Continuous Functions & Epsilon-Delta Definition

Definition 1.

Let $f (x)$ be defined on an interval containing $x_{0}$ , and assume now that $f (x_{0})$ is defined. Then we say that $f (x)$ is continuous at $x_{0}$ if

lim_{x \to x_{0}} f (x) = f (x_{0}),

or in other words, if for every $ϵ > 0$ we can find a $δ > 0$ such that whenever $| x - x_{0} | < δ$ , then $| f (x) - f (x_{0}) | < ϵ \cdot$

This is the precise way of saying " $f (x)$ is near $f (x_{0})$ when $x$ is near $x_{0}$ ", or that "a small change in $x$ produces only a small change in $f (x)$ ." Note, however, that it is not the change in $x$ that we start with, but rather the number $ϵ$ , which is to be a bound for the change in $f (x)$ . Then we find an interval about $x_{0}$ , of length $2 δ$ , such that if $x$ is in this interval, i.e., $x$ is not farther than $δ$ away from $x_{0}$ , then $f (x)$ is not farther away from $f (x_{0})$ than the given bound $ϵ$ . It is only by this type of definition that we can be sure of meeting the challenge of a man who says, "I don't call $f (x)$ near $f (x_{0})$ until it is within $10^{- 4}$ ", a second one who says "Near means to me within $10^{- 8}$ ", and so on. Thus in order to show that a given function $f (x)$ is continuous at $x_{0}$ , we must start with a symbol $ϵ$ , representing a general positive number, and then show that we can find a positive number $δ$ , which will usually depend on $ϵ$ , such that the inequality $| x - x_{0} | < δ$ gives us some information about the expression $| f (x) - f (x_{0}) |$ , namely that it is less than $ϵ$ . To prove that $f (x)$ is not continuous at $x_{0}$ , it would suffice to demonstrate just one $ϵ$ for which we could not find a suitable $δ$ . (Of course, then we could not find a suitable $δ$ for any smaller $ϵ$ .)

Let us illustrate by discussing some examples:

Example 1.

$f (x) = x^{2}, x_{0} = 1$ . If you recheck our proof that a differentiable function is continuous, you will see that it is completely rigorous in terms of the definitions and the theorems about limits which we have now given. However, let us prove that $x^{2}$ is continuous at $1$ using only the definition of continuity.

Let $ϵ > 0$ be given. Then we must study $| f (x) - f (x_{0}) | = | x^{2} - 1 |$ . Let us write $h = x - 1, x = 1 + h$ . Then

| f (x) - f (x_{0}) | = | 1 + 2 h + h^{2} - 1 | = | 2 h + h^{2} | = | h | | 2 + h | \cdot

Now let us take $δ = Min (1, \frac{ϵ}{3})$ . Then if $| h | < δ$ ,

| 2 + h | \leq 2 + | h | < 3, and | h | | 2 + h | < 3 | h | < ϵ \cdot

Since $h = x - 1 = x - x_{0}$ , we see that whenever $| x - x_{0} | < δ$ , we have $| f (x) - f (x_{0}) | < ϵ$ , and thus that $f (x)$ is continuous at $x_{0} \cdot$

Example 2.

Our second example will seem somewhat more bizarre. Let $f (x)$ be defined for $0 \leq x \leq 1$ as follows:
If $x$ is irrational, $f (x) = 0$ .
If $x = \frac{p}{q}$ in lowest terms ( $q > 0$ , $p \geq 0$ , $p$ and $q$ integers), then

f (x) = \frac{1}{q} \cdot

For example,
$f (\frac{\sqrt{2}}{2}) = 0$ , $f (\frac{π}{4}) = 0$ , $f (\frac{6}{7}) = \frac{1}{7}$ , $f (\frac{1}{3}) = \frac{1}{3}$ , $f (\frac{2}{3}) = \frac{1}{3};$

f (0) = 1 (0 = \frac{0}{1} in lowest terms), f (1) = 1 \cdot

First of all, $f (x)$ is not continuous at $x_{0}$ if $x_{0}$ is a rational number; for if $x_{0} = \frac{p}{q}$ in lowest terms, $f (x_{0}) = \frac{1}{q} > 0$ . In every interval about $x_{0}$ we have irrational numbers $x$ , where $f (x) = 0$ . Therefore, if we tried $ϵ = \frac{1}{2 q}$ , we could not find a suitable $δ$ . Now we show that $f (x)$ is continuous at $x_{0}$ if $x_{0}$ is any irrational number; then $f (x_{0}) = 0$ . Let $ϵ > 0$ be given, and find an integer $n > 0$ such that $\frac{1}{n} < ϵ \cdot$ Now there are in the interval $0 \leq x \leq 1$ only a finite number of rational numbers which, in lowest terms, have denominators less than $n$ ; there are two with denominator $1$ , one with denominator $2$ , two with denominator $3$ , two with denominator $4$ , four with denominator $5, \dots \cdot$ In general, except for denominator $1$ , there at most $(k - 1)$ with denominator $k$ (and this is only achieved if $k$ is prime). Therefore those with denominator less than $n$ are at most

2 + 1 + 2 + \dots + (n - 2) = 2 + \frac{(n - 1) (n - 2)}{2}

in number. Some one of these, say $r$ , is therefore closest to $x_{0}$ . Let $δ = | r - x_{0} | \cdot δ > 0$ , since $x_{0}$ is irrational. Now let $| x - x_{0} | < δ$ . If $x$ is irrational, then $| f (x) - f (x_{0}) | = | 0 - 0 | = 0 < ϵ$ . If $x = \frac{p}{q}$ is rational, we know that $q \geq n$ , or $f (x) = \frac{1}{q} \leq \frac{1}{n} < ϵ$ . Thus $| f (x) - f (x_{0}) | = \frac{1}{q} < ϵ$ , and we have proved that $f (x)$ is continuous at $x_{0}$ .

EXERCISES

Exercise 1.

Prove that $| x |$ is continuous at $x = 0$ .

Exercise 2.

Prove that $x^{2 / 3}$ is continuous at $x = 0$ .

Exercise 3.

Prove that

f(x) = \begin{cases} x \sin \dfrac{1}{x}, & x \neq 0 \\ 0, & x = 0 \end{cases}

is continuous at $x = 0$ .

Exercise 4.

Prove that $[x]$ is continuous at $x_{0}$ if $x_{0}$ is not an integer, and that it is not continuous at $x_{0}$ if $x_{0}$ is an integer.

Exercise 5.

Prove the following, assuming that $f (x)$ and $g (x)$ are continuous at $x_{0}$ :
a) $f (x) + g (x)$ is continuous at $x_{0}$ .
b) $c f (x)$ is continuous at $x_{0}$ , where $c$ is a constant.
c) $f (x) g (x)$ is continuous at $x_{0}$ .
d) If $f (x_{0}) \neq 0, \frac{1}{f (x)}$ is continuous at $x_{0}$ .

Exercise 6.

Write out in formal language the proof that if f'(x_0) exists, then $f (x)$ is continuous at $x_{0}$ .

Continuous Functions

EXERCISES

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