4.4 Limits of functions

The $\varepsilon$ - $\delta$ definition of continuity bears a very close resemblance to the $\varepsilon$ - $N$ definition of a limit of a sequence. In this section, we develop this connection by introducing limits of functions.

The examples in Figure 4.15 show different behaviours that are possible, and are helpful for building up intuition about limits of functions as $x\to a$ . In each case, we describe what we would intuitively expect our definition of a limit to satisfy: the functions in (a), (b) and (c) should all converge to $\ell$ as $x\to a$ , but the function in (d) should not converge as $x\to a$ .

(a)

f(x)

converges to

f(a)

x\to a

(b)

g(x)

converges to

\ell

x\to a

(c)

h(x)

converges to

\ell

x\to a

(d)

k(x)

does not converge as

x\to a

Figure 4.15: Examples of functions with different behaviours at the point

a

One curious feature is that Figure 4.15 (b) and (c) suggest that our definition of “the limit of $f(x)$ as $x\to a$ ” should work whether or not $f$ is defined at $a$ : the value of the limit should depend on the behaviour of the function around $a$ , but not actually at $a$ . This observation turns out to be important and useful.

With the above examples in mind, we turn to the formal definition.

Definition 4.41 (Limit of a function).

Let $I\subseteq\mathbb{R}$ be an interval and $a\in I$ . Let $f\colon E\to\mathbb{R}$ where either $E=I$ or $E=I\setminus\{a\}$ .

We say $f(x)$ converges to $\ell\in\mathbb{R}$ as $x$ tends to $a$ if for all $\varepsilon>0$ , there exists some $\delta>0$ (which in general depends on $\varepsilon$ ) such that

\text{if $x\in I$ satisfies $0<|x-a|<\delta$, then}\quad|f(x)-\ell|<\varepsilon.

In this case, we write $\displaystyle\lim_{x\to a}f(x)=\ell$ or $f(x)\to\ell$ as $x\to a$ .

As we shall see, Definition 4.41 turns out to be precisely what is needed to formalise our intuition from Figure 4.15.

We can reinterpret continuity in terms of the notion of a limit of a function. Indeed, for the functions in Figure 4.15, we can see that the only example which is continuous at $a$ is the first one, where the limit at $a$ agrees with $f(a)$ . For the functions that are not continuous, either $f$ is undefined at $a$ , or the limit at $a$ does not exist. The following lemma formalises this observation.

Lemma 4.42 (Limit characterisation of continuity).

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

f\text{ is continuous at }a\in I\quad\text{if and only if}\quad\lim_{x\to a}f(% x)=f(a).

Proof.

This is immediate from the definitions of continuity (Definition 4.20) and limits (Definition 4.41). ∎

Example 4.43.

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

(4.7) (4.7)

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

we illustrate the graph of $\chi_{\mathbb{Z}}$ in Figure 4.16.

Claim: $\displaystyle\lim_{x\to 0}\chi_{\mathbb{Z}}(x)=0$ .

Proof.

Let $\varepsilon>0$ be given and choose $\delta:=1$ . If $x\in\mathbb{R}$ satisfies $0<|x|<\delta$ , then $x\notin\mathbb{Z}$ and so $\chi_{\mathbb{Z}}(x)=0$ . Hence,

|\chi_{\mathbb{Z}}(x)-0|=|0-0|=0<\varepsilon.

Thus, by the $\varepsilon$ - $\delta$ definition of a limit, $\lim_{x\to 0}\chi_{\mathbb{Z}}(x)=0$ . ∎

Note that while $\lim_{x\to 0}\chi_{\mathbb{Z}}(x)=0$ , we have $\chi_{\mathbb{Z}}(0)=1$ . Since $\lim_{x\to 0}\chi_{\mathbb{Z}}(x)\neq\chi_{\mathbb{Z}}(0)$ , it follows from the limit characterisation of continuity from Lemma 4.42 that $\chi_{\mathbb{Z}}$ is discontinuous at $0$ .

Figure 4.16: The function

\chi_{\mathbb{Z}}\colon\mathbb{R}\to\mathbb{R}

from Example 4.43 satisfies

\displaystyle\lim_{x\to 0}\chi_{\mathbb{Z}}(x)=0

The definition of the limit $\lim_{x\to a}f(x)$ also has the additional flexibility that it does not require $f$ to be defined at $a$ . We illustrate why this can be useful in the following example.

(a) The function

\mathrm{sinc}^{*}

is defined on

\mathbb{R}\setminus\{0\}

. There is a conspicuous ‘hole’ in the graph.

(b) The function

\mathrm{sinc}

is formed by extending the definition of

\mathrm{sinc}^{*}

x=0

by ‘filling in the hole’.

Figure 4.17: The function

\mathrm{sinc}^{*}

from Example 4.44 and its continuous extension

\mathrm{sinc}

Example 4.44.

Consider the function

\mathrm{sinc}^{*}\colon\mathbb{R}\setminus\{0\}\to\mathbb{R};\qquad\mathrm{% sinc}^{*}(x):=\frac{\sin x}{x}\qquad\text{for all $x\in\mathbb{R}\setminus\{0% \}$;}

see Figure 4.17. Although we have not defined $\mathrm{sinc}^{*}$ at $x=0$ , we can nevertheless define the limit of $\mathrm{sinc}^{*}$ as $x\to 0$ . In particular, we claim that $\displaystyle\lim_{x\to 0}\mathrm{sinc}^{*}(x)=1$ .

Proof.

Let $x\in(0,\pi/2]$ and recall from Lemma 4.32 that $\sin x\leq x\leq\tan x$ . Taking reciprocals and multiplying through by $\sin x>0$ , we obtain

0\leq\cos x\leq\frac{\sin x}{x}\leq 1.

This implies that

(4.8) (4.8)

\Big{|}\frac{\sin x}{x}-1\Big{|}\leq|\cos x-1|.

A similar argument shows that (4.8) also holds for $x\in[-\pi/2,0)$ .

By Lemma 4.33, the function $\cos$ is continuous and so given $\varepsilon>0$ , there exists some $\delta_{0}>0$ such that if $|x|<\delta_{0}$ , then $|\cos x-1|<\varepsilon$ . Set $\delta:=\min\{\delta_{0},\pi/2\}$ . If $0<|x|<\delta$ , then (4.8) implies that

|\mathrm{sinc}^{*}(x)-1|=\Big{|}\frac{\sin x}{x}-1\Big{|}<\varepsilon.

Hence, by the $\varepsilon$ - $\delta$ definition of a limit, $\displaystyle\lim_{x\to 0}\mathrm{sinc}^{*}(x)=1$ . ∎

Example 4.45.

In light of Example 4.44, it makes sense to extend $\mathrm{sinc}^{*}\colon\mathbb{R}\setminus\{0\}\to\mathbb{R}$ to a function on $\mathrm{sinc}\colon\mathbb{R}\to\mathbb{R}$ defined on the whole real line and given by

\mathrm{sinc}\colon\mathbb{R}\to\mathbb{R};\qquad\mathrm{sinc}(x):=\begin{% cases}\displaystyle\frac{\sin x}{x}&\text{if $x\neq 0$,}\\ 1&\text{if $x=0$.}\end{cases}

Intuitively, we form $\mathrm{sinc}$ by ‘filling the hole’ in the graph of $\mathrm{sinc}^{*}$ . Moreover, by choosing the value $\mathrm{sinc}(0)=1$ , we fill in the hole in such a way that $\mathrm{sinc}\colon\mathbb{R}\to\mathbb{R}$ is continuous: see Figure 4.17.

Exercise 4.46.

Show that $\displaystyle\lim_{x\to 0}\frac{\cos x-1}{x}=0$ .

Exercise 4.47.

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

f(x):=\begin{cases}\sin(\pi x)&\text{if $x<1$;}\\ 1066&\text{if $x=1$;}\\ x^{2}-1&\text{if $x>1$.}\end{cases}

Determine whether the limit $\displaystyle\lim_{x\to 1}f(x)$ exists. If the limit does exist, then compute its value. Is $f$ continuous at $1$ ?

Exercise 4.48.

Recall that the boundedness test for sequences states that if a limit of a sequence exists, then the sequence is bounded.

(i)

Show that the analogue of this result does not hold for limits of functions. In particular, find a function $f\colon\mathbb{R}\to\mathbb{R}$ such that $\displaystyle\lim_{x\to a}f(x)$ exists for some $a\in\mathbb{R}$ , but $f$ is unbounded.
(ii)

However, a local version of the boundedness test does hold. Let $I\subseteq\mathbb{R}$ be an interval, $a\in I$ and $f\colon I\setminus\{a\}\to\mathbb{R}$ and suppose $\lim_{x\to a}f(x)$ exists. Show there exists some $\delta_{0}>0$ and $M>0$ such that

$|f(x)|<M\qquad\text{for all $x\in(a-\delta_{0},a+\delta_{0})\cap I\setminus\{a% \}$.}$