Limits

Saying “arbitrarily close” precisely with ε–δ.

about 7 min

Start from a problem

Problem

$a_n=\dfrac{n}{n+1}$ never reaches $1$ , yet we want to say it "tends to $1$ ". Right now that sentence has no testable meaning: how close counts as close? From which term on?

Without a precise definition, "tends to" is a feeling; it can be neither proved nor refuted. We need one sentence that turns "arbitrarily close" into a statement that can be checked.

Observe

Look at the sequence $a_n = \dfrac{n}{n+1}$ : $\tfrac12,\ \tfrac23,\ \tfrac34,\ \tfrac45,\ \dots$

It never reaches $1$ , yet it gets ever closer. We want to say " $a_n$ tends to $1$ ". But what does "ever closer" mean? $b_n = 1-\tfrac1n$ also gets closer; $c_n = 1+\tfrac{(-1)^n}{n}$ jumps from side to side and also gets closer. What do these three kinds of "approaching" have in common?

Conjecture

This: however small an error you allow, from some term onward every term is within that error of $1$ .

"However small" becomes an arbitrary positive number $\varepsilon$ ; "from some term onward" becomes an index $N$ . That is what we will turn into a definition.

Definition

Definition 1.1Limit of a sequence

Let $(a_n)$ be a real sequence and $L\in\mathbb R$ . If for every $\varepsilon>0$ there exists $N\in\mathbb N$ such that $|a_n-L|<\varepsilon$ whenever $n>N$ , we say $(a_n)$ converges to $L$ and write $\lim_{n\to\infty}a_n=L$ .

Mind the order of quantifiers: first an arbitrary $\varepsilon$ is given, then we find $N$ . $N$ may depend on $\varepsilon$ (usually smaller $\varepsilon$ forces larger $N$ ). Swapping the order, "there exists $N$ such that for all $\varepsilon$ ", would demand $a_n=L$ exactly from some point on, which is a different statement.

Definition 1.2Limit of a function

Let $f$ be defined on a punctured neighbourhood of $x_0$ . If for every $\varepsilon>0$ there exists $\delta>0$ such that $|f(x)-L|<\varepsilon$ whenever $0<|x-x_0|<\delta$ , we write $\lim_{x\to x_0}f(x)=L$ .

$\varepsilon$ controls the output error, $\delta$ controls the input window. The definition says: however accurate you want the output, you can find how accurate the input must be.

Derivation: verifying a limit from the definition

Example 1.1

Show $\lim_{n\to\infty}\dfrac{n}{n+1}=1$ .

Let $\varepsilon>0$ . We need $\left|\dfrac{n}{n+1}-1\right|=\dfrac1{n+1}<\varepsilon$ , which holds when $n+1>\tfrac1\varepsilon$ . Take $N=\lceil\tfrac1\varepsilon\rceil$ ; then for $n>N$ , $\tfrac1{n+1}<\tfrac1N\le\varepsilon$ .

The pattern is always the same: simplify $|a_n-L|$ to an expression in $n$ , then solve for how large $n$ must be.

Theorems and proofs

Theorem 1.1Uniqueness of the limit

If $\lim a_n=L$ and $\lim a_n=M$ , then $L=M$ .

Proof

Suppose $L\ne M$ and set $\varepsilon=\tfrac{|L-M|}2>0$ . There are $N_1,N_2$ with $|a_n-L|<\varepsilon$ for $n>N_1$ and $|a_n-M|<\varepsilon$ for $n>N_2$ . For $n>\max(N_1,N_2)$ , $|L-M|\le|L-a_n|+|a_n-M|<2\varepsilon=|L-M|,$ a contradiction.

Theorem 1.2Algebra of limits

If $a_n\to L$ and $b_n\to M$ , then $a_n+b_n\to L+M$ , $a_nb_n\to LM$ , and if $M\ne0$ , $a_n/b_n\to L/M$ .

Proof

We prove the product rule, the most instructive case. Write $a_nb_n-LM=(a_n-L)b_n+L(b_n-M).$ A convergent sequence is bounded: there is $B$ with $|b_n|\le B$ for all $n$ . Given $\varepsilon>0$ , choose $N$ so that for $n>N$ , $|a_n-L|<\dfrac{\varepsilon}{2B}$ and $|b_n-M|<\dfrac{\varepsilon}{2(|L|+1)}$ . Then $|a_nb_n-LM|\le|a_n-L|\,B+|L|\,|b_n-M|<\tfrac\varepsilon2+\tfrac\varepsilon2=\varepsilon.$

Theorem 1.3Squeeze theorem

If $a_n\le c_n\le b_n$ for all large $n$ , and $a_n\to L$ , $b_n\to L$ , then $c_n\to L$ .

Proof

Given $\varepsilon>0$ , pick $N$ so that for $n>N$ , $L-\varepsilon<a_n$ and $b_n<L+\varepsilon$ . Then $L-\varepsilon<a_n\le c_n\le b_n<L+\varepsilon$ , i.e. $|c_n-L|<\varepsilon$ .

Common mistake

The quantifiers do not commute. "For every $\varepsilon$ there is $N$ " and "there is $N$ such that for every $\varepsilon$ " are different sentences: the second demands that $a_n$ equal $L$ from some term on. Also, whether a limit exists has nothing to do with whether $a_n$ ever equals $L$ ; $\frac n{n+1}$ never equals $1$ .

Application: a fundamental limit

Application

$\lim_{x\to0}\frac{\sin x}{x}=1.$

For $0<x<\tfrac\pi2$ compare three areas in the unit circle: the triangle $\tfrac12\sin x$ , the sector $\tfrac12x$ , the larger triangle $\tfrac12\tan x$ . Hence $\sin x<x<\tan x$ , so $\cos x<\dfrac{\sin x}{x}<1$ . Since $\cos x\to1$ , the squeeze theorem finishes the job; for $x<0$ use that $\dfrac{\sin x}x$ is even.

This single limit is the entire source of $\sin'x=\cos x$ in The derivative.

Remark

A limit is not "the value finally reached" but "the value that can be approached arbitrarily closely". It does not matter that $a_n$ never equals $1$ . For the same reason $0.999\ldots=1$ : it is a limit, not a number that is "slightly short".

Exercises

Compute

\displaystyle\lim_{x\to 0}\frac{\sin 3x}{x}

Let

f(x)=2x+1

. Which

\delta

guarantees

|f(x)-3|<\varepsilon

whenever

0<|x-1|<\delta