldots

The familiar idea must be reinterpreted in a more precise way.

Definition

Sequence in a set

Let $X$ be a set. A sequence in $X$ is a function

N\to X.

If the image of $n\in N$ is written $x_n$ , then the sequence is denoted $(x_n)$ .

So a sequence is not required to come from a simple formula. The only formal requirement is that each natural number n is assigned an element $x_n$ of the target set.

Definition

Rational and real sequences

A sequence of rational numbers is a map $N\to Q$ . A sequence of real numbers is a map $N\to R$ .

This functional viewpoint matters later. It reminds you that a sequence has a domain, a codomain, and an indexing variable. The dots $\ldots$ are only informal shorthand.

Why limits of sequences come before limits of functions

For a sequence, approaching the limit means going further and further out in the index n. There is only one direction to move: toward larger natural numbers.

That is much simpler than the function-limit situation, where x can approach a point a from the left and the right and through infinitely many real values in between.

So chapter 5 begins with the discrete version:

choose a candidate limit $L$ ,
choose an error tolerance $\varepsilon>0$ ,
ask whether the terms eventually stay inside the band of radius $\varepsilon$ around $L$ .

The formal definition

Definition

Limit of a real sequence

A sequence $(x_n)$ of real numbers has limit $L\in R$ if for every positive real number $\varepsilon>0$ , there exists $N\in N$ such that for all $n>N$ ,

|L-x_n|<\varepsilon.

In that case we write

L=\lim_{n\to\infty}x_n

or say that $(x_n)$ converges to $L$ .

A natural exercise is to write the definition symbolically. Here it is in compact form:

\lim_{n\to\infty}x_n=L \iff \forall \varepsilon>0\ \exists N\in N\ \forall n>N,\ |x_n-L|<\varepsilon.

How to read the quantifiers correctly

The definition is easier once you separate its jobs.

$\forall \varepsilon>0$ : you do not get to choose only one error bar. The sequence must eventually fit inside every positive tolerance band.
$\exists N\in N$ : after seeing the chosen $\varepsilon$ , you are allowed to pick a tail-starting point $N$ .
$\forall n>N$ : once you pass that point, every later term must stay inside the band.

So convergence is an eventual-tail statement. The first few terms may behave badly. That does not matter. What matters is the behavior far enough out in the sequence.

Common mistake

N may depend on epsilon, but not on n

When proving convergence, you are allowed to choose $N$ after the tolerance $\varepsilon$ is given. But once $N$ is fixed, the inequality must hold for every $n>N$ . You are not allowed to choose a different $N$ for each later term.

Common mistake

Convergence is not about early terms

Changing finitely many initial terms of a sequence never changes whether the sequence converges, because the definition only cares about the tail $n>N$ .

First worked example: `1/n \to 0`

A basic example is that

\frac11,\frac12,\frac13,\frac14,\ldots

has limit 0. Let us write the proof in full.

Worked example

Proving that `\lim_{n\to\infty} 1/n = 0`

Let $\varepsilon>0$ be given. We want |1/n-0|<\varepsilon, that is,

\frac1n<\varepsilon.

This is guaranteed once

n>\frac1\varepsilon.

So choose a natural number $N$ with N>1/\varepsilon. Then every $n>N$ satisfies

\left|\frac1n-0\right|=\frac1n<\varepsilon.

Therefore

\lim_{n\to\infty}\frac1n=0.

This proof is typical. Start from the quantity you need to make small, simplify it, and then choose $N$ large enough.

Second worked example: $\frac{5n+2}{3n-7} \to \frac53$

A slightly richer example uses a rational-function sequence and proves its limit.

Worked example

Proving that $\lim_{n\to\infty}\frac{5n+2}{3n-7}=\frac53$

We compute

\left|\frac{5n+2}{3n-7}-\frac53\right| = \left|\frac{3(5n+2)-5(3n-7)}{3(3n-7)}\right| = \left|\frac{41}{3(3n-7)}\right|.

For large n, the denominator is positive, so

\left|\frac{41}{3(3n-7)}\right|=\frac{41}{3(3n-7)}.

Now pick $N$ large enough that $3n-7\ge 2n$ whenever $n>N$ . Then

\frac{41}{3(3n-7)}\le \frac{41}{6n}.

So it is enough to ensure

\frac{41}{6n}<\varepsilon.

Choose $N$ so large that both $3n-7\ge 2n$ and n>41/(6\varepsilon) hold for all $n>N$ . Then

\left|\frac{5n+2}{3n-7}-\frac53\right|<\varepsilon.

Hence

\lim_{n\to\infty}\frac{5n+2}{3n-7}=\frac53.

The algebra looks longer, but the logical structure is exactly the same as for 1/n: rewrite the error term and make it small by forcing n to be large.

A simple but important special case

For the constant sequence $x_n=0$ , the limit is immediate.

Worked example

Constant sequences converge to their constant value

Let $x_n=0$ for all n. Then for every $\varepsilon>0$ ,

|x_n-0|=0<\varepsilon

for every natural number n.

So any natural number can serve as $N$ , and

\lim_{n\to\infty}x_n=0.

This small example is worth keeping in mind because it shows what the definition looks like when the error term is already zero.

What convergence means geometrically

If $\lim_{n\to\infty}x_n=L$ , then no matter how narrow a band you draw around $L$ , the tail of the sequence eventually stays inside it.

Equivalently:

you may reject finitely many early terms;
afterward the sequence cannot keep escaping the chosen $\varepsilon$ -band.

That is why convergence is stronger than “having many terms near $L$ ”. The definition requires all sufficiently late terms to be near $L$ .

A sequence tail entering an epsilon-band

Figure. Convergence means more than seeing some terms near $L$ . After a large enough index $N$ , the whole tail must remain inside the chosen $\varepsilon$ -band.

Compare tails interactively

Use the explorer below to change the candidate sequence and the tolerance $\varepsilon$ . The useful quantity is the first index $N$ after which the shown tail stays inside the $\varepsilon$ -band around the proposed limit.

Read and try

Test whether a sequence tail stays inside an epsilon-band

The explorer compares convergent and non-convergent sequences by asking whether the tail can be trapped inside a chosen epsilon-band around the candidate limit.

The terms shrink steadily, so once you choose epsilon you can eventually trap every later term inside the band around 0.

Candidate limit L

Epsilon ε

0.2

Tail index N

Term index n	Term x_n	Band check \|x_n - L\| < ε
1	1	No
2	0.5	No
3	0.3333	No
4	0.25	No
5	0.2	No
6	0.1667	Yes
7	0.1429	Yes
8	0.125	Yes
9	0.1111	Yes
10	0.1	Yes
11	0.0909	Yes
12	0.0833	Yes

Verdict: Starting after N = 5, every checked term stays inside the band around L.

Quick checks

Quick check

In the definition of $\lim_{n\to\infty}x_n=L$ , what is the role of N?

Answer in terms of the tail of the sequence.

Solution

Answer

Quick check

Why does changing the first ten terms of a convergent sequence never destroy convergence?

Look at which indices the definition actually controls.

Solution

Answer

Quick check

For the sequence $x_n=0$ , can the same N work for every epsilon?

Use the fact that the error term is always zero.

Solution

Answer

Exercises

Quick check

Write the sequence-limit definition entirely in symbols.

Do not omit the order of the quantifiers.

Solution

Guided solution

Quick check

Show directly from the definition that `\lim_{n\to\infty} 1/(2n)=0`.

Compare it with the proof for 1/n.

Solution

Guided solution

Quick check

Why is it not enough for infinitely many terms of a sequence to lie close to L?

Contrast “infinitely many” with “all sufficiently late”.

Solution

Guided solution

Read this after 4.6 Decimal expansions and irrational numbers and 4.3 Completeness and gaps in Q. Then continue to 5.2 Cauchy sequences and another model of the reals.

5.1 Sequences and epsilon-N limits

MATH1090: Set theory

A sequence is a function, not just a pattern

Sequence in a set

Rational and real sequences

Why limits of sequences come before limits of functions

The formal definition

Limit of a real sequence

How to read the quantifiers correctly

N may depend on epsilon, but not on n

Convergence is not about early terms

First worked example: `1/n \to 0`

Proving that `\lim_{n\to\infty} 1/n = 0`

Second worked example: $\frac{5n+2}{3n-7} \to \frac53$

Proving that $\lim_{n\to\infty}\frac{5n+2}{3n-7}=\frac53$

A simple but important special case

Constant sequences converge to their constant value

What convergence means geometrically

Compare tails interactively

Test whether a sequence tail stays inside an epsilon-band

Quick checks

In the definition of $\lim_{n\to\infty}x_n=L$ , what is the role of N?

Answer

Why does changing the first ten terms of a convergent sequence never destroy convergence?

Answer

For the sequence $x_n=0$ , can the same N work for every epsilon?

Answer

Exercises

Write the sequence-limit definition entirely in symbols.

Guided solution

Show directly from the definition that `\lim_{n\to\infty} 1/(2n)=0`.

Guided solution

Why is it not enough for infinitely many terms of a sequence to lie close to L?

Guided solution

Prerequisites

Key terms in this unit

More notes in this series

5.1 Sequences and epsilon-N limits

MATH1090: Set theory

A sequence is a function, not just a pattern

Sequence in a set

Rational and real sequences

Why limits of sequences come before limits of functions

The formal definition

Limit of a real sequence

How to read the quantifiers correctly

N may depend on epsilon, but not on n

Convergence is not about early terms

First worked example: 1/n \to 0

Proving that \lim_{n\to\infty} 1/n = 0

Second worked example: 5n+23n−7→53\frac{5n+2}{3n-7} \to \frac533n−75n+2​→35​

Proving that lim⁡n→∞5n+23n−7=53\lim_{n\to\infty}\frac{5n+2}{3n-7}=\frac53limn→∞​3n−75n+2​=35​

A simple but important special case

Constant sequences converge to their constant value

What convergence means geometrically

Compare tails interactively

Test whether a sequence tail stays inside an epsilon-band

Quick checks

In the definition of lim⁡n→∞xn=L\lim_{n\to\infty}x_n=Llimn→∞​xn​=L, what is the role of N?

Answer

Why does changing the first ten terms of a convergent sequence never destroy convergence?

Answer

For the sequence xn=0x_n=0xn​=0, can the same N work for every epsilon?

Answer

Exercises

Write the sequence-limit definition entirely in symbols.

Guided solution

Show directly from the definition that \lim_{n\to\infty} 1/(2n)=0.

Guided solution

Why is it not enough for infinitely many terms of a sequence to lie close to L?

Guided solution

Related notes

Prerequisites

Key terms in this unit

More notes in this series

First worked example: `1/n \to 0`

Proving that `\lim_{n\to\infty} 1/n = 0`

Second worked example: $\frac{5n+2}{3n-7} \to \frac53$

Proving that $\lim_{n\to\infty}\frac{5n+2}{3n-7}=\frac53$

In the definition of $\lim_{n\to\infty}x_n=L$ , what is the role of N?

For the sequence $x_n=0$ , can the same N work for every epsilon?

Show directly from the definition that `\lim_{n\to\infty} 1/(2n)=0`.