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 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

$$1,2,3,4,\ldots$$

or

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

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 $\varepsilon\gt0$,
• 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 $\varepsilon\gt0$, there exists $N\in N$ such that for all $n\gt N$,

$$|L-x_n|\lt\varepsilon.$$

In that case we write

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

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

The quantifiers make the order of choices explicit:

$$\lim_{n\to\infty}x_n=L \iff \forall \varepsilon\gt0\ \exists N\in N\ \forall n\gt N,\ |x_n-L|\lt\varepsilon.$$

How to read the quantifiers correctly

The definition is easier once you separate its jobs.

1. $\forall \varepsilon\gt0$: you do not get to choose only one error bar. The sequence must eventually fit inside every positive tolerance band.
2. $\exists N\in N$: after seeing the chosen $\varepsilon$, you are allowed to pick a tail-starting point $N$.
3. $\forall n\gt N$: 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 $n\gt N$. 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 $n\gt N$.

The strict inequality in the definition is $n\gt N$. If an estimate is proved for $n≥K$, where $K$ is an integer with $K≥1$, choosing $N=K$ is safe because $n\gt N$ implies $n≥K+1≥K$. If the estimate needs $n\gt K$, choose an integer $N$ with $N≥K$; the choice of $N$ is made after $\varepsilon$ but before the arbitrary later index.

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 $\lim_{n\to\infty} 1/n = 0$

Here $n≥1$, so the reciprocal is defined even if the course convention includes 0 in $N$. Let $\varepsilon\gt0$ be given. We want $|1/n-0|\lt\varepsilon$, that is,

$$\frac1n\lt\varepsilon.$$

This is guaranteed once

$$n\gt\frac1\varepsilon.$$

So choose a natural number $N$ with $N\gt1/\varepsilon$. Then every $n\gt N$ satisfies

$$\left|\frac1n-0\right|=\frac1n\lt\varepsilon.$$

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 $n\gt N$. Then

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

So it is enough to ensure

$$\frac{41}{6n}\lt\varepsilon.$$

Choose $N$ so large that both $3n-7\ge 2n$ and $n\gt41/(6\varepsilon)$ hold for all $n\gt N$. Then

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

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.

Designing an epsilon threshold

The rational example requires two conditions on the same tail: the denominator estimate $3n-7≥2n$ needs $n≥7$, and the error estimate needs $n>41/(6\varepsilon)$. Given $\varepsilon>0$, choose an integer

$$N\ge\max\left(7,\frac{41}{6\varepsilon}\right).$$

Then every $n>N$ satisfies both conditions, so $41/(6n)<\varepsilon$. This is the dependency order of the proof: derive the bound, choose one threshold for all restrictions, then verify it for an arbitrary later index.

The index convention also matters. The expression $1/n$ is defined for $n≥1$. To make it a sequence on $\mathbb N=\{0,1,2,\ldots\}$, specify the value at $n=0$ separately. A finite initial change does not affect convergence.

Worked example

A finite prefix and an identical tail

Let $x_0=100$ and $x_n=1/n$ for $n≥1$. Given $\varepsilon>0$, choose an integer $N>1/\varepsilon$ with $N≥1$. Then every $n>N$ satisfies $|x_n|=1/n<\varepsilon$; the exceptional first term lies before the threshold. Thus $x_n→0$.

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 $\varepsilon\gt0$,

$$|x_n-0|=0\lt\varepsilon$$

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 $L$.

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

In the figure below, choose a sequence and a tolerance $\varepsilon$, then move $N$ to test a proposed tail. The plotted signed errors $x_n-L$ show which terms lie in the open band $-\varepsilon\lt x_n-L\lt\varepsilon$. Check the terms with $n\gt N$, and compare the figure with the accompanying algebraic explanation: a finite plot illustrates the tail, but the estimate must justify every later term.

1/n → 0(5n+2)/(3n−7) → 5/3(−1)ⁿ, L = 0

ε = 0.5ε = 0.2ε = 0.1ε = 0.05

N = 5

Open circles precede the tail; arrows mark errors beyond the plotted scale.

The selected N works. A sufficient threshold is N=5.

$n>\lfloor1/\varepsilon\rfloor\implies 1/n<\varepsilon$.

The plot shows the signed error $x_n-L$, so the open band is $-\varepsilon<x_n-L<\varepsilon$. Terms with $n>N$ form the selected tail. A finite picture illustrates the condition; the algebraic bound establishes it for every later index. Here the convention is $n>N$ (strictly after $N$).

Proof tools for sequence limits

The definition controls a tail, so several useful consequences follow by making that tail explicit.

Theorem

Limits are unique

A real sequence cannot converge to two different limits. Suppose that $x_n→L$ and $x_n→M$ with $L≠M$. Choose

$$\varepsilon=\frac{|L-M|}{3}\gt0.$$

For sufficiently large $n$, both $|x_n-L|\lt\varepsilon$ and $|x_n-M|\lt\varepsilon$. The triangle inequality then gives

$$|L-M|\le |L-x_n|+|x_n-M|\lt2\varepsilon=\frac{2|L-M|}{3},$$

which is impossible. Thus the limit, when it exists, is unique.

Theorem

Every convergent sequence is bounded

If $x_n→L$, then there are real numbers $m$ and $M$ such that $m≤x_n≤M$ for every index. Use $\varepsilon=1$ to choose $N$ so that $n\gt N$ implies $L-1\lt x_n\lt L+1$. The terms with $n≤N$ form a finite set, so they have a minimum $m_0$ and maximum $M_0$. Taking

$$m=\min(m_0,L-1),\qquad M=\max(M_0,L+1)$$

bounds both the finite initial segment and the tail.

Theorem

Absolute values preserve limits

If $x_n→L$, then $|x_n|→|L|$. The reverse triangle inequality says

$$\bigl||x_n|-|L|\bigr|\le |x_n-L|.$$

Given $\varepsilon\gt0$, use the same index supplied for $|x_n-L|\lt\varepsilon$. The displayed inequality transfers that estimate directly to the absolute values.

Worked example

Derive both triangle inequalities

The triangle inequality applied to $(x-y)+(y-z)$ gives

$$|x-z|=|(x-y)+(y-z)|\le |x-y|+|y-z|.$$

For the reverse triangle inequality, apply the triangle inequality to $x=(x-y)+y$ and then rearrange:

$$|x|\le |x-y|+|y|\quad\Longrightarrow\quad |x|-|y|\le |x-y|.$$

Interchanging $x$ and $y$ gives $|y|-|x|≤|x-y|$; combining the two inequalities is exactly $\bigl||x|-|y|\bigr|≤|x-y|$.

Algebra of sequence limits

Once the basic examples are established, the limit laws let us combine them. Their proofs show why the same threshold must control every error.

Theorem

Sum and product laws for sequences

If $x_n\to L$ and $y_n\to M$, then $x_n+y_n\to L+M$ and $x_ny_n\to LM$.

Proof. For the sum, make each error smaller than $\varepsilon/2$ and use the triangle inequality on the common tail. For the product, first use $x_n\to L$ with tolerance $1$ to obtain the fixed bound $|x_n|<|L|+1=:A$. Put $B=\max(1,|M|)$. Given $\varepsilon>0$, choose a common tail on which

$$|y_n-M|<\frac{\varepsilon}{2A},\qquad |x_n-L|<\frac{\varepsilon}{2B}.$$

All bounds now hold for the same arbitrary $n$ beyond the threshold, so

$$|x_ny_n-LM| \le |x_n||y_n-M|+|M||x_n-L| <\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon.$$

The preliminary bound prevents the factor $x_n$ from magnifying a small error without control; using $B$ also covers $M=0$.

Theorem

Reciprocals of nonzero limits

If $x_n\to L$ and $L\ne0$, then $1/x_n\to1/L$ on the tail where the reciprocal is defined.

Proof. Choose a tail with $|x_n-L|<|L|/2$. Then $|x_n|>|L|/2$, and

$$\left|\frac1{x_n}-\frac1L\right| =\frac{|x_n-L|}{|x_n||L|} \le\frac{2|x_n-L|}{|L|^2}.$$

Given $\varepsilon>0$, enlarge the threshold so that $|x_n-L|<\varepsilon|L|^2/2$ also holds. The error is then below $\varepsilon$, and every division is justified by the earlier lower bound.

Worked example

Apply the sequence laws

For $n\ge1$, the sum law gives $1+1/n\to1$. The product law then gives $(1+1/n)/n\to0$ by writing the expression as $(1+1/n)(1/n)$. Both conclusions use the proved limit $1/n\to0$ with its stated domain.

Further examples: comparison, oscillation, and growth

Worked example

Compare two rational-expression limits

For

$$x_n=\frac{5n^2+n+7}{n^3+3n^2+3n+1},$$

we have, for $n≥1$,

$$0\le x_n\le\frac{13n^2}{n^3}=\frac{13}{n},$$

because the denominator is at least $n^3$ and $5n^2+n+7≤13n^2$. Given $\varepsilon\gt0$, choose an integer $N\gt13/\varepsilon$. Then $n\gt N$ implies $|x_n|\lt\varepsilon$, so $x_n→0$.

Now consider a numerator with an oscillating term:

$$y_n=\frac{n^2-(-1)^n n-13}{n^2+n+1}.$$

Subtracting 1 gives

$$y_n-1=\frac{-(-1)^n n-n-14}{n^2+n+1}.$$

Using $|(-1)^n|=1$, for $n≥1$ we obtain

$$|y_n-1|\le\frac{2n+14}{n^2}\le\frac{16}{n}.$$

Choose $N\gt16/\varepsilon$; then $y_n→1$.

Worked example

The sequence $2^n$ has no finite limit

For every $n≥0$, induction gives $2^n≥n+1$: it is true at $n=0$, and $2^{n+1}=2\cdot2^n≥2(n+1)≥n+2$. Thus $2^n$ is unbounded. A convergent real sequence is bounded, so the sequence $2^n$ cannot converge to a real number.

Worked example

Rationalize $\sqrt{n+1}-\sqrt n$

For $n≥1$,

$$0\lt\sqrt{n+1}-\sqrt n =\frac{1}{\sqrt{n+1}+\sqrt n} \le\frac{1}{\sqrt n}.$$

Given $\varepsilon\gt0$, choose an integer $N\gt1/\varepsilon^2$. Then $n\gt N$ implies $1/\sqrt n\lt\varepsilon$, proving $\sqrt{n+1}-\sqrt n\to0$.

Worked example

The discrete limit $n^2/4^n$

For $n≥4$, induction shows $n≤2^{n/2}$. The base case is $4≤4$; if the inequality holds at $n≥4$, then $(n+1)/n≤5/4\lt\sqrt2$, so $n+1≤2^{(n+1)/2}$. Consequently,

$$0\le\frac{n^2}{4^n}\le\frac{2^n}{4^n}=2^{-n}.$$

Given $\varepsilon\gt0$, choose $N$ larger than both 4 and $\log_2(1/\varepsilon)$. Then $n\gt N$ gives $2^{-n}\lt\varepsilon$, hence $n^2/4^n\to0$.

Worked example

Finite changes do not change a sequence limit

If two sequences agree for all $n\gt N_0$ and one converges to $L$, choose the limit threshold for a given $\varepsilon$ and enlarge it to exceed $N_0$. The two tails then have identical terms and the same estimate, so the other sequence also converges to $L$.

Quick checks

Checkpoint

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

$N$ marks the point after which every later term must lie within the chosen $\varepsilon$-band around $L$. It is a tail-starting index.

Checkpoint

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

Look at which indices the definition actually controls.

Solution · Answer

Because convergence only asks for a property of all terms with $n\gt N$ for some large $N$. Changing finitely many initial terms does not change the eventual tail behavior.

Exercises

Checkpoint

Write the sequence-limit definition entirely in symbols.

Do not omit the order of the quantifiers.

Solution · Guided solution

The symbolic form is

$$\lim_{n\to\infty}x_n=L \iff \forall \varepsilon\gt0\ \exists N\in N\ \forall n\gt N,\ |x_n-L|\lt\varepsilon.$$

The order matters: first the tolerance is chosen, then $N$, then every later term is checked.

Checkpoint

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

Compare it with the proof for $1/n$.

Solution · Guided solution

Let $\varepsilon\gt0$. We want

$$\left|\frac{1}{2n}-0\right|=\frac{1}{2n}\lt\varepsilon.$$

This is true once $n\gt1/(2\varepsilon)$. Choose a natural number $N$ with $N\gt1/(2\varepsilon)$. Then every $n\gt N$ satisfies $1/(2n)\lt\varepsilon$, so $1/(2n)\to 0$.

Checkpoint

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

Because convergence requires the whole tail to stay close to $L$, not merely a subcollection of terms. A sequence can visit points near $L$ infinitely often and still keep escaping far away at later indices.

Prerequisites and continuation

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.