4.3 The $\varepsilon$ - $\delta$ definition of continuity

Intuitively, a function is continuous if its graph contains no ‘abrupt jumps’ or ‘breaks’, so that it can be drawn without lifting the pen off the page. Many familiar functions such as polynomials and trigonometric functions are continuous. For contrast, we illustrate two functions which do not have this property in Figure 4.9: in both cases, the graph is ‘broken’ at $x=a$ . Whilst the mechanical ‘without lifting the pen’ idea of continuity is helpful for building intuition, it is not useful for mathematical reasoning. We therefore need to formulate a precise mathematical definition, which encapsulates our intuition.

Figure 4.9: Functions with jumps. In both cases, we can find values of

x

slightly to the right of

a

for which

|f(x)-f(a)|>1/2

. In particular, the point

(x,f(x))

lies outside the yellow corridor around the line

y=1

We first need to have a better understanding of what it means for the graph to have an ‘abrupt jump’. Let’s consider the functions illustrated in Figure 4.9. From the picture, we see that each function has a jump at $x=a$ . If we take any point $x$ slightly to the right of $a$ , then we see from the sketch that $f(x)\geq 3/2$ but $f(a)=1$ . Thus, even if $x$ is really close to $a$ , the values of $f(x)$ and $f(a)$ are far apart: they are separated by at least $1/2$ . Visually, the point $(x,f(x))$ lies outside the horizontal corridor of radius $1/2$ , centred around the line $y=1$ .

With the above example in mind, we can start to understand what it means for a function to be continuous at a point $a\in\mathbb{R}$ . In particular, we need to rule out the behaviour observed in Figure 4.9. Thus, continuity should mean

by taking

x

close to

a

, we can make

f(x)

as close as we like to

f(a)

Another way to express this is:

(4.3) (4.3) For any measure of closeness, if

x

is near enough to

a

, then

f(x)

is close to

f(a)

We can turn (4.3) into a formal mathematical definition in a similar way to how we defined the limit of a sequence. Note that, since we want to consider $x$ that are ‘close’ to $a$ , we need $f$ to be defined on an interval around $a$ . For simplicity, the way we will ensure this is by simply requiring that the domain of $f$ is an interval.

Definition 4.20 (

\varepsilon

\delta

definition of continuity).

Let $I\subseteq\mathbb{R}$ be an interval, $f\colon I\to\mathbb{R}$ and $a\in I$ .

1

We say $f$ is continuous at $a$ if for all $\varepsilon>0$ , there exists some $\delta>0$ (which in general depends on $\varepsilon$ ) such that

$\text{if $x\in I$ satisfies $|x-a|<\delta$, then}\quad|f(x)-f(a)|<\varepsilon.$
2

We say $f$ is discontinuous at $a$ if it is not continuous at $a$ .

(a) Given

\varepsilon>0

, we consider a horizontal

\varepsilon

-corridor around

f(a)

(b) We choose

\delta>0

so that the graph of

f

over the interval

(a-\delta,a+\delta)

lies entirely within the

\varepsilon

-corridor.

Figure 4.10: The

\varepsilon

\delta

definition of continuity.

We illustrate the idea behind this definition in Figure 4.10, which shows the graph of a function $f$ which is continuous at $a$ . Given $\varepsilon>0$ , in Figure 4.10(a) we consider the horizontal ‘ $\varepsilon$ -corridor’ around $f(a)$ : that is, all points $(x,y)$ in the plane with $y$ values satisfying $f(a)-\varepsilon<y<f(a)+\varepsilon$ . We are free to choose $\varepsilon>0$ as small as we like, which means we can take the corridor to be as thin as we like. In Figure 4.10(b) we see that, no matter how thin we make the corridor²² 2 You may find the applet at https://www.desmos.com/calculator/kycrjq0aqj helpful to see this: it is an interactive version of Figure 4.10(b), where you can use the sliders to vary $\varepsilon$ and $\delta$ and see the effect this has on the shaded regions., it is always possible choose $\delta>0$ such that the graph of $f$ over the interval $(a-\delta,a+\delta)$ lies entirely within the corridor. Thus, if $|x-a|<\delta$ , so that $x\in(a-\delta,a+\delta)$ , then the point $(x,f(x))$ lies in the $\varepsilon$ -corridor, which means precisely that $|f(x)-f(a)|<\varepsilon$ : see Figure 4.11.

We can also compare how the formal definition matches with our intuitive notion of continuity.

$\varepsilon$ - $\delta$ definition (Definition 4.20)	Informal idea (4.3)
For all $\varepsilon>0$	For any measure of closeness,
if $x\in I$ satisfies $\|x-a\|<\delta$ ,	if $x$ is near enough to $a$ ,
then $\|f(x)-f(a)\|<\varepsilon$ .	then $f(x)$ is close to $f(a)$ .

Figure 4.11: The

\varepsilon

\delta

definition of continuity. If

|x-a|<\delta

, then

(x,f(x))

lies in the

\varepsilon

-corridor about

f(a)

and so

|f(x)-f(a)|<\varepsilon

Definition 4.20 is very reminiscent of the $\varepsilon$ - $N$ definition of the limit of a sequence. For sequences, the parameter $N$ is used to describe ‘terms far enough along the sequence’. Here the parameter $\delta$ plays a similar role: it is used to describe ‘ $x$ near enough to $a$ ’. For this reason, many of the skills you acquired when working with sequences can be applied to understand continuity of functions.

Example 4.21.

The function $p_{2}\colon\mathbb{R}\to\mathbb{R}$ given by $p_{2}(x):=x^{2}$ for all $x\in\mathbb{R}$ is continuous at $1$ .

Proof.

For $x\in\mathbb{R}$ , we have

|p_{2}(x)-p_{2}(1)|=|x^{2}-1|=|x+1||x-1|.

Suppose $|x-1|<1$ . Then, by the triangle inequality,

|x+1|=|x-1+2|\leq|x-1|+2<1+2=3

and so

|p_{2}(x)-p_{2}(1)|<3|x-1|.

Let $\varepsilon>0$ be given and set $\delta:=\min\{\varepsilon/3,1\}$ . If $x\in\mathbb{R}$ satisfies $|x-1|<\delta$ , then we certainly have $|x-1|<1$ and so it follows by our earlier observations that

|p_{2}(x)-p_{2}(1)|<3|x-1|<3\delta\leq\varepsilon.

Thus, by the $\varepsilon$ - $\delta$ definition of continuity, $p_{2}$ is continuous at $1$ . ∎

Example 4.22.

Let $g\colon\mathbb{R}\to\mathbb{R}$ be the function from Example 4.1, which is given by

g(x):=\begin{cases}-x&\text{if $x<0$,}\\ \sqrt{x}&\text{if $x\geq 0$.}\end{cases}

We sketched the graph of $g$ in Figure 4.5, which shows that there is a rough ‘bend’ at $x=0$ . Nevertheless, we shall prove $g$ is continuous at $0$ .

Proof.

Let $\varepsilon>0$ be given and choose $\delta:=\min\{\varepsilon^{2},\varepsilon\}$ . If $0\leq x<\delta$ , then

|g(x)-g(0)|=|\sqrt{x}-0|=\sqrt{x}<\sqrt{\delta}\leq\varepsilon.

On the other hand, if $-\delta<x<0$ , then

|g(x)-g(0)|=|-x-0|=|x|<\delta\leq\varepsilon.

Combining these cases, we see that if $|x|<\delta$ , then $|g(x)-g(0)|<\varepsilon$ . Hence, by the $\varepsilon$ - $\delta$ definition of continuity, $g$ is continuous at $0$ . ∎

Exercise 4.23.

Consider the linear function $\ell\colon\mathbb{R}\to\mathbb{R}$ given by $\ell(x):=5x+6$ for all $x\in\mathbb{R}$ . Show that $\ell$ is continuous at $1$ .

Exercise 4.24.

Consider the function $f\colon\mathbb{R}\to\mathbb{R}$ given by $f(x):=x^{3}$ for all $x\geq 0$ and $f(x):=x$ for $x<0$ . Sketch the graph of $f$ and show that it is continuous at $0$ .

So far we have only talked about continuity at a point. We can also define what it means for the entire function to be continuous.

Definition 4.25.

Let $I\subseteq\mathbb{R}$ be an interval and $f\colon I\to\mathbb{R}$ . We say $f$ is continuous if $f$ is continuous at $a$ for all $a\in I$ . We say $f$ is discontinuous if it is not continuous.

Example 4.26.

The function $p_{2}\colon\mathbb{R}\to\mathbb{R}$ given by $p_{2}(x):=x^{2}$ for all $x\in\mathbb{R}$ is continuous.

Proof.

Let $a\in\mathbb{R}$ . For $x\in\mathbb{R}$ , we have

|p_{2}(x)-p_{2}(a)|=|x^{2}-a^{2}|=|x+a||x-a|.

Suppose $|x-a|<1$ . Then, by the triangle inequality,

|x+a|=|x-a+2a|\leq|x-a|+2|a|<1+2|a|

and so

|p_{2}(x)-p_{2}(a)|<(1+2|a|)|x-a|.

Let $\varepsilon>0$ be given and set $\delta:=\min\{\varepsilon/(1+2|a|),1\}$ . If $x\in\mathbb{R}$ satisfies $|x-a|<\delta$ , then $|x-1|<1$ and so it follows by our earlier observations that

|p_{2}(x)-p_{2}(a)|<(1+2|a|)|x-a|<(1+2|a|)\delta\leq\varepsilon.

Thus, by the $\varepsilon$ - $\delta$ definition of continuity, $p_{2}$ is continuous at $a$ for all $a\in\mathbb{R}$ , therefore $p_{2}$ is continuous. ∎

Exercise 4.27.

Show that the function $p_{3}\colon\mathbb{R}\to\mathbb{R}$ given by $p_{3}(x):=x^{3}$ for all $x\in\mathbb{R}$ is continuous. Hint: $x^{3}-a^{3}=(x-a)(x^{2}+ax+a^{2})$ .

Exercise 4.28.

Show that the reciprocal function $r\colon(0,\infty)\to\mathbb{R}$ given by $r(x):=1/x$ is continuous at every point $a\in(0,\infty)$ .

It is equally important to be able to show certain functions fail to be continuous. Let $I\subseteq\mathbb{R}$ be an interval, $f\colon I\to\mathbb{R}$ and $a\in I$ . By negating the definition of continuous at $a$ , we obtain the following.

	The function $f\colon I\to\mathbb{R}$ is discontinuous at $a$ if
(4.4)	(4.4)	there exists some $\varepsilon>0$ such that for all $\delta>0$ ,
	there exists some $x\in I$ with $\|x-a\|<\delta$ such that $\|f(x)-f(a)\|\geq\varepsilon$ .

Exercise 4.29.

Give an informal explanation of what it means for $f\colon I\to\mathbb{R}$ to be discontinuous at $a$ and match it to the precise definition in (4.4) by filling in the following table.

$\varepsilon$ - $\delta$ definition (4.4)	Informal idea
There exists some $\varepsilon>0$
such that for all $\delta>0$
there exists some $x\in I$ with $\|x-a\|<\delta$
such that $\|f(x)-f(a)\|\geq\varepsilon$ .

Example 4.30.

Let $h\colon\mathbb{R}\to\mathbb{R}$ be the function from Example 4.1, given by

h(x):=\begin{cases}x^{2}&\text{for $x\leq 1$,}\\ x^{2}+1&\text{for $x>1$.}\end{cases}

We sketched the graph of $h$ in Figure 4.5, which shows that there is a ‘jump’ or ‘break’ in the graph at $x=1$ . Consequently, we expect $h$ fails to be continuous at $1$ . We prove this is indeed the case. (You may find the applet at https://www.desmos.com/calculator/5lsgjmwbdl helpful to illustrate the proof.)

Proof.

Let $\varepsilon:=1$ and $x>1$ . It follows that

|h(x)-h(1)|=|x^{2}+1-1|=x^{2}>1=\varepsilon.

In particular, given any $\delta>0$ , if we choose $x\in(1,1+\delta)$ , then it follows that $|x-1|<\delta$ and $|h(x)-h(1)|\geq\varepsilon$ . Thus, $h$ is discontinuous at $1$ . ∎

Exercise 4.31.

Consider the function $f\colon\mathbb{R}\to\mathbb{R}$ given by

f(x):=\begin{cases}0&\text{if $x\leq 0$,}\\ 1&\text{if $x>0$,}\end{cases}

which is graphed in Figure 4.12. Show that $f$ is discontinuous at $0$ .

Figure 4.12: The function

f\colon\mathbb{R}\to\mathbb{R}

as defined in Exercise 4.31.

Trigonometric functions

Here we begin an in-depth study of the familiar trigonometric functions such as $\sin$ and $\cos$ . Intuitively, from our knowledge of the graphs of these functions we expect them to be continuous. Our goal is to verify that this is indeed the case.

To achieve this goal, we will rely on certain inequalities for trigonometric functions, which can be proved using geometric arguments. Since these inequalities will also be useful later in the course (and in other courses!), we give them their own lemma.

Lemma 4.32 (Trigonometric inequalities).

For all $\theta\in(-\pi/2,\pi/2)$ we have

|\sin\theta|\leq|\theta|\leq|\tan\theta|\qquad\text{and}\qquad|1-\cos\theta|% \leq\frac{\theta^{2}}{2}.

We shall return to prove these results after achieving our goal of establishing the continuity of $\sin$ and $\cos$ .

Lemma 4.33.

The functions $\sin\colon\mathbb{R}\to\mathbb{R}$ and $\cos\colon\mathbb{R}\to\mathbb{R}$ are continuous.

Proof.

Since $\cos\theta=\sin(\theta+\pi/2)$ for all $\theta\in\mathbb{R}$ , it suffices to show the result for the sine function only.

Let $\varepsilon>0$ be given and fix $a\in\mathbb{R}$ . To establish that $\sin$ is continuous at $a$ , we need to choose $\delta$ so that whenever $|x-a|<\delta$ , we have $|\sin x-\sin a|<\varepsilon$ . To make progress, we write $x=a+(x-a)$ and use the angle summation formulæ:

$\displaystyle\|\sin x-\sin a\|$	$\displaystyle=\left\|\sin\left(a+(x-a)\right)-\sin a\right\|$
	$\displaystyle=\left\|\sin a\left(\cos(x-a)-1\right)+\cos a\sin(x-a)\right\|$
	$\displaystyle\leq\left\|\sin a\right\|\left\|1-\cos(x-a)\right\|+\|\cos a\|\|\sin(x-a% )\|.$

Let $\theta:=x-a$ . Since $|\theta|=|x-a|<\delta$ , if we ensure $\delta\leq\pi/2$ then Lemma 4.32 will certainly apply. Thus,

|\sin x-\sin a|\leq 1\cdot\frac{\theta^{2}}{2}+1\cdot|\theta|\leq 2|\theta|,

where we have also used the fact that $|\sin a|\leq 1$ and $|\cos a|\leq 1$ .

So, choose $\delta:=\min\{\varepsilon/2,\pi/2\}$ . From the above observations, it follows that if $x\in\mathbb{R}$ satisfies $|x-a|<\delta$ , then

\displaystyle|\sin x-\sin a|\leq 2|\theta|<2\delta\leq\varepsilon.

Thus, $\sin$ is continuous at $a$ by the $\varepsilon$ - $\delta$ definition of continuity. Since our choice of $a$ was arbitrary, we conclude that $\sin$ is continuous on $\mathbb{R}$ . ∎

We now return to prove the borrowed Lemma.

Proof (of Lemma 4.32).

Since $\sin$ and $\tan$ are odd functions and $\cos$ is an even function, it suffices to consider the case $\theta\in[0,\pi/2)$ . Since the trigonometric functions arise in geometry, our argument is geometric.

We shall prove the inequality $\sin\theta\leq\theta$ and leave the remaining inequalities as exercises (see Exercise 4.34 below).

On the one hand, the area of the triangle $OAB$ in Figure 4.13(a) is $(\sin\theta)/2$ . On the other hand, the area of the sector $OAB$ is $\theta/2$ . Since the sector contains the triangle, $(\sin\theta)/2\leq\theta/2$ and the result follows. ∎

(a)

\sin\theta\leq\theta

for

\theta\in[0,\pi/2]

(b)

\theta\leq\tan\theta

for

\theta\in[0,\pi/2]

Figure 4.13: Geometric observations behind the trigonometric inequalities.

Exercise 4.34.

Let $\theta\in[0,\pi/2)$ . Complete the proof of Lemma 4.32 as follows.

(i)

Using Figure 4.13(b), apply a geometric argument to show that $|\theta|\leq|\tan\theta|$ .
(ii)

Use $|\sin\theta|\leq|\theta|$ and the double angle formula to show that $|1-\cos\theta|\leq\theta^{2}/2$ .

The exponential and logarithm

We also consider the exponential and natural logarithm functions, introduced in Definition 4.5 and Definition 4.7.

Lemma 4.35.

The functions $\exp\colon\mathbb{R}\to(0,\infty)$ and $\log\colon(0,\infty)\to\mathbb{R}$ as continuous.

For now will ‘borrow’ this result and leave the proof until later. At the end of this chapter, we shall see that continuity of $\exp$ implies continuity of $\log$ . In the next chapter, we shall show $\exp$ is continuous, and therefore complete the proof.

Further examples

Here we present some more ‘exotic’ examples which really push the boundaries of our intuition.

The following example demonstrates a new kind of behaviour which results in a discontinuity: oscillation.

Example 4.36.

Let $f\colon\mathbb{R}\to\mathbb{R}$ be defined by

(4.5) (4.5)

f(x):=\begin{cases}\sin(1/x)&\text{if $x\neq 0$,}\\ 0&\text{if $x=0$.}\end{cases}

We recall the graph of $f$ in Figure 4.14(a). Recall that, for $x$ close to $0$ , the function $\sin(1/x)$ will oscillate very rapidly. In particular, there will be infinitely many ‘peaks’ where $f(x)=1$ . Using this observation, we can show $f$ is not continuous at $0$ .

Proof.

Choose $\varepsilon:=1$ . For any $\delta>0$ , there exists some $n\in\mathbb{N}$ such that $0<x_{n}<\delta$ where

x_{n}:=\frac{2}{\pi(4n+1)}\qquad\text{satisfies}\qquad\frac{1}{x_{n}}=2\pi n+% \frac{\pi}{2}.

In particular, by the periodicity of the $\sin$ function,

f(x_{n})=\sin(1/x_{n})=\sin(2\pi n+\pi/2)=\sin(\pi/2)=1.

Thus, $|x_{n}-0|<\delta$ and $|f(x_{n})-f(0)|=1\geq\varepsilon$ . Since $\delta>0$ was chosen arbitrarily, this shows that $f$ is not continuous at $0$ . ∎

(a)

f\colon\mathbb{R}\to\mathbb{R}

defined in (4.5).

(b)

g\colon\mathbb{R}\to\mathbb{R}

defined in (4.6).

Figure 4.14: Highly oscillatory functions.

The next example demonstrates even more extreme oscillatory behaviour.

Example 4.37.

Let $\chi_{\mathbb{Q}}\colon\mathbb{R}\to\mathbb{R}$ be the Dirichlet function from Example 4.3, defined by

\chi_{\mathbb{Q}}(x):=\begin{cases}1&\text{if $x\in\mathbb{Q}$,}\\ 0&\text{if $x\in\mathbb{R}\setminus\mathbb{Q}$.}\end{cases}

Then $\chi_{\mathbb{Q}}$ is discontinuous at every point $a\in\mathbb{R}$ .

Proof.

Fix $a\in\mathbb{R}$ . We will consider the case where $a\in\mathbb{R}\setminus\mathbb{Q}$ is irrational; the remaining case is left as an exercise (see Exercise 4.38).

Choose $\varepsilon:=1$ . For any $\delta>0$ , by the density of the rationals (see Section 1.6), there exists some $x\in\mathbb{Q}$ such that $0<|x-a|<\delta$ . In this case, $|\chi_{\mathbb{Q}}(x)-\chi_{\mathbb{Q}}(a)|=|1-0|=1\geq\varepsilon$ . Since $\delta>0$ was chosen arbitrarily, this shows that $\chi_{\mathbb{Q}}$ is not continuous at $a$ . ∎

Exercise 4.38.

Let $\chi_{\mathbb{Q}}\colon\mathbb{R}\to\mathbb{R}$ be the function from Example 4.37 and $a\in\mathbb{Q}$ be a rational point. Show that $\chi_{\mathbb{Q}}$ is discontinuous at $a$ .

Exercise 4.39.

Let $g\colon\mathbb{R}\to\mathbb{R}$ be defined by

(4.6) (4.6)

g(x):=\begin{cases}x\sin(1/x)&\text{if $x\neq 0$,}\\ 0&\text{if $x=0$.}\end{cases}

We sketch the graph of $g$ in Figure 4.14(b). Is $g$ continuous at $0$ ?

Exercise 4.40.

Let $h\colon\mathbb{R}\to\mathbb{R}$ be defined by

h(x):=\begin{cases}x&\text{if $x\in\mathbb{Q}$,}\\ 0&\text{if $x\in\mathbb{R}\setminus\mathbb{Q}$.}\end{cases}

Give a rough sketch of the graph of this function. For which values of $a\in\mathbb{R}$ is $h$ continuous at $a$ ?

$\displaystyle\|\sin x-\sin a\|$	$\displaystyle=\left\|\sin\left(a+(x-a)\right)-\sin a\right\|$
	$\displaystyle=\left\|\sin a\left(\cos(x-a)-1\right)+\cos a\sin(x-a)\right\|$
	$\displaystyle\leq\left\|\sin a\right\|\left\|1-\cos(x-a)\right\|+\|\cos a\|\|\sin(x-a% )\|.$

4.3 The ε\varepsilon-δ\delta definition of continuity

Trigonometric functions

The exponential and logarithm

Further examples

4.3 The $\varepsilon$ - $\delta$ definition of continuity