§17 Continuous Functions

Most of calculus is the study of continuous functions. In this chapter we study continuous and uniformly continuous functions.

Why this matters

Continuity is what lets limits pass through a function: if $x_n \to x_0$, then $f(x_n) \to f(x_0)$. The big theorems of the next section (maximum values, the Intermediate Value Theorem) and later of differentiation and integration all depend on it. As always, the vague idea (“the graph has no breaks”) has to be replaced by something precise enough to prove things with.

Recall the salient features of a function $f$:

1. the set on which $f$ is defined, called the domain of $f$ and written $\operatorname{dom}(f)$;
2. the assignment, rule or formula specifying the value $f(x)$ of $f$ at each $x$ in $\operatorname{dom}(f)$.

We consider functions with $\operatorname{dom}(f) \subseteq \mathbb{R}$ that are real-valued, i.e. $f(x) \in \mathbb{R}$ for all $x \in \operatorname{dom}(f)$. Strictly, $f$ is the function and $f(x)$ its value at $x$. A function is often given just by its values; then the domain is understood to be the natural domain, the largest subset of $\mathbb{R}$ on which the formula is a well-defined real-valued function. Thus “the function $f(x) = \frac1x$” is shorthand for “the function $f$ given by $f(x) = \frac1x$ with natural domain $\{x \in \mathbb{R} : x \ne 0\}$.” Similarly the natural domain of $g(x) = \sqrt{4 - x^2}$ is $[-2, 2]$, and that of $\csc x = \frac1{\sin x}$ is the set of $x$ not of the form $n\pi$, $n \in \mathbb{Z}$.

In keeping with the approach of this book, we define continuity in terms of sequences, and then show the definition is equivalent to the usual ε–δ definition.

17.1 Definition.

Let $f$ be a real-valued function whose domain is a subset of $\mathbb{R}$. The function $f$ is continuous at $x_0$ in $\operatorname{dom}(f)$ if, for every sequence $(x_n)$ in $\operatorname{dom}(f)$ converging to $x_0$, we have $\lim_n f(x_n) = f(x_0)$. If $f$ is continuous at each point of a set $S \subseteq \operatorname{dom}(f)$, then $f$ is continuous on $S$. The function $f$ is continuous if it is continuous on $\operatorname{dom}(f)$.

Used in Exercise 17.10, Exercise 17.15, §18

Our definition says the values $f(x)$ are close to $f(x_0)$ when $x$ is close to $x_0$. The next theorem says this another way: condition (1) is the ε–δ definition of continuity found in many calculus books.

17.2 Theorem.

Let $f$ be a real-valued function whose domain is a subset of $\mathbb{R}$. Then $f$ is continuous at $x_0$ in $\operatorname{dom}(f)$ if and only if

$$\begin{gathered} \text{for each } \varepsilon > 0 \text{ there exists } \delta > 0 \text{ such that} \\ x \in \operatorname{dom}(f) \text{ and } |x - x_0| < \delta \text{ imply } |f(x) - f(x_0)| < \varepsilon. \end{gathered} \tag{1}$$

Used in Exercise 17.9, Theorem 18.5

Proof

Suppose (1) holds, and let $(x_n)$ be a sequence in $\operatorname{dom}(f)$ with $\lim x_n = x_0$. Let $\varepsilon > 0$. By (1) there is $\delta > 0$ such that $x \in \operatorname{dom}(f)$ and $|x - x_0| < \delta$ imply $|f(x) - f(x_0)| < \varepsilon$. Since $\lim x_n = x_0$, there is $N$ such that $n > N$ implies $|x_n - x_0| < \delta$. Hence $n > N$ implies $|f(x_n) - f(x_0)| < \varepsilon$, which proves $\lim f(x_n) = f(x_0)$.

Now assume $f$ is continuous at $x_0$ but (1) fails. Then there is $\varepsilon > 0$ for which the implication

$$\text{“}x \in \operatorname{dom}(f) \text{ and } |x - x_0| < \delta \text{ imply } |f(x) - f(x_0)| < \varepsilon\text{”}$$

fails for every $\delta > 0$. In particular it fails for $\delta = \frac1n$, for each $n \in \mathbb{N}$: there is $x_n \in \operatorname{dom}(f)$ with $|x_n - x_0| < \frac1n$ and yet $|f(x_0) - f(x_n)| \ge \varepsilon$. Thus $\lim x_n = x_0$, but $\lim f(x_n) = f(x_0)$ is impossible. So $f$ is not continuous at $x_0$, contrary to assumption.

∎

Intuitiontwo readings of the same idea

Definition 17.1 says that $f$ commutes with limits: $\lim f(x_n) = f(\lim x_n)$ for every sequence in the domain converging to $x_0$. Condition (1) says that any output tolerance can be met by some input tolerance: to get $f(x)$ within $\varepsilon$ of $f(x_0)$, it is enough to take $x$ within $\delta$ of $x_0$. In general $\delta$ depends on both $\varepsilon$ and the point $x_0$. When one $\delta$ works for every point at once, the function is uniformly continuous (§19).

Let $f(x) = 0$ for $x < 0$ and $f(x) = 1$ for $x \ge 0$. At $x_0 = 0$, choose the band half-width $\varepsilon = \tfrac12$ around $f(0) = 1$. Is there a $\delta > 0$ that keeps the graph inside the band for all $|x| < \delta$?

Yes, take $\delta$ small enoughNo, no $\delta$ works

Try it: the ε–δ game

Your opponent picks ε, which sets the blue band around $f(x_0)$. You must answer with a δ so that the whole graph above the orange strip stays inside the band. For a continuous function you can always win; for Example 3 or the step function, once ε is small, no δ does.

ε (band half-height)0.300Your δ0.500

Find the largest δUse the book’s δ = min{1, ε/(2(2|x₀|+1))} = min{1, ε/6}

band f(x₀) ± εstrip x₀ ± δgraph inside the bandgraph escaping the band

✗ δ = 0.500 fails: x = 1.500 is within δ of x₀ but |f(x) − f(x₀)| ≈ 2.50 ≥ ε. The largest δ that works is about 0.0724. Continuous. The book’s δ is safe but not the largest possible.

Try it: defeat the ε

The game above lets you adjust ε and δ freely. Here the roles are fixed, as in the definition. The page plays the opponent: it announces an ε, you commit to a δ, and then it searches the interval $(x_0-\delta, x_0+\delta)$ for a point where $|f(x)-f(x_0)| \ge \varepsilon$. Survive all seven challenges and the function is behaving continuously at $x_0$; if the page keeps breaking you however small you make δ, you have found a discontinuity.

Try Thomae’s function at a rational and at an irrational point, and $\sin(1/x)$ at $0$ and at $0.2$. Notice that surviving seven rounds is not a proof: a proof must give a δ for every ε.

band f(x₀) ± εstrip x₀ ± δinside the bandescaping the band

Your δ0.125

Submit δHalve δNew gameReveal: is f continuous at x₀?

Round 1 of 7 · survived 0. The page challenges you with ε = 0.5. Choose a δ and press “Submit δ”: the page will search for a point x with |x − x₀| < δ and |f(x) − f(x₀)| ≥ ε.

As the next example shows, the sequential definition is sometimes easier to work with than the ε–δ property. But it is important to get comfortable with ε–δ, partly because uniform continuity (§19) is more closely related to it.

Example 1.

Let $f(x) = 2x^2 + 1$ for $x \in \mathbb{R}$. Prove $f$ is continuous on $\mathbb{R}$ (a) using the definition, and (b) using the ε–δ property of Theorem 17.2.

Solution

(a) Suppose $\lim x_n = x_0$. Then

$$\lim f(x_n) = \lim [2x_n^2 + 1] = 2[\lim x_n]^2 + 1 = 2x_0^2 + 1 = f(x_0),$$

where the second equality uses the limit Theorems 9.2–9.4. Hence $f$ is continuous at each $x_0 \in \mathbb{R}$.

(b) Let $x_0 \in \mathbb{R}$ and $\varepsilon > 0$. We want $|f(x) - f(x_0)| < \varepsilon$ whenever $|x - x_0|$ is less than some $\delta$. Observe

$$|f(x) - f(x_0)| = |2x^2 + 1 - (2x_0^2 + 1)| = |2x^2 - 2x_0^2| = 2|x - x_0| \cdot |x + x_0|.$$

We need a bound for $|x + x_0|$ that does not depend on $x$. If $|x - x_0| < 1$, say, then $|x| < |x_0| + 1$ and hence $|x + x_0| \le |x| + |x_0| < 2|x_0| + 1$. Thus

$$|f(x) - f(x_0)| \le 2|x - x_0|(2|x_0| + 1) \qquad\text{provided } |x - x_0| < 1.$$

To make $2|x - x_0|(2|x_0| + 1) < \varepsilon$, it suffices to have $|x - x_0| < \frac{\varepsilon}{2(2|x_0| + 1)}$ and also $|x - x_0| < 1$. So put

$$\delta = \min\left\{1, \frac{\varepsilon}{2(2|x_0| + 1)}\right\}.$$

The work above shows $|x - x_0| < \delta$ implies $|f(x) - f(x_0)| < \varepsilon$.

Intuitionthe “first restrict, then shrink” trick

Part (b) uses a technique that recurs constantly. The difference $|f(x) - f(x_0)|$ factors as $|x - x_0|$ times a factor $|x + x_0|$ that depends on $x$. First restrict to $|x - x_0| < 1$ to get a fixed bound on that factor, then shrink further so that the product is below $\varepsilon$. Taking $\delta$ to be the $\min$ of the two requirements enforces both. Notice that this $\delta$ gets smaller as $|x_0|$ grows: the graph of $2x^2 + 1$ gets steeper, so you must stay closer to $x_0$.

Solution (a) is much easier because the careful analysis was already done in proving the limit theorems of §9.

Example 2.

Let $f(x) = x^2 \sin\left(\frac1x\right)$ for $x \ne 0$ and $f(0) = 0$. Its graph looks continuous; prove $f$ is continuous at $0$.

Let $\varepsilon > 0$. Clearly $|f(x) - f(0)| = |f(x)| \le x^2$ for all $x$. We want this less than $\varepsilon$, so set $\delta = \sqrt\varepsilon$. Then $|x - 0| < \delta$ implies $x^2 < \delta^2 = \varepsilon$, so

$$|x - 0| < \delta \quad\text{implies}\quad |f(x) - f(0)| < \varepsilon.$$

Hence the ε–δ property holds and $f$ is continuous at $0$. It would have been equally easy to show $\lim x_n = 0$ implies $\lim f(x_n) = 0$. This $f$ is also continuous at every other point of $\mathbb{R}$; see Example 4.

Example 3.

Let $f(x) = \frac1x \sin\left(\frac1{x^2}\right)$ for $x \ne 0$ and $f(0) = 0$. Show $f$ is discontinuous, i.e. not continuous, at $0$.

It suffices to find a sequence $(x_n)$ converging to $0$ such that $f(x_n)$ does not converge to $f(0) = 0$. We arrange $\frac1{x_n} \sin\left(\frac1{x_n^2}\right) = \frac1{x_n}$, i.e. $\sin\left(\frac1{x_n^2}\right) = 1$: take $\frac1{x_n^2} = 2\pi n + \frac\pi2$, that is,

$$x_n = \frac1{\sqrt{2\pi n + \frac\pi2}}.$$

Then $\lim x_n = 0$ while $\lim f(x_n) = \lim \frac1{x_n} = +\infty$.

Common pitfallone sequence is not enough to prove continuity

To show $f$ is discontinuous at $x_0$, one bad sequence is enough, as in Example 3. To show $f$ is continuous, Definition 17.1 needs every sequence converging to $x_0$; checking a single convenient one proves nothing. For example, let $g(x) = \sin\frac{\pi}{x}$ for $x \ne 0$ and $g(0) = 0$. Along $x_n = \frac1n$ we get $g(x_n) = \sin(n\pi) = 0 \to g(0)$. But along $x_n = \frac{2}{4n+1}$ we get $g(x_n) = \sin\left(2n\pi + \frac\pi2\right) = 1$ for every $n$. So $g$ is not continuous at $0$.

Common pitfallcontinuity only makes sense at points of the domain

$f(x) = \frac1x$ is continuous: it is continuous at every point of its domain, and $0$ is not in its domain. The graph has a break, but Definition 17.1 says nothing about points where $f$ is not defined. Whether $f$ can be extended continuously to $0$ is a different question, about limits of functions, taken up in §20.

Combining continuous functions

For a real-valued function $f$ and $k \in \mathbb{R}$, $kf$ is the function $(kf)(x) = kf(x)$ for $x \in \operatorname{dom}(f)$, and $|f|$ is the function $|f|(x) = |f(x)|$. For example, if $f(x) = \sqrt x - 4$ for $x \ge 0$, then $(3f)(x) = 3\sqrt x - 12$ and $|f|(x) = |\sqrt x - 4|$, both for $x \ge 0$.

17.3 Theorem.

Let $f$ be a real-valued function with $\operatorname{dom}(f) \subseteq \mathbb{R}$. If $f$ is continuous at $x_0$ in $\operatorname{dom}(f)$, then $|f|$ and $kf$, $k \in \mathbb{R}$, are continuous at $x_0$.

Used in Exercise 17.7, Exercise 17.8, Exercise 18.11

Proof

Let $(x_n)$ be a sequence in $\operatorname{dom}(f)$ converging to $x_0$. Since $f$ is continuous at $x_0$, $\lim f(x_n) = f(x_0)$, and Theorem 9.2 gives $\lim kf(x_n) = kf(x_0)$. So $kf$ is continuous at $x_0$.

For $|f|$ we need $\lim |f(x_n)| = |f(x_0)|$. This follows from

$$\big||f(x_n)| - |f(x_0)|\big| \le |f(x_n) - f(x_0)| \tag{1}$$

(Exercise 3.5). In detail: given $\varepsilon > 0$ there is $N$ such that $n > N$ implies $|f(x_n) - f(x_0)| < \varepsilon$, so by (1) $n > N$ implies $\big||f(x_n)| - |f(x_0)|\big| < \varepsilon$.

∎

Real-valued functions $f$ and $g$ combine into new functions:

$$\begin{aligned} (f + g)(x) &= f(x) + g(x), & fg(x) &= f(x)g(x), \\ (f/g)(x) &= \frac{f(x)}{g(x)}, & g \circ f(x) &= g(f(x)), \\ \max(f, g)(x) &= \max\{f(x), g(x)\}, & \min(f, g)(x) &= \min\{f(x), g(x)\}. \end{aligned}$$

The function $g \circ f$ is the composition of $g$ and $f$. Each new function is defined exactly where it makes sense: $f + g$, $fg$, $\max(f, g)$ and $\min(f, g)$ have domain $\operatorname{dom}(f) \cap \operatorname{dom}(g)$; $f/g$ has domain $\operatorname{dom}(f) \cap \{x \in \operatorname{dom}(g) : g(x) \ne 0\}$; and $g \circ f$ has domain $\{x \in \operatorname{dom}(f) : f(x) \in \operatorname{dom}(g)\}$. Note $f + g = g + f$ and $fg = gf$, but in general $f \circ g \ne g \circ f$.

These new functions are continuous if $f$ and $g$ are.

17.4 Theorem.

Let $f$ and $g$ be real-valued functions that are continuous at $x_0$ in $\mathbb{R}$. Then

1. $f + g$ is continuous at $x_0$;
2. $fg$ is continuous at $x_0$;
3. $f/g$ is continuous at $x_0$ if $g(x_0) \ne 0$.

Used in Exercise 17.3, Exercise 17.5, Exercise 17.6, Exercise 17.12

Proof

We are given $x_0 \in \operatorname{dom}(f) \cap \operatorname{dom}(g)$. Let $(x_n)$ be a sequence in $\operatorname{dom}(f) \cap \operatorname{dom}(g)$ converging to $x_0$. Then $\lim f(x_n) = f(x_0)$ and $\lim g(x_n) = g(x_0)$, and Theorem 9.3 gives

$$\lim (f + g)(x_n) = \lim [f(x_n) + g(x_n)] = \lim f(x_n) + \lim g(x_n) = f(x_0) + g(x_0) = (f + g)(x_0),$$

so $f + g$ is continuous at $x_0$. Likewise Theorem 9.4 shows $fg$ is continuous at $x_0$.

For $f/g$, assume $g(x_0) \ne 0$ and let $(x_n)$ be a sequence in $\operatorname{dom}(f) \cap \{x \in \operatorname{dom}(g) : g(x) \ne 0\}$ converging to $x_0$. Theorem 9.6 gives

$$\lim \left(\frac fg\right)(x_n) = \lim \frac{f(x_n)}{g(x_n)} = \frac{f(x_0)}{g(x_0)} = \left(\frac fg\right)(x_0).$$∎

17.5 Theorem.

If $f$ is continuous at $x_0$ and $g$ is continuous at $f(x_0)$, then the composite function $g \circ f$ is continuous at $x_0$.

Used in Exercise 17.3, Exercise 17.7

Proof

We are given $x_0 \in \operatorname{dom}(f)$ and $f(x_0) \in \operatorname{dom}(g)$. Let $(x_n)$ be a sequence in $\{x \in \operatorname{dom}(f) : f(x) \in \operatorname{dom}(g)\}$ converging to $x_0$. Since $f$ is continuous at $x_0$, $\lim f(x_n) = f(x_0)$. Since $(f(x_n))$ converges to $f(x_0)$ and $g$ is continuous at $f(x_0)$, also $\lim g(f(x_n)) = g(f(x_0))$; that is, $\lim g \circ f(x_n) = g \circ f(x_0)$.

∎

Example 4.

Accept as known that polynomials and $\sin x$, $\cos x$, $e^x$ are continuous on $\mathbb{R}$. Then $4e^x$ and $|\sin x|$ are continuous on $\mathbb{R}$ by Theorem 17.3; $\sin x + 4e^x + x^3$ by 17.4(i); and $x^4 \sin x$ by 17.4(ii). By 17.4(iii), $\tan x = \frac{\sin x}{\cos x}$ is continuous wherever $\cos x \ne 0$, i.e. at all $x$ not of the form $n\pi + \frac\pi2$, $n \in \mathbb{Z}$. Theorem 17.5 says $e^{\sin x}$ and $\cos(e^x)$ are continuous on $\mathbb{R}$; for example $\cos(e^x) = g \circ f(x)$ with $f(x) = e^x$ and $g(x) = \cos x$. Several applications of Theorems 17.3–17.5 show $x^2 \sin\left(\frac1x\right)$ and $\frac1x \sin\left(\frac1{x^2}\right)$ are continuous at all nonzero $x$.

Example 5.

Let $f$ and $g$ be continuous at $x_0$ in $\mathbb{R}$. Prove $\max(f, g)$ is continuous at $x_0$.

We could use Definition 17.1 or the ε–δ condition, but instead we illustrate a useful technique: reduce to results already established. Observe

$$\max(f, g) = \tfrac12 (f + g) + \tfrac12 |f - g|,$$

since $\max\{a, b\} = \frac12(a + b) + \frac12|a - b|$ for all $a, b \in \mathbb{R}$ (check the cases $a \ge b$ and $a < b$). By 17.4(i), $f + g$ and $f - g$ are continuous at $x_0$, so $|f - g|$ is by Theorem 17.3. Then $\frac12(f + g)$ and $\frac12|f - g|$ are continuous at $x_0$ by Theorem 17.3, and another application of 17.4(i) shows $\max(f, g)$ is continuous at $x_0$.

Reading alongside Abbott

Abbott’s Sections 4.2 and 4.3 make the opposite choice from Ross: continuity is defined by the ε–δ condition, and the sequence version is proved as a theorem. Theorem 17.2 is exactly why the choice does not matter. Abbott opens Chapter 4 (Section 4.1) with two striking examples: the function that is $1$ on the rationals and $0$ on the irrationals, which is continuous nowhere, and a function that is continuous at every irrational and discontinuous at every rational. Both appear here as Exercises 17.13 and 17.14, and they are well worth working through with the sequence definition.

Practice: fill in the gaps

An $\varepsilon$–$\delta$ proof where the clever part, the choice of $\delta$, is left to you.

Fill in the gaps: x² is continuous at 3

This is the skeleton of the proof. In each gap choose the option that makes the argument work, then check.

1. By Theorem 17.2 it is enough to show: for each $\varepsilon > 0$ there is $\delta > 0$ with $|x - 3| < \delta \Rightarrow |x^2 - 9| < \varepsilon$. We have $|x^2 - 9| = |x - 3|\,|x + 3|$.
2. If $|x - 3| < 1$, then $2 < x < 4$, so $|x + 3| <$ $7$$5$$1$.
3. Given $\varepsilon > 0$, let $\delta =$ $\min\{1, \varepsilon/7\}$$\varepsilon/7$$\min\{1, 7\varepsilon\}$.
4. If $|x - 3| < \delta$, then $|x^2 - 9| < 7 \cdot \frac\varepsilon7 = \varepsilon$.

Practice: spot the error

Continuity is a statement about every sequence, which makes one tempting shortcut.

Spot the error: a sine that oscillates

One line of this argument is wrong. Press the button on the line you think is the mistake.

1. Let $g(x) = \sin\frac\pi x$ for $x \ne 0$ and $g(0) = 0$. We show $g$ is continuous at $0$.
2. Take $x_n = \frac1n$, so $x_n \to 0$. Then $g(x_n) = \sin(n\pi) = 0$ for every $n$.
3. So $g(x_n) \to 0 = g(0)$, and therefore $g$ is continuous at $0$.

Check your understanding

Take the quiz (3 questions)

Which is the ε–δ condition of Theorem 17.2 for continuity at $x_0$?

For each $\varepsilon > 0$ there is $\delta > 0$ such that $x \in \operatorname{dom}(f)$ and $|x - x_0| < \delta$ imply $|f(x) - f(x_0)| < \varepsilon$.There is $\delta > 0$ such that for each $\varepsilon > 0$, $|x - x_0| < \delta$ implies $|f(x) - f(x_0)| < \varepsilon$.For each $\delta > 0$ there is $\varepsilon > 0$ such that $|x - x_0| < \delta$ implies $|f(x) - f(x_0)| < \varepsilon$.

Let $f(x) = 1$ for rational $x$ and $f(x) = 0$ for irrational $x$. Where is $f$ continuous?

NowhereAt every rational $x$At $x = 0$ only

Which result shows directly that $h(x) = \sqrt{1 + x^2}$ is continuous, given that $1 + x^2$ and $\sqrt{x}$ are?

Theorem 17.5 (composition)Theorem 17.3 ($|f|$ and $kf$)Theorem 18.2 (Intermediate Value Theorem)

Exercises

Exercise 17.1

Let $f(x) = \sqrt{4 - x}$ for $x \le 4$ and $g(x) = x^2$ for all $x \in \mathbb{R}$.

1. Give the domains of $f + g$, $fg$, $f \circ g$ and $g \circ f$.
2. Find $f \circ g(0)$, $g \circ f(0)$, $f \circ g(1)$, $g \circ f(1)$, $f \circ g(2)$ and $g \circ f(2)$.
3. Are the functions $f \circ g$ and $g \circ f$ equal?
4. Are $f \circ g(3)$ and $g \circ f(3)$ meaningful?

Solution

(a) $(-\infty, 4]$, $(-\infty, 4]$, $[-2, 2]$, $(-\infty, 4]$. (b) $2, 4, \sqrt3, 3, 0, 2$. (c) No. (d) $f \circ g(3)$ is not; $g \circ f(3) = 1$.

Tags: Warm-upComputation

Exercise 17.2

Let $f(x) = 4$ for $x \ge 0$, $f(x) = 0$ for $x < 0$, and $g(x) = x^2$ for all $x$. Thus $\operatorname{dom}(f) = \operatorname{dom}(g) = \mathbb{R}$.

1. Determine $f + g$, $fg$, $f \circ g$, $g \circ f$. Be sure to specify their domains.
2. Which of $f$, $g$, $f + g$, $fg$, $f \circ g$, $g \circ f$ is continuous?

Solution

All have domain $\mathbb{R}$. $f \circ g \equiv 4$; $g \circ f = 16$ for $x \ge 0$ and $0$ for $x < 0$. (b) $g$ and $f \circ g$ are continuous; the others jump at $0$.

Tags: Warm-upComputation

Exercise 17.3

Accept on faith that the following are continuous on their domains: $\sin x$, $\cos x$, $e^x$, $2^x$, $\log_e x$ for $x > 0$, and $x^p$ for $x > 0$ ($p$ any real number). Use these facts and the theorems of this section to prove the following functions are also continuous.

1. $\log_e(1 + \cos^4 x)$
2. $[\sin^2 x + \cos^6 x]^\pi$
3. $2^{x^2}$
4. $8^x$
5. $\tan x$ for $x \ne$ odd multiple of $\frac\pi2$
6. $x \sin\left(\frac1x\right)$ for $x \ne 0$
7. $x^2 \sin\left(\frac1x\right)$ for $x \ne 0$
8. $\frac1x \sin\left(\frac1{x^2}\right)$ for $x \ne 0$

Solution

Throughout, the identity $x$ and constants are continuous, and sums, products, quotients (with nonzero denominator) and composites of continuous functions are continuous (Theorems 17.4 and 17.5).

1. $\cos^4 x$ is a product of continuous functions, so $1 + \cos^4 x$ is continuous, and it is at least $1 > 0$. Composing with $\log_e$, which is continuous on $(0, \infty)$, gives a continuous function on all of $\mathbb{R}$.
2. $\sin^2 x + \cos^6 x$ is continuous. It is positive everywhere: it is nonnegative, and if it were $0$ then both $\sin x = 0$ and $\cos x = 0$, contradicting $\sin^2 x + \cos^2 x = 1$. Composing with $t \mapsto t^\pi$, continuous on $(0, \infty)$, finishes it.
3. $2^{x^2}$ is the composite of $x \mapsto x^2$ and $t \mapsto 2^t$.
4. $8^x = (2^3)^x = 2^{3x}$ is the composite of $x \mapsto 3x$ and $t \mapsto 2^t$.
5. $\tan x = \sin x / \cos x$ is a quotient of continuous functions, and the denominator is nonzero off the odd multiples of $\frac\pi2$.
6. On $x \neq 0$ the function $1/x$ is continuous (a quotient), so $\sin(1/x)$ is a composite of continuous functions, and multiplying by $x$ keeps it continuous.
7. Same as (f), using the product of $x^2$ and $\sin(1/x)$.
8. $1/x^2$ is continuous for $x \ne 0$, so $\sin(1/x^2)$ is too, and the product with $1/x$ is continuous on $x \neq 0$.

Tags: Warm-upSequences & continuity

Exercise 17.4

Prove the function $\sqrt x$ is continuous on its domain $[0, \infty)$.

Hint

Apply Example 5 in §8.

Solution

Let $x_0 \ge 0$ and let $(x_n)$ be a sequence in $[0, \infty)$ with $x_n \to x_0$. We show $\sqrt{x_n} \to \sqrt{x_0}$.

Case $x_0 = 0$. Given $\varepsilon > 0$, choose $N$ so that $x_n < \varepsilon^2$ for $n > N$. Then $\sqrt{x_n} < \varepsilon$, so $\sqrt{x_n} \to 0$.

Case $x_0 > 0$. Multiplying and dividing by the conjugate,

$$|\sqrt{x_n} - \sqrt{x_0}| = \frac{|x_n - x_0|}{\sqrt{x_n} + \sqrt{x_0}} \le \frac{|x_n - x_0|}{\sqrt{x_0}} \to 0 .$$

Hence $\sqrt{x_n} \to \sqrt{x_0}$, and $\sqrt{x}$ is continuous at every point of $[0,\infty)$.

Tags: Standardε–N / ε–δ

Used in §18

Exercise 17.5

1. Prove that if $m \in \mathbb{N}$, then $f(x) = x^m$ is continuous on $\mathbb{R}$.
2. Prove every polynomial $p(x) = a_0 + a_1 x + \cdots + a_n x^n$ is continuous on $\mathbb{R}$.

Hint

Induction on $m$ with Theorem 17.4(ii); then 17.3 and 17.4(i).

Solution

1. Induct on $m$. For $m = 1$, $f(x) = x$ is continuous: if $x_n \to x_0$ then $f(x_n) = x_n \to x_0$. If $x^m$ is continuous, then $x^{m+1} = x \cdot x^m$ is a product of continuous functions, hence continuous by Theorem 17.4(ii).
2. A constant function is continuous (every sequence of values is constant). So each term $a_k x^k$ is a product of continuous functions, and $p$ is a finite sum of them. Induction on the number of terms with Theorem 17.4(i) shows $p$ is continuous on $\mathbb{R}$.

Tags: StandardSequences & continuityInduction

Used in Exercise 17.6, Exercise 17.7, §18, Exercise 18.9

Exercise 17.6

A rational function is a function $f = p/q$ with $p, q$ polynomials; its domain is $\{x \in \mathbb{R} : q(x) \ne 0\}$. Prove every rational function is continuous.

Hint

Use Exercise 17.5.

Solution

Let $f = p/q$. By Exercise 17.5, $p$ and $q$ are continuous on $\mathbb{R}$. At any $x_0$ with $q(x_0) \neq 0$, that is, any point of the domain of $f$, Theorem 17.4(iii) says $p/q$ is continuous at $x_0$. So $f$ is continuous at every point of its domain.

Tags: StandardSequences & continuity

Exercise 17.7

1. Observe that if $k \in \mathbb{R}$, then $g(x) = kx$ is continuous by Exercise 17.5.
2. Prove $|x|$ is a continuous function on $\mathbb{R}$.
3. Use (a), (b) and Theorem 17.5 to give another proof of Theorem 17.3.

Solution

1. $g(x) = kx$ is the product of the constant function $k$ and the identity, both continuous by Exercise 17.5 (take $m = 1$ and the polynomial $k$).
2. Let $x_n \to x_0$. The reverse triangle inequality gives $\bigl||x_n| - |x_0|\bigr| \le |x_n - x_0| \to 0$, so $|x_n| \to |x_0|$.
3. Let $f$ be continuous at $x_0$. Then $|f|$ is the composite of $f$ with the continuous function $|\cdot|$ from (b), so it is continuous at $x_0$ by Theorem 17.5. Likewise $kf$ is the composite of $f$ with $x \mapsto kx$ from (a).

Tags: Warm-upSequences & continuity

Exercise 17.8

Let $f$ and $g$ be real-valued functions.

1. Show $\min(f, g) = \frac12(f + g) - \frac12|f - g|$.
2. Show $\min(f, g) = -\max(-f, -g)$.
3. Use (a) or (b) to prove that if $f$ and $g$ are continuous at $x_0$, then $\min(f, g)$ is continuous at $x_0$.

Solution

1. If $f \ge g$ at a point, then $|f - g| = f - g$ there and $\frac12(f+g) - \frac12(f-g) = g = \min(f,g)$. If $f < g$, then $|f-g| = g - f$ and $\frac12(f+g) - \frac12(g-f) = f = \min(f,g)$.
2. Negating reverses order: the larger of $-f(x), -g(x)$ is the negative of the smaller of $f(x), g(x)$. So $\max(-f,-g) = -\min(f,g)$, which is the claim.
3. If $f, g$ are continuous at $x_0$, then so are $f + g$ and $f - g$, hence $|f - g|$ (Theorem 17.3) and the scalar multiples $\frac12(f+g)$ and $\frac12|f-g|$, and finally their difference. By (a) that difference is $\min(f,g)$.

Tags: StandardSequences & continuity

Exercise 17.9

Prove each function is continuous at $x_0$ by verifying the ε–δ property of Theorem 17.2. (Try (a)–(c) in the widget above.)

1. $f(x) = x^2$, $x_0 = 2$;
2. $f(x) = \sqrt x$, $x_0 = 0$;
3. $f(x) = x \sin\left(\frac1x\right)$ for $x \ne 0$ and $f(0) = 0$, $x_0 = 0$;
4. $g(x) = x^3$, $x_0$ arbitrary.

Hint

(a) $\delta = \min\{1, \varepsilon/5\}$. (b) $\delta = \varepsilon^2$. (c) $\delta = \varepsilon$. (d) $x^3 - x_0^3 = (x - x_0)(x^2 + x_0 x + x_0^2)$; bound the second factor assuming $|x - x_0| < 1$.

Solution

In each part, fix $\varepsilon > 0$ and show the stated $\delta$ works.

1. Take $\delta = \min\{1, \varepsilon/5\}$. If $|x - 2| < \delta$ then $|x| < 3$, so $|x + 2| < 5$ and $|x^2 - 4| = |x-2||x+2| < 5 \cdot \tfrac\varepsilon5 = \varepsilon$.
2. Take $\delta = \varepsilon^2$. If $x \ge 0$ and $|x - 0| < \varepsilon^2$ then $|\sqrt x - 0| = \sqrt x < \varepsilon$.
3. Take $\delta = \varepsilon$. If $x \neq 0$ then $|f(x) - f(0)| = |x||\sin(1/x)| \le |x| < \varepsilon$ when $|x| < \delta$; if $x = 0$ the difference is $0$.
4. Put $M = 3x_0^2 + 3|x_0| + 1$ and $\delta = \min\{1, \varepsilon/M\}$. If $|x - x_0| < \delta$, then $|x| < |x_0| + 1$, so $|x^2 + x_0 x + x_0^2| \le (|x_0|+1)^2 + |x_0|(|x_0|+1) + x_0^2 = M$, and $|x^3 - x_0^3| = |x - x_0|\,|x^2 + x_0x + x_0^2| < \tfrac\varepsilon M \cdot M = \varepsilon$.

Tags: Standardε–N / ε–δ

Exercise 17.10

Prove the following functions are discontinuous at the indicated points, using either Definition 17.1 or the ε–δ property.

1. $f(x) = 1$ for $x > 0$ and $f(x) = 0$ for $x \le 0$, $x_0 = 0$;
2. $g(x) = \sin\left(\frac1x\right)$ for $x \ne 0$ and $g(0) = 0$, $x_0 = 0$;
3. $\operatorname{sgn}(x) = -1$ for $x < 0$, $\operatorname{sgn}(x) = 1$ for $x > 0$, $\operatorname{sgn}(0) = 0$, $x_0 = 0$. The function $\operatorname{sgn}$ is the signum function; note $\operatorname{sgn}(x) = \frac{x}{|x|}$ for $x \ne 0$.

Hint

(a), (c): use $x_n = \frac1n$. (b): use $x_n = \frac1{2\pi n + \pi/2}$, where $g(x_n) = 1$.

Solution

In each part we exhibit a sequence $x_n \to 0$ along which $f(x_n) \not\to f(0)$, which contradicts Definition 17.1.

1. With $x_n = \frac1n$ we have $f(x_n) = 1$ for all $n$, so $f(x_n) \to 1 \ne 0 = f(0)$.
2. With $x_n = \frac{1}{2\pi n + \pi/2}$ we have $x_n \to 0$ and $g(x_n) = \sin(2\pi n + \tfrac\pi2) = 1$, so $g(x_n) \to 1 \neq 0 = g(0)$.
3. With $x_n = \frac1n$ we have $\operatorname{sgn}(x_n) = 1 \to 1 \neq 0 = \operatorname{sgn}(0)$.

Tags: Standardε–N / ε–δExamples & counterexamples

Used in Exercise 18.12

Exercise 17.11

Let $f$ be real-valued with $\operatorname{dom}(f) \subseteq \mathbb{R}$. Prove $f$ is continuous at $x_0$ if and only if, for every monotonic sequence $(x_n)$ in $\operatorname{dom}(f)$ converging to $x_0$, $\lim f(x_n) = f(x_0)$.

Hint

Don’t forget Theorem 11.4.

Solution

($\Rightarrow$) If $f$ is continuous at $x_0$, then $f(x_n) \to f(x_0)$ for every sequence in the domain converging to $x_0$, in particular for monotonic ones.

($\Leftarrow$) Suppose $f$ is not continuous at $x_0$. Then there is a sequence $(x_n)$ in $\operatorname{dom}(f)$ with $x_n \to x_0$ but $f(x_n) \not\to f(x_0)$. So for some $\varepsilon > 0$ there are infinitely many $n$ with $|f(x_n) - f(x_0)| \ge \varepsilon$; these form a subsequence $(x_{n_k})$ that still converges to $x_0$. By Theorem 11.4 this subsequence has a monotonic subsequence $(y_j)$. It converges to $x_0$ and satisfies $|f(y_j) - f(x_0)| \ge \varepsilon$ for all $j$, so $f(y_j) \not\to f(x_0)$. That contradicts the hypothesis, so $f$ is continuous at $x_0$.

Tags: ChallengeSequences & continuitySubsequences

Exercise 17.12

1. Let $f$ be a continuous real-valued function with domain $(a, b)$. Show that if $f(r) = 0$ for each rational $r$ in $(a, b)$, then $f(x) = 0$ for all $x \in (a, b)$.
2. Let $f$ and $g$ be continuous on $(a, b)$ with $f(r) = g(r)$ for each rational $r$ in $(a, b)$. Prove $f(x) = g(x)$ for all $x \in (a, b)$.

Hint

(a) Every $x$ is a limit of rationals in $(a, b)$ (denseness, Theorem 4.7). (b) Apply (a) to $f - g$.

Solution

1. Let $x \in (a,b)$. For each $n$ choose a rational $r_n$ with $x < r_n < \min\{x + \tfrac1n, b\}$, which is possible since the rationals are dense. Then $r_n \in (a,b)$ and $r_n \to x$. By continuity, $f(x) = \lim f(r_n) = \lim 0 = 0$.
2. The function $f - g$ is continuous on $(a,b)$ (Theorem 17.4) and vanishes at each rational point of $(a,b)$. By (a), $f - g = 0$ everywhere, so $f = g$.

Tags: StandardSequences & continuity

Exercise 17.13

1. Let $f(x) = 1$ for rational $x$ and $f(x) = 0$ for irrational $x$. Show $f$ is discontinuous at every $x$ in $\mathbb{R}$.
2. Let $h(x) = x$ for rational $x$ and $h(x) = 0$ for irrational $x$. Show $h$ is continuous at $x = 0$ and at no other point.

Solution

Both parts use that every real number is a limit of a sequence of rationals and also of a sequence of irrationals (density).

1. Fix $x_0$. Take rationals $p_n \to x_0$ and irrationals $q_n \to x_0$. Then $f(p_n) = 1$ and $f(q_n) = 0$ for all $n$, so $f(p_n) \to 1$ and $f(q_n) \to 0$. These cannot both equal $f(x_0)$, so $f$ is discontinuous at $x_0$.
2. At $x_0 = 0$: $|h(x)| \le |x|$ for all $x$, so if $x_n \to 0$ then $|h(x_n)| \le |x_n| \to 0$, and $h(x_n) \to 0 = h(0)$. Now let $x_0 \neq 0$. If $x_0$ is rational, take irrationals $q_n \to x_0$: then $h(q_n) = 0 \to 0 \neq x_0 = h(x_0)$. If $x_0$ is irrational, take rationals $p_n \to x_0$: then $h(p_n) = p_n \to x_0 \neq 0 = h(x_0)$. Either way $h$ is discontinuous at $x_0$.

Tags: StandardSequences & continuityExamples & counterexamples

Exercise 17.14

For each nonzero rational $x$, write $x = \frac pq$ with $p, q$ integers having no common factors and $q > 0$, and define $f(x) = \frac1q$. Also define $f(0) = 1$ and $f(x) = 0$ for all $x \in \mathbb{R} \setminus \mathbb{Q}$. Thus $f(x) = 1$ for each integer, $f\left(\frac12\right) = f\left(-\frac12\right) = f\left(\frac32\right) = \cdots = \frac12$, etc. Show $f$ is continuous at each point of $\mathbb{R} \setminus \mathbb{Q}$ and discontinuous at each point of $\mathbb{Q}$.

Hint

At an irrational $x_0$, given $\varepsilon$, only finitely many rationals in $(x_0 - 1, x_0 + 1)$ have denominator $q \le 1/\varepsilon$; choose $\delta$ to avoid them all.

Solution

Rational points. If $x_0 \in \mathbb{Q}$ then $f(x_0) > 0$ (it is $1/q$, or $1$ at $0$). Choose irrationals $q_n \to x_0$; then $f(q_n) = 0 \not\to f(x_0)$. So $f$ is discontinuous at $x_0$.

Irrational points. Let $x_0$ be irrational, so $f(x_0) = 0$, and let $\varepsilon > 0$. Fix an integer $N > 1/\varepsilon$. Let $F$ be the set of rationals in $(x_0 - 1, x_0 + 1)$ that can be written with a denominator $q \le N$. For each $q \le N$ there are only finitely many integers $p$ with $p/q \in (x_0 - 1, x_0 + 1)$, so $F$ is finite. Since $x_0 \notin F$, the number $\delta = \min\bigl(\{1\} \cup \{|r - x_0| : r \in F\}\bigr)$ is positive. Now let $|x - x_0| < \delta$. If $x$ is irrational, $f(x) = 0$. If $x$ is rational, then $x \in (x_0 - 1, x_0+1)$ but $x \notin F$, so its reduced denominator satisfies $q > N$, and $f(x) = 1/q < 1/N < \varepsilon$. In all cases $|f(x) - f(x_0)| < \varepsilon$, so $f$ is continuous at $x_0$.

Tags: ChallengeExamples & counterexamplesε–N / ε–δ

Exercise 17.15

Let $f$ be real-valued with $\operatorname{dom}(f) \subseteq \mathbb{R}$. Show $f$ is continuous at $x_0$ in $\operatorname{dom}(f)$ if and only if, for every sequence $(x_n)$ in $\operatorname{dom}(f) \setminus \{x_0\}$ converging to $x_0$, we have $\lim f(x_n) = f(x_0)$.

Solution

($\Rightarrow$) A sequence in $\operatorname{dom}(f) \setminus \{x_0\}$ is in particular a sequence in $\operatorname{dom}(f)$, so Definition 17.1 applies.

($\Leftarrow$) Let $(x_n)$ be any sequence in $\operatorname{dom}(f)$ with $x_n \to x_0$, and let $\varepsilon > 0$. Let $A = \{n : x_n \neq x_0\}$. If $A$ is finite, then $f(x_n) = f(x_0)$ for all large $n$ and we are done. Otherwise $(x_n)_{n \in A}$ is a subsequence in $\operatorname{dom}(f) \setminus \{x_0\}$ converging to $x_0$, so by hypothesis there is $N$ with $|f(x_n) - f(x_0)| < \varepsilon$ for $n \in A$, $n > N$. For $n \notin A$ we have $f(x_n) = f(x_0)$ exactly. Hence $|f(x_n) - f(x_0)| < \varepsilon$ for all $n > N$, and $f(x_n) \to f(x_0)$.

Tags: StandardSequences & continuity

Exercise 17.16

The postage-stamp function $P$ is defined by $P(x) = A$ for $0 \le x < 1$ and $P(x) = A + Bn$ for $n \le x < n + 1$. So $P$ takes the value $A$ on $[0, 1)$, $A + B$ on $[1, 2)$, $A + 2B$ on $[2, 3)$, and so on: $A$ is the postage for the first ounce and $B$ for each additional ounce. Show $P$ is discontinuous at every positive integer. (Because postage rates tend to increase over time, $A$ and $B$ are actually functions.)

Solution

We assume $B \neq 0$ (if $B = 0$ then $P \equiv A$ is constant and, of course, continuous). Fix an integer $n \ge 1$. Then $P(n) = A + Bn$. Let $x_k = n - \frac1k$ for $k \ge 1$; then $x_k \to n$ and $n - 1 \le x_k < n$, so $P(x_k) = A + B(n-1)$. Therefore $P(x_k) \to A + Bn - B \neq A + Bn = P(n)$. So $P$ is discontinuous at $n$.

Tags: Warm-upComputation