Contents / Real Analysis / Limits & Continuity

Chapter 3

Limits & Continuity

Limits of functions by epsilon-delta with the laws derived rather than quoted; the sequential characterisation of limits and continuity; the classification of discontinuities; uniform continuity, Lipschitz and Heine-Cantor; the Intermediate Value Theorem proved from completeness; and the Extreme Value Theorem with compactness, each hypothesis shown necessary by counterexample.

Introduction

The previous chapter built $R$ and established what a sequence does: convergence, the algebra of limits, the Cauchy criterion, and the least-upper-bound property that makes all of it work. A sequence, though, is a function on $N$ , and its index marches to infinity in one direction only. This chapter replaces that index by a real variable, which may approach a point from either side, from inside a gap in the domain, or from nowhere in particular, and it asks what survives.

Almost everything does, but three things genuinely change. The "index" $x$ can approach $a$ along infinitely many routes, so a limit is a statement about all of them at once — and the sequential characterisation proved below is the bridge that lets us borrow sequence theorems. The point $a$ is one where $f$ may or may not be defined, and whether $lim_{x \to a} f (x)$ agrees with $f (a)$ is no longer automatic: that agreement is continuity. And because $x$ ranges over a set rather than a list, the same $ε$ - $δ$ promise can be demanded uniformly over that set, which turns out to depend on the shape of the set rather than on the function.

The chapter has five movements: the $ε$ - $δ$ definition and the limit laws proved from it; continuity and its sequential characterisation, which does most of the real work; uniform continuity and the Heine–Cantor theorem, the first place compactness pays for itself; the Intermediate Value Theorem, which is the completeness of $R$ in disguise; and the Extreme Value Theorem, which is compactness in disguise. In each of the last two, the counterexample that appears the moment a hypothesis is dropped is exhibited alongside the proof.

Notation is fixed once. Throughout, $f$ and $g$ are real-valued functions on a set $D \subseteq R$ , the letter $a$ denotes a limit point of $D$ , and $L$ , $M$ denote candidate limits. The results Triangle inequality and Upper bound, supremum, infimum from The Real Number System are used without comment, as are Uniqueness of limits, the Squeeze Theorem and the Cauchy criterion for sequences from Sequences and Series. The Bolzano–Weierstrass theorem — every bounded real sequence has a convergent subsequence — is likewise taken as proved there.

3.1Approaching a point: one-sided limits

The informal sentence is: $lim_{x \to a} f (x) = L$ when $f (x)$ can be made as close to $L$ as we please by taking $x$ close enough to $a$ , but not equal to $a$ . That last clause is the whole reason limits are worth having — the derivative is a limit of difference quotients undefined at the point in question. Before making the sentence precise it pays to split it, since approaching $a$ from the left and from the right are different experiments and a function may pass one and fail the other.

Definition 3.1 (One-sided limits). Let $f$ be defined on an interval to the left of $a$ . Then $lim_{x \to a^{-}} f (x) = L$ means: for every $ε > 0$ there is $δ > 0$ such that

a - δ < x < a ⟹ ∣ f (x) - L ∣ < ε .

The right-hand limit $lim_{x \to a^{+}} f (x) = L$ is the same statement with $a < x < a + δ$ .

Each one-sided limit ignores half the picture on purpose: the left-hand limit is computed from the values of $f$ strictly below $a$ and nothing else — not from $f (a)$ , not from anything to the right. The two-sided limit is exactly the conjunction of the two.

Theorem 3.2 (Two-sided limits from one-sided limits). Suppose $f$ is defined on a punctured interval around $a$ . Then $lim_{x \to a} f (x) = L$ if and only if both one-sided limits exist and

x \to a^{-} lim f (x) = x \to a^{+} lim f (x) = L .

Proof. Suppose first that $lim_{x \to a} f (x) = L$ , and let $ε > 0$ . The two-sided definition supplies $δ > 0$ with $∣ f (x) - L ∣ < ε$ whenever $0 < ∣ x - a ∣ < δ$ . The condition $a - δ < x < a$ implies $0 < ∣ x - a ∣ < δ$ , so the same $δ$ witnesses the left-hand limit; the condition $a < x < a + δ$ likewise implies $0 < ∣ x - a ∣ < δ$ , so it witnesses the right-hand one. Both one-sided limits therefore exist and equal $L$ .

Conversely, suppose both one-sided limits equal $L$ , and let $ε > 0$ . Choose $δ_{-} > 0$ from the left-hand definition and $δ_{+} > 0$ from the right-hand one, and set $δ = min (δ_{-}, δ_{+}) > 0$ — a minimum of two positive numbers is positive, which is the only reason this step is legal. If $0 < ∣ x - a ∣ < δ$ then $x \neq = a$ , so either $a - δ \leq a - δ_{-} < x < a$ or $a < x < a + δ_{+}$ , and in both cases the corresponding one-sided estimate gives $∣ f (x) - L ∣ < ε$ .∎

The theorem earns its keep in the contrapositive: to show a limit fails to exist, produce the two one-sided limits and observe that they disagree. That is a finite piece of work, whereas verifying "no $L$ satisfies the definition" directly would be an argument about every real number at once.

Intuition. Walk toward the point $a$ along the $x$ -axis and watch the height of the graph. If the heights settle on a single number $L$ whichever direction you come from, that number is the limit. The height at $a$ is not part of the experiment — the function may be undefined there, or defined with an entirely unrelated value, and neither affects the answer.

Example 3.3 (A limit that exists where the function does not). Find $lim_{x \to 1} \frac{x ^{2} - 1}{x - 1}$ , noting that the expression is undefined at $x = 1$ .

Solution.

Factor the numerator: $x^{2} - 1 = (x - 1) (x + 1)$ , so the quotient is $\frac{( x - 1 ) ( x + 1 )}{x - 1}$ .
Cancel. The cancellation is legitimate precisely because the limit never evaluates at $x = 1$ : for every $x$ in a punctured neighbourhood of $1$ we have $x - 1 \neq = 0$ , and there the function is literally equal to $x + 1$ .
Two functions that agree on a punctured neighbourhood of $a$ have the same limit at $a$ , since the definition only ever inspects such a neighbourhood. So the limit is $lim_{x \to 1} (x + 1) = 2$ .
Sanity check: $f (0.99) = 1.99$ and $f (1.01) = 2.01$ , bracketing $2$ symmetrically. The limit is $2$ although $f (1)$ does not exist — a removable discontinuity, since defining $f (1) = 2$ would repair it and nothing else needs changing.

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A second example shows the theorem doing its destructive work. For $g (x) = ∣ x ∣/ x$ , defined for $x \neq = 0$ , we have $g (x) = 1$ identically on $(0, \infty)$ and $g (x) = - 1$ identically on $(- \infty, 0)$ , so the one-sided limits at $0$ are $1$ and $- 1$ . They exist but disagree, so by the theorem Two-sided limits from one-sided limits the two-sided limit does not exist — and note that this rules out every candidate $L$ at once, since no number is within $1/2$ of both $1$ and $- 1$ . This is a jump discontinuity, and unlike the previous one no redefinition at $0$ repairs it.

Figure 3.4. The two failure modes side by side. On the left,

(x^{2} - 1) / (x - 1)

agrees with the line

y = x + 1

everywhere except at

x = 1

, where it is undefined: both one-sided limits are

2

, so the hole is removable. On the right,

∣ x ∣/ x

sits at height

1

for positive

x

and at height

- 1

for negative

x

; the one-sided limits exist but disagree, so no value at

0

can repair the jump.

Exercises for §3.1 →

3.2The precise $ε$ - $δ$ definition

The phrase "as close as we please" hides a game between two players, and the definition is nothing but a transcript of that game with the moves in a fixed order.

Definition 3.5 (Limit of a function). $lim_{x \to a} f (x) = L$ means: for every $ε > 0$ there exists $δ > 0$ such that

0 < ∣ x - a ∣ < δ ⟹ ∣ f (x) - L ∣ < ε .

Two features of the ordering carry all the content. The $ε$ is quantified first, so $δ$ may depend on it — and normally must. The variable $x$ is quantified last, inside the implication, so $δ$ may not depend on $x$ : one $δ$ serves the whole punctured neighbourhood. A "proof" producing a $δ$ that still mentions $x$ has proved nothing, and that is the single most common error in a first course. The clause $0 < ∣ x - a ∣$ excludes $a$ itself, which is what made the previous section's cancellation legal; nothing demands that $f$ be defined at $a$ , or that $f (a)$ be near $L$ .

Intuition. Think of $ε$ as a horizontal band of half-width $ε$ drawn around the height $L$ , and $δ$ as a vertical strip of half-width $δ$ drawn around $a$ . Your opponent picks the band, as narrow as they like; you must answer with a strip so narrow that the whole graph inside it — with the single point above $a$ deleted — lies inside the band. You lose if there is any $ε$ for which no strip works.

Example 3.6 (Two limits from the definition). Prove from the definition that (a) $lim_{x \to 3} (2 x + 1) = 7$ , and (b) $lim_{x \to 3} x^{2} = 9$ .

Solution.

(a) Scratch work, which is where $δ$ is discovered and which is not itself the proof. We need $∣ (2 x + 1) - 7∣ < ε$ ; simplifying, $∣ (2 x + 1) - 7∣ = ∣2 x - 6∣ = 2∣ x - 3∣$ , so the requirement is $∣ x - 3∣ < ε /2$ .
The proof, written forwards. Let $ε > 0$ be given and put $δ = ε /2 > 0$ . If $0 < ∣ x - 3∣ < δ$ then $∣ (2 x + 1) - 7∣ = 2∣ x - 3∣ < 2 δ = ε$ . Since $ε$ was arbitrary, the limit is $7$ . With $ε = 0.1$ the recipe gives $δ = 0.05$ ; the linear relation $δ = ε /2$ records that a slope- $2$ map doubles input errors.
(b) The quadratic case is harder because $∣ f (x) - L ∣$ is not a constant multiple of $∣ x - 3∣$ . Factor: $∣ x^{2} - 9∣ = ∣ x - 3∣ ∣ x + 3∣$ , where the first factor is controlled and the second is a nuisance depending on $x$ .
Pre-restrict. Insist from the outset that $δ \leq 1$ . Then $∣ x - 3∣ < 1$ gives $2 < x < 4$ , hence $5 < x + 3 < 7$ and $∣ x + 3∣ < 7$ , so on that neighbourhood $∣ x^{2} - 9∣ < 7∣ x - 3∣$ .
To force $7∣ x - 3∣ < ε$ it suffices that $∣ x - 3∣ < ε /7$ , so both constraints hold at once for

δ = min (1, \frac{ε}{7}) > 0.

Verification: if $0 < ∣ x - 3∣ < δ$ then $∣ x - 3∣ < 1$ , making the bound $∣ x + 3∣ < 7$ available, and $∣ x - 3∣ < ε /7$ , so $∣ x^{2} - 9∣ < 7 \cdot ε /7 = ε$ .
Sanity check with $ε = 0.7$ : $δ = min (1, 0.1) = 0.1$ , and $2.9 < x < 3.1$ gives $8.41 < x^{2} < 9.61$ , so $∣ x^{2} - 9∣ < 0.61 < 0.7$ . The estimate is not tight, and need not be — $δ$ has only to be positive and adequate.
Repeating step 4 at a general base point $a$ instead of $3$ gives $δ = min (1, ε / (2∣ a ∣ + 1))$ , which proves $lim_{x \to a} x^{2} = a^{2}$ for every $a$ — and which visibly shrinks as $∣ a ∣$ grows. No single $δ$ serves every base point, a defect the section Uniform continuity: moving the quantifier is entirely about.

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The linear case is easy because $∣ f (x) - L ∣$ is an exact constant multiple of $∣ x - a ∣$ . The moment $f$ is not linear, the factor multiplying $∣ x - a ∣$ itself depends on $x$ , and a second idea is needed: restrict $δ$ in advance so that the nuisance factor is bounded, then shrink further.

Example 3.7 (A quadratic limit, and the "restrict $δ$ first" device). Prove $lim_{x \to 3} x^{2} = 9$ .

Solution.

Factor the target: $∣ x^{2} - 9∣ = ∣ x - 3∣ ∣ x + 3∣$ . The first factor is the one we control directly; the second is the nuisance.
Pre-restrict. Insist from the outset that $δ \leq 1$ . Then $∣ x - 3∣ < 1$ gives $2 < x < 4$ , hence $5 < x + 3 < 7$ and so $∣ x + 3∣ < 7$ . On that neighbourhood, $∣ x^{2} - 9∣ < 7∣ x - 3∣$ .
Now finish as in the linear case: to force $7∣ x - 3∣ < ε$ it suffices that $∣ x - 3∣ < ε /7$ . Both constraints hold at once if we take

δ = min (1, \frac{ε}{7}) > 0.

Verification. Let $ε > 0$ and take this $δ$ . If $0 < ∣ x - 3∣ < δ$ then $∣ x - 3∣ < 1$ , so the bound $∣ x + 3∣ < 7$ is available, and $∣ x - 3∣ < ε /7$ , so $∣ x^{2} - 9∣ < 7 \cdot ε /7 = ε$ .
Sanity check with $ε = 0.7$ : $δ = min (1, 0.1) = 0.1$ , and $2.9 < x < 3.1$ gives $8.41 < x^{2} < 9.61$ , so $∣ x^{2} - 9∣ < 0.61 < 0.7$ . The estimate is not tight — $δ$ never has to be the largest that works, only positive and adequate.

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Figure 3.8. The definition drawn for

f (x) = x^{2}

a = 3

with

ε = 1

. The horizontal band is

8 < y < 10

; the vertical strip is the interval

(8, 10)

, whose half-widths on the two sides differ, so the symmetric

δ

must be the smaller of them, about

0.162

. Inside the strip the curve never leaves the band, which is exactly what the implication asserts.

Exercises for §3.2 →

3.3Infinite limits, limits at infinity, and the laws in outline

Two more variants of the definition cover the cases where either the output or the input is allowed to run off the end of the line. Neither introduces a new idea: each replaces one of the two tolerances by a threshold to be beaten.

Definition 3.9 (Infinite limits and limits at infinity). $lim_{x \to a} f (x) = + \infty$ means that for every $M > 0$ there is $δ > 0$ with

0 < ∣ x - a ∣ < δ ⟹ f (x) > M,

and symmetrically for $- \infty$ . Separately, $lim_{x \to \infty} f (x) = L$ means that for every $ε > 0$ there is $N$ with

x > N ⟹ ∣ f (x) - L ∣ < ε .

Notation. Writing $lim_{x \to a} f (x) = + \infty$ is a statement of a particular way of failing to converge, not a claim that the limit exists as a real number. When a text says "the limit does not exist" of $1/ x^{2}$ at $0$ it is not contradicting the display above; it is declining to work in the extended reals. We will use the symbol freely and say "diverges to $+ \infty$ " when precision matters.

The limit laws are the reason nobody computes limits from the definition in practice. They are collected here as a single statement and then proved one at a time in the section Proving the limit laws, because each proof is a distinct technique worth owning.

The limit laws are the reason nobody computes limits from the definition in practice: sums, products, quotients and constant multiples of convergent functions converge to the corresponding combination of the limits. Each is proved from the $ε$ - $δ$ definition in the section Proving the limit laws — the sum in the theorem Sum rule for limits, the product in Product rule for limits, the quotient in Quotient rule for limits — and the constant multiple is the product rule applied with the constant function, whose limit is that constant because $∣ c - c ∣ = 0 < ε$ for every $δ$ whatsoever.

A remark on the hypotheses, which are stronger than they look. The laws require both limits to exist before they say anything about the combination, and the converse direction is false: $f (x) = ⌊ x ⌋$ and $g (x) = - ⌊ x ⌋$ have no limits at $0$ , yet $f + g$ is identically zero and converges perfectly well. Knowing that a sum converges tells you nothing about the summands.

Example 3.10 (An infinite limit from the definition). Prove $lim_{x \to 0^{+}} \frac{1}{x} = + \infty$ .

Solution.

Unpack the goal: given an arbitrary threshold $M > 0$ , we must force $1/ x > M$ for all $x$ close enough to $0$ on the right.
Solve the target inequality. For $x > 0$ , $\frac{1}{x} > M ⟺ x < \frac{1}{M}$ , the direction of the inequality reversing because both sides are positive.
Choose $δ = 1/ M > 0$ . If $0 < x < δ$ then $x < 1/ M$ , hence $1/ x > M$ .
Sanity check: to beat $M = 1000$ the recipe gives $δ = 0.001$ , and every $x$ in $(0, 0.001)$ indeed has $1/ x > 1000$ . Note that the two-sided limit is not $+ \infty$ : approaching from the left gives $1/ x \to - \infty$ .

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The laws also settle limits at infinity by the standard device of dividing by the highest power. For instance

\frac{3 x ^{2} + 2 x}{x ^{2} - 5} = \frac{3 + 2/ x}{1 - 5/ x ^{2}},

legitimately for $x \neq = 0$ ; the facts $lim_{x \to \infty} 1/ x = 0$ and $lim_{x \to \infty} 1/ x^{2} = 0$ are immediate from the definition with $N = 1/ ε$ and $N = 1/ ε$ , so the numerator tends to $3$ and the denominator to $1$ . The quotient law applies because that denominator limit is nonzero, giving the value $3$ . At $x = 100$ the expression is $30200/9995 \approx 3.021$ , approaching $3$ from above as the $+ 2/ x$ term predicts.

Exercises for §3.3 →

3.4The definition in full, and what it forces

Written with every quantifier explicit and the domain named, the definition reads as follows. This is the form to check an argument against when something feels wrong.

Definition 3.11 (The $ε$ - $δ$ definition of a limit). Let $f : D \subseteq R \to R$ and let $a$ be a limit point of $D$ . Then $lim_{x \to a} f (x) = L$ if and only if

\forall ε > 0 \exists δ > 0 such that \forall x \in D, 0 < ∣ x - a ∣ < δ ⟹ ∣ f (x) - L ∣ < ε .

Definition 3.12 (Limit point). A point $a$ is a limit point of $D$ if every punctured neighbourhood of $a$ meets $D$ : for every $δ > 0$ there is some $x \in D$ with $0 < ∣ x - a ∣ < δ$ .

The limit-point hypothesis is not decoration. If $a$ were isolated in $D$ — say $D = [0, 1] \cup {5}$ and $a = 5$ — then $δ = 1$ makes the antecedent false for every $x \in D$ , so the implication holds vacuously for every $L$ and "the limit" would be simultaneously $0$ , $7$ and $- π$ . Requiring $a$ to be a limit point rules that out, and the next theorem is the pay-off.

Theorem 3.13 (Uniqueness of the limit). If $a$ is a limit point of $D$ and $lim_{x \to a} f (x) = L$ and $lim_{x \to a} f (x) = L^{'}$ , then $L = L^{'}$ .

Proof. Suppose $L \neq = L^{'}$ and put $ε = ∣ L - L^{'} ∣/2 > 0$ . Applying the definition twice gives $δ_{1} > 0$ with $∣ f (x) - L ∣ < ε$ whenever $x \in D$ and $0 < ∣ x - a ∣ < δ_{1}$ , and $δ_{2} > 0$ with $∣ f (x) - L^{'} ∣ < ε$ under the corresponding condition. Let $δ = min (δ_{1}, δ_{2}) > 0$ .

Because $a$ is a limit point of $D$ , there exists at least one $x_{0} \in D$ with $0 < ∣ x_{0} - a ∣ < δ$ — this is the only place the hypothesis is used, and without it the argument would have no point at which to evaluate $f$ . For that $x_{0}$ both estimates hold, so by the triangle inequality

∣ L - L^{'} ∣ \leq ∣ L - f (x_{0}) ∣ + ∣ f (x_{0}) - L^{'} ∣ < ε + ε = ∣ L - L^{'} ∣,

which asserts $∣ L - L^{'} ∣ < ∣ L - L^{'} ∣$ , a contradiction. Hence $L = L^{'}$ .∎

That argument is the function-limit twin of Uniqueness of limits for sequences, and the proof is verbatim the same once "for all $n \geq N$ " becomes "for all $x$ with $0 < ∣ x - a ∣ < δ$ ". The next theorem explains why: the two notions are interchangeable.

Theorem 3.14 (Sequential characterisation of limits). Let $a$ be a limit point of $D$ . Then $lim_{x \to a} f (x) = L$ if and only if for every sequence $(x_{n})$ in $D ∖ {a}$ with $x_{n} \to a$ we have $f (x_{n}) \to L$ .

Proof. ( $\Rightarrow$ ) Assume the limit is $L$ and let $(x_{n})$ be such a sequence. Given $ε > 0$ , take $δ$ from the definition of the limit, then take $N$ from the convergence $x_{n} \to a$ so that $∣ x_{n} - a ∣ < δ$ for all $n \geq N$ . Since $x_{n} \neq = a$ , we have $0 < ∣ x_{n} - a ∣ < δ$ for those $n$ , hence $∣ f (x_{n}) - L ∣ < ε$ . As $ε$ was arbitrary, $f (x_{n}) \to L$ .

( $\Leftarrow$ ) We prove the contrapositive. Suppose $lim_{x \to a} f (x) = L$ fails. Negating the definition, there is some $ε_{0} > 0$ such that no $δ$ works: for every $δ > 0$ there is $x \in D$ with $0 < ∣ x - a ∣ < δ$ and $∣ f (x) - L ∣ \geq ε_{0}$ . Apply this with $δ = 1/ n$ for each $n \in N$ and call the resulting point $x_{n}$ . Then $x_{n} \in D ∖ {a}$ and $∣ x_{n} - a ∣ < 1/ n \to 0$ , so $x_{n} \to a$ ; but $∣ f (x_{n}) - L ∣ \geq ε_{0}$ for every $n$ , so $f (x_{n}) \neq \to L$ . This exhibits a sequence violating the right-hand condition, which is what the contrapositive required.∎

The device in the second half — feed $δ = 1/ n$ to a failed $ε$ - $δ$ statement and harvest a sequence — recurs three more times below, and each time it converts a negated quantifier statement into a concrete object that is far easier to argue with.

Read in contrapositive form, that equivalence is a divergence test: if two sequences $x_{n} \to a$ and $y_{n} \to a$ in $D ∖ {a}$ give $f (x_{n})$ and $f (y_{n})$ different limits, then no $L$ can satisfy the definition, since the forward direction would force both to equal $L$ and Uniqueness of limits for sequences forbids that.

Example 3.15 (Divergence by two routes). Show that $lim_{x \to 0} sin (1/ x)$ does not exist.

Solution.

First route, chosen to make the sine vanish: $x_{n} = \frac{1}{nπ}$ is nonzero and tends to $0$ , with $sin (1/ x_{n}) = sin (nπ) = 0$ , so $f (x_{n}) \to 0$ .
Second route, chosen to make it equal $1$ : $y_{n} = \frac{1}{2 πn + π /2}$ , again nonzero and tending to $0$ , with $sin (1/ y_{n}) = 1$ , so $f (y_{n}) \to 1$ .
The route-limits $0$ and $1$ differ, so by the theorem Sequential characterisation of limits no limit exists.
Sanity check: $f$ is bounded, so this is not a blow-up — the function completes infinitely many oscillations in every neighbourhood of $0$ , hitting every value of $[- 1, 1]$ infinitely often. It is the standard essential (oscillatory) discontinuity.

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Exercises for §3.4 →

3.5Strategies for proving a limit from the definition

There are only four moves, and every $ε$ - $δ$ exercise in a first course is some combination of them.

Method 3.16 (Proving a limit from the definition).

Factor out $∣ x - a ∣$ . Manipulate $∣ f (x) - L ∣$ into the form (something) $\times ∣ x - a ∣$ .
Pre-restrict $δ$ to some convenient bound, usually $δ \leq 1$ , so that the "something" is bounded by a constant $K$ on that neighbourhood.
Solve $K ∣ x - a ∣ < ε$ , giving $δ = min (1, ε / K)$ .
For quotients, bound the denominator away from zero on a small neighbourhood first, so that dividing by it does not destroy the estimate.

Step 4 is the one with real content, and it deserves a statement of its own because the same fact is needed again when we prove the quotient law.

Lemma 3.17 (A nonzero limit keeps the function away from zero). If $lim_{x \to a} g (x) = M \neq = 0$ , then there is $δ_{0} > 0$ such that

0 < ∣ x - a ∣ < δ_{0} ⟹ ∣ g (x) ∣ > \frac{∣ M ∣}{2} .

Proof. Apply the definition of the limit with the particular tolerance $ε = ∣ M ∣/2$ , which is positive because $M \neq = 0$ . It supplies $δ_{0} > 0$ with $∣ g (x) - M ∣ < ∣ M ∣/2$ whenever $0 < ∣ x - a ∣ < δ_{0}$ . By the reverse triangle inequality,

∣ g (x) ∣ \geq ∣ M ∣ - ∣ M - g (x) ∣ > ∣ M ∣ - \frac{∣ M ∣}{2} = \frac{∣ M ∣}{2} .

∎

Proposition 3.18 (Local boundedness). If $lim_{x \to a} f (x) = L$ exists and is finite, then $f$ is bounded on some punctured neighbourhood of $a$ : there are $δ_{0} > 0$ and $C$ with $∣ f (x) ∣ \leq C$ whenever $0 < ∣ x - a ∣ < δ_{0}$ .

Proof. Apply the definition with $ε = 1$ to get $δ_{0} > 0$ such that $∣ f (x) - L ∣ < 1$ on the punctured $δ_{0}$ -neighbourhood. Then $∣ f (x) ∣ \leq ∣ f (x) - L ∣ + ∣ L ∣ < 1 + ∣ L ∣$ there, so $C = 1 + ∣ L ∣$ works.∎

Local boundedness is why a convergent function cannot oscillate with growing amplitude near $a$ , and it is what makes the product rule's stray factor manageable — all for one application of the definition at a fixed tolerance.

Example 3.19 (Squeezing an oscillation). Prove $lim_{x \to 0} x sin (1/ x) = 0$ , even though $sin (1/ x)$ itself has no limit at $0$ .

Solution.

Bound the bad factor: $∣ sin (1/ x) ∣ \leq 1$ for every $x \neq = 0$ , whatever $1/ x$ happens to be. Hence $∣ x sin (1/ x) - 0∣ = ∣ x ∣ ∣ sin (1/ x) ∣ \leq ∣ x ∣$ .
Let $ε > 0$ and put $δ = ε$ . If $0 < ∣ x ∣ < δ$ then $∣ x sin (1/ x) ∣ \leq ∣ x ∣ < ε$ .
Sanity check: at $x = 0.01$ the factor $sin (100)$ lies in $[- 1, 1]$ , so the product is at most $0.01$ in absolute value however wild the oscillation. The Squeeze Theorem applied to $- ∣ x ∣ \leq x sin (1/ x) \leq ∣ x ∣$ gives the same conclusion.
Note the contrast with the previous example: multiplying an unbounded oscillation by a factor tending to zero repairs it. The oscillation is still infinitely rapid; it is the amplitude that the definition sees.

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Step 4 of the recipe is what the lemma above is for. To prove $lim_{x \to 2} 1/ x = 1/2$ , write $\frac{1}{x} - \frac{1}{2} = \frac{∣ x - 2∣}{2∣ x ∣}$ ; the nuisance factor $1/∣ x ∣$ blows up at $0$ , so pre-restrict $δ \leq 1$ , whence $1 < x < 3$ , $∣ x ∣ > 1$ and $\frac{1}{2∣ x ∣} < \frac{1}{2}$ . Then $δ = min (1, 2 ε)$ finishes it. Sanity check at $ε = 0.05$ : $δ = 0.1$ and the worst point $x = 1.9$ gives $∣1/1.9 - 0.5∣ = 0.0263 < 0.05$ — worst because $1/ x$ is largest at the left end, which is exactly why the bound $∣ x ∣ > 1$ was the binding one.

Pitfall. A $δ$ that still contains $x$ is not a $δ$ . If your scratch work ends at " $∣ x - a ∣ < ε /∣ x + 3∣$ ", you have not finished: the quantifier order forbids $δ$ from depending on $x$ . The fix is always step 2 of the recipe — bound the offending factor by a constant on a pre-restricted neighbourhood, then use that constant.

Exercises for §3.5 →

3.6Proving the limit laws

Each law is proved from the definition. Read them as a catalogue of techniques: splitting the error budget, adding and subtracting a convenient term, and bounding a denominator.

Theorem 3.20 (Sum rule for limits). If $lim_{x \to a} f (x) = L$ and $lim_{x \to a} g (x) = M$ then $lim_{x \to a} (f + g) (x) = L + M$ .

Proof. Let $ε > 0$ . Since the two limits exist, apply each definition with the tolerance $ε /2$ : there are $δ_{1}, δ_{2} > 0$ with

0 < ∣ x - a ∣ < δ_{1} \Rightarrow ∣ f (x) - L ∣ < \frac{ε}{2}, 0 < ∣ x - a ∣ < δ_{2} \Rightarrow ∣ g (x) - M ∣ < \frac{ε}{2} .

Put $δ = min (δ_{1}, δ_{2}) > 0$ . For $0 < ∣ x - a ∣ < δ$ both estimates hold at once, so by the triangle inequality

∣ (f + g) (x) - (L + M) ∣ = ∣ (f (x) - L) + (g (x) - M) ∣ \leq ∣ f (x) - L ∣ + ∣ g (x) - M ∣ < \frac{ε}{2} + \frac{ε}{2} = ε .

∎

Nothing forces the split to be even — $ε /3$ and $2 ε /3$ serve equally. What matters is that the pieces are positive and sum to at most $ε$ , so each limit can be invoked at a tolerance of its own.

Theorem 3.21 (Product rule for limits). If $lim_{x \to a} f (x) = L$ and $lim_{x \to a} g (x) = M$ then $lim_{x \to a} (f g) (x) = L M$ .

Proof. The difference $f g - L M$ does not obviously split, so manufacture a split by adding and subtracting $f (x) M$ :

f (x) g (x) - L M = f (x) (g (x) - M) + M (f (x) - L),

whence $∣ f (x) g (x) - L M ∣ \leq ∣ f (x) ∣ ∣ g (x) - M ∣ + ∣ M ∣ ∣ f (x) - L ∣$ .

The stray factor $∣ f (x) ∣$ is handled by the proposition Local boundedness: there is $δ_{0} > 0$ and a constant $C = 1 + ∣ L ∣$ with $∣ f (x) ∣ \leq C$ for $0 < ∣ x - a ∣ < δ_{0}$ .

Now allocate. Let $ε > 0$ . Choose $δ_{1}$ so that $∣ g (x) - M ∣ < \frac{ε}{2 C}$ on the punctured $δ_{1}$ -neighbourhood, and $δ_{2}$ so that $∣ f (x) - L ∣ < \frac{ε}{2 ( ∣ M ∣ + 1 )}$ on the punctured $δ_{2}$ -neighbourhood; the $+ 1$ in the denominator costs nothing and covers the case $M = 0$ , where dividing by $∣ M ∣$ would be illegal. Set $δ = min (δ_{0}, δ_{1}, δ_{2}) > 0$ . For $0 < ∣ x - a ∣ < δ$ all three estimates hold simultaneously, and

∣ f (x) g (x) - L M ∣ \leq C \cdot \frac{ε}{2 C} + ∣ M ∣ \cdot \frac{ε}{2 ( ∣ M ∣ + 1 )} < \frac{ε}{2} + \frac{ε}{2} = ε .

∎

Theorem 3.22 (Quotient rule for limits). If $lim_{x \to a} f (x) = L$ and $lim_{x \to a} g (x) = M$ with $M \neq = 0$ , then $lim_{x \to a} \frac{f ( x )}{g ( x )} = \frac{L}{M}$ .

Proof. By the theorem Product rule for limits it suffices to prove the special case $lim_{x \to a} 1/ g (x) = 1/ M$ and then multiply by $f$ . So consider

\frac{1}{g ( x )} - \frac{1}{M} = \frac{∣ M - g ( x ) ∣}{∣ g ( x ) ∣ ∣ M ∣} .

The danger is the factor $1/∣ g (x) ∣$ , and the lemma A nonzero limit keeps the function away from zero removes it: there is $δ_{0} > 0$ with $∣ g (x) ∣ > ∣ M ∣/2$ on the punctured $δ_{0}$ -neighbourhood, so there

\frac{1}{g ( x )} - \frac{1}{M} < \frac{∣ M - g ( x ) ∣}{( ∣ M ∣/2 ) ∣ M ∣} = \frac{2}{∣ M ∣ ^{2}} ∣ g (x) - M ∣.

Given $ε > 0$ , choose $δ_{1}$ so that $∣ g (x) - M ∣ < \frac{∣ M ∣ ^{2} ε}{2}$ on the punctured $δ_{1}$ -neighbourhood, and put $δ = min (δ_{0}, δ_{1})$ . Then for $0 < ∣ x - a ∣ < δ$ ,

\frac{1}{g ( x )} - \frac{1}{M} < \frac{2}{∣ M ∣ ^{2}} \cdot \frac{∣ M ∣ ^{2} ε}{2} = ε .

Finally $f / g = f \cdot (1/ g)$ , and the product rule gives $L \cdot (1/ M) = L / M$ . Note that the hypothesis $M \neq = 0$ was used twice: once to make $∣ M ∣/2$ a legal tolerance, and once to divide by $∣ M ∣^{2}$ .∎

Theorem 3.23 (Squeeze theorem for functions). Suppose $g (x) \leq f (x) \leq h (x)$ for all $x$ in some punctured neighbourhood of $a$ , and $lim_{x \to a} g (x) = lim_{x \to a} h (x) = L$ . Then $lim_{x \to a} f (x) = L$ .

Proof. Let $ε > 0$ , and let $δ_{0}$ be a radius on which the inequalities $g \leq f \leq h$ hold. Choose $δ_{1}, δ_{2}$ so that $∣ g (x) - L ∣ < ε$ and $∣ h (x) - L ∣ < ε$ on the respective punctured neighbourhoods, and set $δ = min (δ_{0}, δ_{1}, δ_{2})$ . For $0 < ∣ x - a ∣ < δ$ all three conditions hold, so

L - ε < g (x) \leq f (x) \leq h (x) < L + ε,

which is exactly $∣ f (x) - L ∣ < ε$ .∎

The squeeze theorem is the only law that concludes existence from an inequality rather than from other limits, which makes it the tool of choice when $f$ itself is too unruly to estimate — as with $x sin (1/ x)$ , or $lim_{x \to 0} \frac{s i n x}{x}$ , whose classical proof squeezes the ratio between $cos x$ and $1$ .

Pitfall. "Substitute the inner limit" is not a theorem. If $lim_{x \to a} g (x) = b$ and $lim_{y \to b} f (y) = L$ , the composite need not tend to $L$ : take $g$ identically $0$ and $f (y) = 1$ for $y \neq = 0$ with $f (0) = 0$ . Then $lim_{x \to 0} g = 0$ and $lim_{y \to 0} f (y) = 1$ , yet $f (g (x)) = f (0) = 0$ for every $x$ . The composition law needs a side condition — either $g (x) \neq = b$ near $a$ , or $f$ continuous at $b$ — and the second is the one that always holds in practice, which is why composition is stated for continuous functions in the theorem Continuity of compositions.

Exercises for §3.6 →

3.7Continuity at a point

Continuity is the assertion that the limit and the value agree. Everything technical is already in place; what is new is that the point $a$ is no longer excluded, and the number $L$ is no longer free to be chosen — it is forced to be $f (a)$ .

Definition 3.24 (Continuity at a point). Let $f : D \to R$ and $a \in D$ . Then $f$ is continuous at $a$ if for every $ε > 0$ there is $δ > 0$ such that

x \in D, ∣ x - a ∣ < δ ⟹ ∣ f (x) - f (a) ∣ < ε .

Three differences from the limit definition matter. The point $a$ must belong to $D$ , since $f (a)$ appears in the conclusion. The condition $0 < ∣ x - a ∣$ has become $∣ x - a ∣$ , admitting $x = a$ — harmless, since there the conclusion reads $0 < ε$ . And $a$ need not be a limit point: at an isolated point of $D$ the definition holds vacuously, so every function is continuous at an isolated point of its domain.

Proposition 3.25 (Continuity as a limit). If $a \in D$ is a limit point of $D$ , then $f$ is continuous at $a$ if and only if $lim_{x \to a} f (x)$ exists and equals $f (a)$ .

Proof. ( $\Rightarrow$ ) Suppose $f$ is continuous at $a$ and let $ε > 0$ ; take the $δ$ from the definition. Any $x$ with $0 < ∣ x - a ∣ < δ$ in particular satisfies $∣ x - a ∣ < δ$ , so $∣ f (x) - f (a) ∣ < ε$ . That is the limit definition with $L = f (a)$ .

( $\Leftarrow$ ) Suppose the limit exists and equals $f (a)$ , and let $ε > 0$ with corresponding $δ$ . For $x \in D$ with $∣ x - a ∣ < δ$ there are two cases: if $x \neq = a$ then $0 < ∣ x - a ∣ < δ$ and the limit estimate gives $∣ f (x) - f (a) ∣ < ε$ ; if $x = a$ then $∣ f (x) - f (a) ∣ = 0 < ε$ . Either way the continuity condition holds.∎

Intuition. Continuity says three separate things at once, and a discontinuity is a failure of any one of them: $f (a)$ must be defined, $lim_{x \to a} f (x)$ must exist, and the two must be equal. Checking a suspicious point means checking all three in that order — most textbook "discontinuities" fail at the second, and the interesting ones fail only at the third.

Theorem 3.26 (Algebra of continuous functions). If $f$ and $g$ are continuous at $a$ , then so are $f + g$ , $f g$ and $c f$ for any constant $c$ ; and $f / g$ is continuous at $a$ provided $g (a) \neq = 0$ .

Proof. If $a$ is isolated in the common domain there is nothing to prove, since every function is continuous there. Otherwise $a$ is a limit point, and by the proposition Continuity as a limit we have $lim_{x \to a} f = f (a)$ and $lim_{x \to a} g = g (a)$ . The theorem Limit laws then gives

x \to a lim (f + g) = f (a) + g (a) = (f + g) (a), x \to a lim (f g) = f (a) g (a) = (f g) (a),

and likewise for $c f$ . For the quotient, the hypothesis $g (a) \neq = 0$ is exactly the nonvanishing required by Quotient rule for limits, and it gives $lim_{x \to a} f / g = f (a) / g (a)$ . In each case the limit equals the value, which by Continuity as a limit read backwards is continuity at $a$ .∎

Example 3.27 (A function continuous at exactly one point). Let $f (x) = x$ for $x$ rational and $f (x) = 0$ for $x$ irrational. Determine where $f$ is continuous.

Solution.

At $a = 0$ : for any $x$ , $f (x)$ is either $x$ or $0$ , so $∣ f (x) - f (0) ∣ \leq ∣ x ∣$ , and $δ = ε$ works. So $f$ is continuous at $0$ .
At any $a \neq = 0$ : take rationals $q_{n} \to a$ and irrationals $t_{n} \to a$ , both available by density. Then $f (q_{n}) = q_{n} \to a$ while $f (t_{n}) = 0 \to 0$ , and these differ because $a \neq = 0$ , so by the theorem Sequential characterisation of limits no limit exists at $a$ and continuity fails.
Sanity check: the two branches of $f$ — the line $y = x$ and the line $y = 0$ — meet at exactly one point, and that is exactly where continuity survives. A function can be continuous at a single point and nowhere else, so continuity at $a$ says nothing whatever about a neighbourhood of $a$ .

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Exercises for §3.7 →

3.8Continuity on a set, and how continuous functions are built

Definition 3.28 (Continuity on a set). $f$ is continuous on $S$ if it is continuous at every point of $S$ . It is continuous if it is continuous at every point of its domain.

Almost every function one writes down is continuous, and the reason is not luck: the constants and the identity are continuous, and the building operations preserve continuity. Two results make this precise, covering polynomials, rational functions, and any composite of them.

Theorem 3.29 (Continuity of compositions). If $g$ is continuous at $a$ and $f$ is continuous at $g (a)$ , then $f \circ g$ is continuous at $a$ .

Proof. Write $b = g (a)$ and let $ε > 0$ . Continuity of $f$ at $b$ supplies $η > 0$ such that $∣ y - b ∣ < η$ implies $∣ f (y) - f (b) ∣ < ε$ , for $y$ in the domain of $f$ . Continuity of $g$ at $a$ , applied with tolerance $η$ , supplies $δ > 0$ such that $∣ x - a ∣ < δ$ implies $∣ g (x) - b ∣ < η$ .

Chaining the two: if $∣ x - a ∣ < δ$ then $y = g (x)$ satisfies $∣ y - b ∣ < η$ , hence $∣ f (g (x)) - f (g (a)) ∣ < ε$ .∎

Compare this with the pitfall about substituting an inner limit, which required an awkward side condition. Here there is none, and the reason is visible in the proof: we never needed $∣ y - b ∣ > 0$ , because continuity of $f$ at $b$ covers $y = b$ as well. Continuity is precisely the hypothesis that makes composition unconditional.

Proposition 3.30 (Polynomials and rational functions are continuous). Every polynomial is continuous on $R$ , and every rational function $p / q$ is continuous at every point where $q$ does not vanish.

Proof. The constant function $x \mapsto c$ is continuous everywhere, since $∣ c - c ∣ = 0 < ε$ for any $δ$ ; the identity $x \mapsto x$ is continuous everywhere, since $δ = ε$ works. A monomial $c x^{n}$ is a finite product of these, hence continuous by the theorem Algebra of continuous functions applied $n$ times, and a polynomial is a finite sum of monomials, hence continuous by the same theorem. For a rational function, apply the quotient clause of that theorem at each point where the denominator is nonzero.∎

Example 3.31 (Continuity of a rational function, and the point where it fails). Determine where $f (x) = \frac{x + 1}{x - 2}$ is continuous, and classify the failure.

Solution.

Numerator and denominator are polynomials, continuous on $R$ , and the quotient clause of Polynomials and rational functions are continuous applies wherever $x - 2 \neq = 0$ , so $f$ is continuous at every $a \neq = 2$ .
At $a = 2$ the function is undefined, so continuity is not merely false but meaningless. The limit does not exist either: the numerator tends to $3 \neq = 0$ while the denominator tends to $0$ , so $∣ f (x) ∣ \to \infty$ — concretely $f (2.01) = 301$ and $f (1.99) = - 299$ .
Sanity check: the discontinuity is infinite, not removable, since no finite value assigned at $2$ is close to both $301$ and $- 299$ .

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Composition then covers everything else one writes down. To see that $h (x) = x^{2} + 1$ is continuous on $R$ , write $h = f \circ g$ with $g (x) = x^{2} + 1$ , a polynomial, and $f (y) = y$ , which is continuous at every $b > 0$ because $∣ y - b ∣ = ∣ y - b ∣/ (y + b) \leq ∣ y - b ∣/ b$ , so $δ = ε b$ works. Since $g (x) \geq 1 > 0$ everywhere, the theorem Continuity of compositions applies at every point.

Exercises for §3.8 →

3.9The sequential characterisation, and a taxonomy of discontinuities

The most useful reformulation of continuity trades quantifiers for sequences: it makes disproving continuity a matter of exhibiting one badly behaved sequence, and it imports every theorem about sequences wholesale.

Theorem 3.32 (Sequential characterisation of continuity). Let $f : D \to R$ and $a \in D$ . Then $f$ is continuous at $a$ if and only if for every sequence $(x_{n})$ in $D$ with $x_{n} \to a$ we have $f (x_{n}) \to f (a)$ .

Proof. ( $\Rightarrow$ ) Assume continuity at $a$ and let $x_{n} \to a$ in $D$ . Given $ε > 0$ , take $δ$ from the definition of continuity and then $N$ with $∣ x_{n} - a ∣ < δ$ for $n \geq N$ . For such $n$ , $∣ f (x_{n}) - f (a) ∣ < ε$ . Hence $f (x_{n}) \to f (a)$ . Observe that, unlike in Sequential characterisation of limits, no clause $x_{n} \neq = a$ is needed, because continuity covers $x = a$ .

( $\Leftarrow$ ) Contrapositive. Suppose $f$ is not continuous at $a$ . Negating the definition gives $ε_{0} > 0$ such that for every $δ > 0$ there is $x \in D$ with $∣ x - a ∣ < δ$ and $∣ f (x) - f (a) ∣ \geq ε_{0}$ . Take $δ = 1/ n$ and let $x_{n}$ be such a point. Then $x_{n} \in D$ and $∣ x_{n} - a ∣ < 1/ n$ , so $x_{n} \to a$ , while $∣ f (x_{n}) - f (a) ∣ \geq ε_{0}$ for every $n$ , so $f (x_{n}) \neq \to f (a)$ . This sequence witnesses the failure of the right-hand condition.∎

Definition 3.33 (Classification of discontinuities). Let $f$ fail to be continuous at $a$ , an interior point of an interval in its domain. The discontinuity is

removable if $lim_{x \to a} f (x)$ exists but either differs from $f (a)$ or $f (a)$ is undefined;
a jump (discontinuity of the first kind) if both one-sided limits exist and are finite but differ;
essential (of the second kind) if at least one one-sided limit fails to exist as a finite number — because the function blows up, or oscillates without settling.

The names are honest. A removable discontinuity is repaired by redefining a single value; a jump cannot be repaired by any redefinition, though the function is still tame on each side; an essential discontinuity means at least one side has no limiting behaviour at all. The trichotomy is exhaustive: either both one-sided limits exist finitely — agreeing, hence removable, or disagreeing, hence a jump — or one does not, which is the essential case.

Theorem 3.34 (Monotone functions have only jump discontinuities). Let $f$ be increasing on an interval $I$ and let $a$ be an interior point. Then both one-sided limits exist at $a$ , with

x \to a^{-} lim f (x) = x < a sup f (x) \leq f (a) \leq x > a in f f (x) = x \to a^{+} lim f (x) .

Consequently every discontinuity of a monotone function is a jump.

Proof. Consider the set $A = {f (x) : x \in I, x < a}$ . It is nonempty since $a$ is interior, and bounded above by $f (a)$ because $f$ is increasing. By the least-upper-bound property of $R$ — the definition Upper bound, supremum, infimum — the number $s = sup A$ exists and satisfies $s \leq f (a)$ .

We claim $lim_{x \to a^{-}} f (x) = s$ . Let $ε > 0$ . By the characterisation of the supremum, $s - ε$ is not an upper bound of $A$ , so there is $x_{0} < a$ with $f (x_{0}) > s - ε$ . Put $δ = a - x_{0} > 0$ . For any $x$ with $a - δ < x < a$ we have $x_{0} < x < a$ , so monotonicity gives $s - ε < f (x_{0}) \leq f (x) \leq s$ , hence $∣ f (x) - s ∣ < ε$ .

The right-hand statement is the same argument applied to $B = {f (x) : x > a}$ , which is bounded below by $f (a)$ , with the infimum in place of the supremum. Both one-sided limits therefore exist and are finite, so no essential discontinuity is possible; and if they are equal then, being sandwiched around $f (a)$ , they both equal $f (a)$ and $f$ is continuous there. So a discontinuity forces the two to differ, which is a jump.∎

A short extra argument turns that into a striking finiteness statement: distinct discontinuities of an increasing $f$ carry disjoint open jump-intervals, so choosing one rational from each injects them into $Q$ , and a monotone function therefore has at most countably many discontinuities.

Example 3.35 (Thomae's function). Define $f (x) = 0$ for $x$ irrational and $f (p / q) = 1/ q$ for a rational written in lowest terms with $q > 0$ (so $f (0) = f (0/1) = 1$ ). Show that $f$ is continuous at every irrational and discontinuous at every rational.

Solution.

Discontinuity at a rational $a = p / q$ . Take irrationals $t_{n} \to a$ , available by density. Then $f (t_{n}) = 0 \to 0$ while $f (a) = 1/ q > 0$ , so by the theorem Sequential characterisation of continuity $f$ is discontinuous at $a$ .
Continuity at an irrational $a$ . Let $ε > 0$ and choose $N$ with $1/ N < ε$ . The rationals in $(a - 1, a + 1)$ with denominator at most $N$ are finitely many, and none equals $a$ . Let $δ > 0$ be less than $1$ and less than the distance from $a$ to each of them — a minimum of finitely many positive numbers is positive, which is the crux.
Take any $x$ with $∣ x - a ∣ < δ$ . If $x$ is irrational, $∣ f (x) - f (a) ∣ = 0 < ε$ . If $x = p / q$ is rational then by the choice of $δ$ its denominator exceeds $N$ , so $∣ f (x) - f (a) ∣ = 1/ q < 1/ N < ε$ .
Sanity check on the asymmetry: near any point there are rationals with huge denominators and hence tiny values, so the irrational value $0$ is approachable from everywhere, while a rational's own value $1/ q$ is isolated in the surrounding sea of near-zeros.

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Example 3.36 (Classifying three discontinuities at the origin). Classify the discontinuity at $0$ of (a) $f (x) = \frac{sin x}{x}$ with $f (0) = 0$ , (b) $g (x) = ⌊ x ⌋$ , and (c) $h (x) = sin (1/ x)$ with $h (0) = 0$ .

Solution.

(a) $lim_{x \to 0} \frac{s i n x}{x} = 1$ exists but the assigned value is $0 \neq = 1$ ; both one-sided limits exist and agree, so the discontinuity is removable and redefining $f (0) = 1$ repairs it.
(b) $lim_{x \to 0^{-}} ⌊ x ⌋ = - 1$ and $lim_{x \to 0^{+}} ⌊ x ⌋ = 0$ both exist finitely and differ, so this is a jump of size $1$ , repairable by no redefinition — consistent with Monotone functions have only jump discontinuities, the floor function being increasing.
(c) By the example Divergence by two routes neither one-sided limit of $sin (1/ x)$ exists at $0$ , so the discontinuity is essential, of oscillatory type.
Sanity check: the three cases are distinguished purely by the one-sided limits — both exist and agree, both exist and differ, or one fails — and (a), (b), (c) realise them in order.

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Exercises for §3.9 →

3.10Uniform continuity: moving the quantifier

The whole content of this section is the position of a single quantifier. Continuity on a set is asserted point by point, so the $δ$ that works at one point need not work at another; uniform continuity demands one $δ$ for the entire set.

Definition 3.37 (Pointwise continuity on a set, in full). $f$ is continuous on $S$ if

\forall a \in S \forall ε > 0 \exists δ > 0 \forall x \in S : ∣ x - a ∣ < δ \Rightarrow ∣ f (x) - f (a) ∣ < ε .

The point $a$ is named before $δ$ is chosen, so $δ$ may depend on $a$ as well as on $ε$ .

Definition 3.38 (Uniform continuity). $f$ is uniformly continuous on $S$ if

\forall ε > 0 \exists δ > 0 \forall x, y \in S : ∣ x - y ∣ < δ \Rightarrow ∣ f (x) - f (y) ∣ < ε .

Here $δ$ is chosen before any point of $S$ is named, so it may depend on $ε$ (and on $f$ and $S$ ) alone.

Uniform continuity implies ordinary continuity on $S$ , and the proof is one line: given $a \in S$ and $ε > 0$ , take the uniform $δ$ and apply the condition to the pair $(x, a)$ . The converse is false, as the example One failure and one success shows with $f (x) = x^{2}$ on $R$ .

To disprove uniform continuity one negates the definition; the negated form is what every counterexample actually verifies.

Proposition 3.39 (Negation of uniform continuity). $f$ fails to be uniformly continuous on $S$ if and only if there exist $ε_{0} > 0$ and sequences $(x_{n}), (y_{n})$ in $S$ with

∣ x_{n} - y_{n} ∣ \to 0 but ∣ f (x_{n}) - f (y_{n}) ∣ \geq ε_{0} for all n .

Proof. ( $\Rightarrow$ ) If uniform continuity fails, there is $ε_{0} > 0$ such that no $δ$ works. Apply this to $δ = 1/ n$ : there are $x_{n}, y_{n} \in S$ with $∣ x_{n} - y_{n} ∣ < 1/ n$ and $∣ f (x_{n}) - f (y_{n}) ∣ \geq ε_{0}$ . The first condition forces $∣ x_{n} - y_{n} ∣ \to 0$ .

( $\Leftarrow$ ) Conversely, given such sequences, no $δ > 0$ can work for $ε_{0}$ : choose $n$ with $∣ x_{n} - y_{n} ∣ < δ$ , which is possible since $∣ x_{n} - y_{n} ∣ \to 0$ ; the pair $(x_{n}, y_{n})$ then satisfies the hypothesis of the implication but violates its conclusion.∎

Proposition 3.40 (Lipschitz implies uniformly continuous). If there is $K > 0$ with $∣ f (x) - f (y) ∣ \leq K ∣ x - y ∣$ for all $x, y \in S$ , then $f$ is uniformly continuous on $S$ . The converse is false.

Proof. Given $ε > 0$ , put $δ = ε / K > 0$ , which depends on nothing but $ε$ and $K$ . If $∣ x - y ∣ < δ$ then $∣ f (x) - f (y) ∣ \leq K ∣ x - y ∣ < K \cdot ε / K = ε$ .

For the converse, $f (x) = x$ on $[0, 1]$ is uniformly continuous by the example One failure and one success below, yet no $K$ can satisfy the Lipschitz condition: taking $y = 0$ and $x = 1/ n^{2}$ gives $∣ f (x) - f (y) ∣/∣ x - y ∣ = (1/ n) / (1/ n^{2}) = n$ , which is unbounded.∎

Example 3.41 (One failure and one success). (a) Show $f (x) = x^{2}$ is continuous on $R$ but not uniformly continuous there. (b) Show $g (x) = x$ is uniformly continuous on the equally unbounded $[0, \infty)$ .

Solution.

(a) Continuity came from the example Two limits from the definition, whose last step gave $δ = min (1, ε / (2∣ a ∣ + 1))$ — a formula already carrying the warning, since it tends to $0$ as $∣ a ∣ \to \infty$ .
Apply the proposition Negation of uniform continuity with $ε_{0} = 1$ and the sequences $x_{n} = n$ , $y_{n} = n + \frac{1}{n}$ . The separation vanishes, $∣ x_{n} - y_{n} ∣ = 1/ n \to 0$ , but the image separation does not:

∣ f (y_{n}) - f (x_{n}) ∣ = (n + \frac{1}{n})^{2} - n^{2} = 2 + \frac{1}{n ^{2}} > 2 > ε_{0} .

So no $δ$ works for $ε_{0} = 1$ . The mechanism is visible in $∣ x^{2} - y^{2} ∣ = ∣ x - y ∣ ∣ x + y ∣$ : on $R$ the amplification factor $∣ x + y ∣$ is unbounded.
(b) The key algebraic fact is $x - y \leq ∣ x - y ∣$ for $x, y \geq 0$ . Assuming $x \geq y$ , we have $(x - y)^{2} \leq (x - y) (x + y) = x - y$ , valid because $0 \leq x - y \leq x + y$ ; take square roots.
Given $ε > 0$ put $δ = ε^{2}$ , which mentions no point. If $∣ x - y ∣ < δ$ then $∣ x - y ∣ \leq ∣ x - y ∣ < ε$ .
Sanity check on the contrast: an unbounded domain does not by itself destroy uniform continuity. What destroys it is unbounded amplification of separations, and the square root compresses rather than amplifies. Note too that $g$ is uniformly continuous without being Lipschitz — from $y = 0$ and $x = 1/ n^{2}$ the ratio $∣ g (x) - g (y) ∣/∣ x - y ∣$ equals $n$ , which is unbounded — so the proposition Lipschitz implies uniformly continuous runs in one direction only.

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Exercises for §3.10 →

3.11Heine–Cantor: compactness buys uniformity

On $R$ or on $(0, 1]$ the two notions genuinely differ. On a closed bounded interval they coincide, and the reason is compactness — the same property that drives the Extreme Value Theorem below.

Theorem 3.42 (Heine–Cantor). Let $K \subseteq R$ be closed and bounded and let $f : K \to R$ be continuous. Then $f$ is uniformly continuous on $K$ .

Proof. Suppose not. By the proposition Negation of uniform continuity there are $ε_{0} > 0$ and sequences $(x_{n}), (y_{n})$ in $K$ with $∣ x_{n} - y_{n} ∣ \to 0$ and

∣ f (x_{n}) - f (y_{n}) ∣ \geq ε_{0} for all n . (*)

Since $K$ is bounded, the sequence $(x_{n})$ is bounded, so by Bolzano–Weierstrass it has a convergent subsequence $x_{n_{k}} \to c$ . Since $K$ is closed and every $x_{n_{k}}$ lies in $K$ , the limit $c$ lies in $K$ too — this is where closedness is used, and without it $c$ could escape the set and continuity at $c$ would be unavailable.

The matching subsequence of $(y_{n})$ converges to the same point:

∣ y_{n_{k}} - c ∣ \leq ∣ y_{n_{k}} - x_{n_{k}} ∣ + ∣ x_{n_{k}} - c ∣ ⟶ 0 + 0 = 0.

Now apply the theorem Sequential characterisation of continuity at the point $c \in K$ to both subsequences: $f (x_{n_{k}}) \to f (c)$ and $f (y_{n_{k}}) \to f (c)$ . Hence

∣ f (x_{n_{k}}) - f (y_{n_{k}}) ∣ \leq ∣ f (x_{n_{k}}) - f (c) ∣ + ∣ f (c) - f (y_{n_{k}}) ∣ ⟶ 0,

so the left-hand side is eventually smaller than $ε_{0}$ , contradicting $(*)$ , which asserted it is at least $ε_{0}$ for every index. Therefore $f$ is uniformly continuous.∎

Neither hypothesis is decorative: boundedness produced the convergent subsequence, closedness kept its limit inside $K$ where $f$ is continuous. Drop boundedness and $x^{2}$ on $R$ is a counterexample; drop closedness and $1/ x$ on $(0, 1]$ is one.

Example 3.43 (Heine–Cantor made explicit). Verify by an explicit $δ$ that $f (x) = x^{2}$ , not uniformly continuous on $R$ , is uniformly continuous on $[0, 3]$ .

Solution.

The abstract route takes one line: $[0, 3]$ is closed and bounded and $x^{2}$ is continuous on it, so Heine–Cantor applies, with no computation at all.
The explicit route shows why. For $x, y \in [0, 3]$ , $∣ x^{2} - y^{2} ∣ = ∣ x - y ∣ ∣ x + y ∣ \leq 6∣ x - y ∣$ since $x + y \leq 6$ , so $δ = ε /6$ works and mentions no point of the interval.
Sanity check: with $ε = 0.06$ we get $δ = 0.01$ , and the worst pair sits at the right end, $x = 2.99$ , $y = 3$ : $∣9 - 8.9401∣ = 0.0599 < 0.06$ . The bound is essentially attained, confirming that $∣ x + y ∣ \leq 6$ is the binding constraint.
The moral: capping the domain caps the amplification factor. Here $f$ is in fact Lipschitz with $K = 6$ , so Lipschitz implies uniformly continuous gives the conclusion a third way.

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Exercises for §3.11 →

3.12The Cauchy criterion and continuous extension

Uniform continuity has a second characterisation, in terms of what the function does to Cauchy sequences, and it explains why uniformly continuous maps extend to the closure of their domain while merely continuous ones need not.

Theorem 3.44 (Cauchy-sequence criterion). If $f$ is uniformly continuous on $S$ , then $f$ maps every Cauchy sequence in $S$ to a Cauchy sequence in $R$ .

Proof. Let $(x_{n})$ be Cauchy in $S$ and let $ε > 0$ . Uniform continuity supplies $δ > 0$ such that $∣ x - y ∣ < δ$ implies $∣ f (x) - f (y) ∣ < ε$ for all $x, y \in S$ . Since $(x_{n})$ is Cauchy, there is $N$ with $∣ x_{m} - x_{n} ∣ < δ$ for all $m, n \geq N$ . For such $m, n$ , applying the uniform condition to the pair $(x_{m}, x_{n})$ gives $∣ f (x_{m}) - f (x_{n}) ∣ < ε$ . Hence $(f (x_{n}))$ is Cauchy.∎

Note what the proof needed: one $δ$ good for the pair $(x_{m}, x_{n})$ , neither member being a fixed point of the domain. Ordinary continuity attaches its $δ$ to a point, and a Cauchy sequence need not have its limit in $S$ — so there may be no point to attach it to. That is the gap the next example exploits.

Corollary 3.45 (Extension from a dense subset). Let $S \subseteq R$ and let $f : S \to R$ be uniformly continuous. Then $f$ extends uniquely to a continuous function $\overset{ˉ}{f}$ on the closure $\overline{S}$ , and $\overset{ˉ}{f}$ is uniformly continuous.

Proof. Let $c \in \overline{S}$ and choose $x_{n} \in S$ with $x_{n} \to c$ . A convergent sequence is Cauchy, so by the theorem Cauchy-sequence criterion the image $(f (x_{n}))$ is Cauchy in $R$ , hence convergent by the completeness of $R$ — the Cauchy criterion of the previous chapter. Define $\overset{ˉ}{f} (c) = lim_{n} f (x_{n})$ .

Well defined. If $y_{n} \to c$ is another such sequence, interleave them into $x_{1}, y_{1}, x_{2}, y_{2}, \dots$ , which also converges to $c$ ; its image is Cauchy by the same theorem, hence convergent, and a convergent sequence has all subsequences converging to the same value. So $lim f (x_{n}) = lim f (y_{n})$ .

Extends $f$ . If $c \in S$ , take the constant sequence $x_{n} = c$ to get $\overset{ˉ}{f} (c) = f (c)$ .

Uniformly continuous. Let $ε > 0$ and take $δ$ from the uniform continuity of $f$ for the tolerance $ε /3$ . Let $u, v \in \overline{S}$ with $∣ u - v ∣ < δ /3$ , and pick $s, t \in S$ with $∣ s - u ∣ < δ /3$ , $∣ t - v ∣ < δ /3$ , $∣ f (s) - \overset{ˉ}{f} (u) ∣ < ε /3$ and $∣ f (t) - \overset{ˉ}{f} (v) ∣ < ε /3$ , all possible by the construction of $\overset{ˉ}{f}$ . Then $∣ s - t ∣ \leq ∣ s - u ∣ + ∣ u - v ∣ + ∣ v - t ∣ < δ$ , so $∣ f (s) - f (t) ∣ < ε /3$ and

∣ \overset{ˉ}{f} (u) - \overset{ˉ}{f} (v) ∣ \leq ∣ \overset{ˉ}{f} (u) - f (s) ∣ + ∣ f (s) - f (t) ∣ + ∣ f (t) - \overset{ˉ}{f} (v) ∣ < ε .

Unique. Any continuous extension $g$ must satisfy $g (c) = lim g (x_{n}) = lim f (x_{n}) = \overset{ˉ}{f} (c)$ for $x_{n} \to c$ in $S$ , by Sequential characterisation of continuity.∎

Example 3.46 (One function that extends and two that do not). On $(0, 1]$ consider $u (x) = 1/ x$ , $v (x) = sin (1/ x)$ and $w (x) = x sin (1/ x)$ . Decide in each case whether the function is uniformly continuous and whether it extends continuously to $[0, 1]$ .

Solution.

$u (x) = 1/ x$ . Take $x_{n} = 1/ n$ , which converges in $R$ to $0$ and hence is Cauchy — even though its limit lies outside the domain, which is the whole point. Its image $u (x_{n}) = n$ is unbounded and so not Cauchy, so by the theorem Cauchy-sequence criterion $u$ is not uniformly continuous and no continuous extension to $[0, 1]$ exists.
The direct route agrees: $x_{n} = 1/ n$ , $y_{n} = 1/ (n + 1)$ give $∣ x_{n} - y_{n} ∣ = \frac{1}{n ( n + 1 )} \to 0$ while $∣ u (x_{n}) - u (y_{n}) ∣ = 1$ always, matching Negation of uniform continuity at $ε_{0} = 1$ .
$v (x) = sin (1/ x)$ . By the example Divergence by two routes the limit at $0^{+}$ does not exist, so no value at $0$ extends it continuously. Interleaving the Cauchy sequences $1/ (nπ)$ and $1/ (2 πn + π /2)$ gives a Cauchy sequence whose image oscillates between $0$ and $1$ , so $v$ is not uniformly continuous either — which the corollary Extension from a dense subset, read in contrapositive, already predicted.
$w (x) = x sin (1/ x)$ . By the example Squeezing an oscillation, $lim_{x \to 0^{+}} w = 0$ . Setting $\overset{w}{ˉ} (0) = 0$ gives a function whose limit at $0$ equals its value, hence continuous there by Continuity as a limit and elsewhere by the algebra and composition theorems.
Since $[0, 1]$ is closed and bounded, Heine–Cantor makes $\overset{w}{ˉ}$ uniformly continuous, so $w$ was uniformly continuous on $(0, 1]$ all along. Sanity check: $(0, 1]$ is bounded but not closed, so Heine–Cantor applied to none of the three directly — the compactness had to be manufactured by filling the missing endpoint, which only $w$ permitted.

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Exercises for §3.12 →

3.13The Intermediate Value Theorem

The first of the two great theorems about continuous functions on a closed bounded interval says that such a function cannot get from one value to another without passing through everything in between. It sounds obvious, and it is false over $Q$ — so whatever makes it true is a property of $R$ , not of continuity alone.

Theorem 3.47 (Intermediate Value Theorem). Let $f$ be continuous on $[a, b]$ and let $k$ lie strictly between $f (a)$ and $f (b)$ . Then there exists $c \in (a, b)$ with $f (c) = k$ .

Proof. Assume $f (a) < k < f (b)$ ; the other case follows by applying this one to $- f$ and $- k$ . Define

S = {x \in [a, b] : f (x) \leq k} .

$S$ is nonempty, since $f (a) < k$ puts $a \in S$ , and bounded above by $b$ . By the least-upper-bound property of $R$ , $c = sup S$ exists, and $a \leq c \leq b$ .

We rule out $f (c) < k$ and $f (c) > k$ in turn, each time using continuity at $c$ with a tolerance chosen to keep $f$ on one side of $k$ .

Suppose $f (c) < k$ . Apply continuity at $c$ with $ε = k - f (c) > 0$ : there is $δ > 0$ such that $∣ x - c ∣ < δ$ and $x \in [a, b]$ imply $∣ f (x) - f (c) ∣ < ε$ , hence $f (x) < f (c) + ε = k$ . Since $f (c) < k \leq f (b)$ forces $c \neq = b$ , the interval $(c, min (c + δ, b))$ is nonempty and every point $x$ in it satisfies $f (x) < k$ , so $x \in S$ with $x > c$ — contradicting that $c$ is an upper bound for $S$ .

Suppose instead $f (c) > k$ . Apply continuity at $c$ with $ε = f (c) - k > 0$ , getting $δ > 0$ with $f (x) > f (c) - ε = k$ for all $x \in [a, b]$ within $δ$ of $c$ . Since $f (a) < k < f (c)$ forces $c \neq = a$ , the interval $(c - δ, c]$ meets $[a, b]$ and contains no point of $S$ . Hence $c - δ$ (or $a$ , whichever is larger) is an upper bound for $S$ strictly smaller than $c$ , contradicting that $c$ is the least upper bound.

Both alternatives are impossible, so $f (c) = k$ . Finally $c \neq = a$ and $c \neq = b$ because $f (a) \neq = k \neq = f (b)$ , so $c \in (a, b)$ .∎

Completeness entered at exactly one word: $sup$ . Over $Q$ the set $S$ has no least upper bound, and the theorem is genuinely false there — $f (x) = x^{2} - 2$ is continuous on $[1, 2] \cap Q$ with $f (1) < 0 < f (2)$ , yet no rational $c$ has $c^{2} = 2$ , by Irrationality of the square root of two. The Intermediate Value Theorem is the least-upper-bound property wearing a disguise.

Intuition. Think of $S$ as the set of inputs at which the function is still at or below the level $k$ , and $c$ as the last such input. Just to the right of $c$ the function has committed to being above $k$ ; just to the left it was still at or below. Continuity forbids it from making that switch by a jump, so at $c$ itself it must be sitting exactly on the level.

Example 3.48 (Existence of a square root of 2). Use the Intermediate Value Theorem to prove that some positive real $c$ satisfies $c^{2} = 2$ .

Solution.

Let $f (x) = x^{2} - 2$ , continuous on $[1, 2]$ by Polynomials and rational functions are continuous, with $f (1) = - 1 < 0 < 2 = f (2)$ straddling the level $k = 0$ .
By the Intermediate Value Theorem there is $c \in (1, 2)$ with $f (c) = 0$ , that is $c^{2} = 2$ .
Sanity check: $1. 4^{2} = 1.96 < 2$ and $1. 5^{2} = 2.25 > 2$ , so $c \in (1.4, 1.5)$ , consistent with $2 \approx 1.41421$ . Note what has been delivered — existence, nothing more, since the proof located $c$ as a supremum rather than computing it. Digits are the business of the bisection method below.

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Figure 3.49. The Intermediate Value Theorem for

f (x) = x^{3} + x - 1

[0, 1]

at level

k = 0

. The endpoint values

f (0) = - 1

and

f (1) = 1

sit on opposite sides of the level line, and continuity forces the graph to meet it; the crossing is near

0.6823

. The first bisection midpoint

x = 0.5

has

f (0.5) = - 0.375 < 0

, so the root is trapped in the right half.

Exercises for §3.13 →

3.14Consequences of the Intermediate Value Theorem

Corollary 3.50 (Continuous images of intervals are intervals). If $I$ is an interval and $f$ is continuous on $I$ , then $f (I)$ is an interval.

Proof. A set $J \subseteq R$ is an interval precisely when it contains every point between any two of its points. So take $u, v \in f (I)$ with $u < k < v$ , say $u = f (p)$ and $v = f (q)$ with $p, q \in I$ . The closed interval with endpoints $p$ and $q$ lies inside $I$ , because $I$ is an interval, and $f$ is continuous on it. By the theorem Intermediate Value Theorem applied on that interval, there is $c$ between $p$ and $q$ with $f (c) = k$ , so $k \in f (I)$ .∎

Corollary 3.51 (Odd-degree polynomials have a real root). Every polynomial of odd degree with real coefficients has a real root.

Proof. Let $p (x) = a_{n} x^{n} + \dots + a_{0}$ with $n$ odd and $a_{n} \neq = 0$ ; replacing $p$ by $- p$ if necessary, assume $a_{n} > 0$ . Writing $p (x) = a_{n} x^{n} (1 + \frac{a _{n - 1}}{a _{n} x} + \dots + \frac{a _{0}}{a _{n} x ^{n}})$ , the bracket tends to $1$ as $∣ x ∣ \to \infty$ , so for $∣ x ∣$ large enough the bracket exceeds $1/2$ and $p (x)$ has the sign of $a_{n} x^{n}$ . Because $n$ is odd, $x^{n} > 0$ for large positive $x$ and $x^{n} < 0$ for large negative $x$ . Choose $b$ with $p (b) > 0$ and $a < b$ with $p (a) < 0$ . Then $p$ is continuous on $[a, b]$ and $p (a) < 0 < p (b)$ , so the Intermediate Value Theorem supplies $c \in (a, b)$ with $p (c) = 0$ .∎

The parity hypothesis does real work: $p (x) = x^{2} + 1$ has even degree and no real root, since for even $n$ the leading term has the same sign at both ends.

Corollary 3.52 (One-dimensional fixed point theorem). If $f : [a, b] \to [a, b]$ is continuous, then $f$ has a fixed point: some $c \in [a, b]$ with $f (c) = c$ .

Proof. Let $g (x) = f (x) - x$ , continuous on $[a, b]$ by the theorem Algebra of continuous functions. Since $f$ maps into $[a, b]$ we have $f (a) \geq a$ and $f (b) \leq b$ , so $g (a) \geq 0$ and $g (b) \leq 0$ .

If $g (a) = 0$ then $a$ is a fixed point; if $g (b) = 0$ then $b$ is. Otherwise $g (b) < 0 < g (a)$ , and the Intermediate Value Theorem applied to $g$ on $[a, b]$ with $k = 0$ gives $c \in (a, b)$ with $g (c) = 0$ , that is $f (c) = c$ .∎

Method 3.53 (The bisection method).

Find $a < b$ with $f (a)$ and $f (b)$ of opposite signs; the Intermediate Value Theorem now guarantees a root in $(a, b)$ .
Evaluate $f$ at the midpoint $m = (a + b) /2$ .
Replace $[a, b]$ by whichever half still shows a sign change.
Repeat. After $n$ steps the root is trapped in an interval of length $(b - a) / 2^{n}$ .

Example 3.54 (Bisection in action). Locate the root of $f (x) = x^{3} + x - 1$ to within $0.125$ .

Solution.

Bracket: $f (0) = - 1 < 0$ and $f (1) = 1 > 0$ , so the Intermediate Value Theorem puts a root in $(0, 1)$ .
$f (0.5) = - 0.375 < 0$ , so the sign change moves to $(0.5, 1)$ ; $f (0.75) = 0.171875 > 0$ , so it moves to $(0.5, 0.75)$ ; $f (0.625) = - 0.130859375 < 0$ , so the root lies in $(0.625, 0.75)$ , of length $0.125$ .
Sanity check: the true root is $\approx 0.682328$ , indeed inside $(0.625, 0.75)$ . Each step halves the bracket, so the error after $n$ steps is at most $2^{- n}$ — about three steps per decimal digit. The method needs only sign changes and continuity, never a derivative.

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Exercises for §3.14 →

3.15Why continuity cannot be dropped

The Intermediate Value Theorem has two hypotheses — continuity and an interval domain — and each fails loudly when removed. This section isolates the first.

Example 3.55 (A jump skips every intermediate value). Define $f$ on $[- 1, 1]$ by $f (x) = - 1$ for $- 1 \leq x < 0$ and $f (x) = 1$ for $0 \leq x \leq 1$ . Show that no $c$ has $f (c) = 0$ , although $f (- 1) < 0 < f (1)$ .

Solution.

The endpoint values straddle $k = 0$ : $f (- 1) = - 1 < 0 < 1 = f (1)$ . But the image of $f$ is the two-point set ${- 1, 1}$ , so no value strictly between is attained and $f (c) \neq = 0$ for every $c$ .
Which hypothesis broke: $[- 1, 1]$ is a perfectly good interval, so it is continuity. The one-sided limits at $0$ are $- 1$ and $1$ , so $f$ jumps there and is discontinuous at $0$ and nowhere else.
Sanity check by exhaustion: $c < 0$ gives $f (c) = - 1 \neq = 0$ ; $c \geq 0$ gives $f (c) = 1 \neq = 0$ . Note also that ${- 1, 1}$ is not an interval, exactly the failure predicted by the corollary Continuous images of intervals are intervals read backwards.

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Remark (A function that misses everything in between). The Dirichlet function $D (x) = 1$ for $x \in Q$ and $D (x) = 0$ otherwise is discontinuous at every point: by density there are rationals $q_{n} \to a$ and irrationals $t_{n} \to a$ with $D (q_{n}) \to 1$ and $D (t_{n}) \to 0$ , and since $D (a)$ is one of $0, 1$ one of these contradicts the theorem Sequential characterisation of continuity. Its image on $[0, 1]$ is the two-point set ${0, 1}$ , so it attains $0$ and $1$ and nothing between.

Note what $D$ nevertheless does satisfy: it is bounded on a compact interval and it attains both a maximum and a minimum, so the conclusion of the Extreme Value Theorem happens to hold for it. Attainment of extrema and the intermediate value property are logically independent consequences of continuity, and a discontinuous function may satisfy either without the other.

Remark (The converse fails too). Having the intermediate value property does not imply continuity. The function $h (x) = sin (1/ x)$ with $h (0) = 0$ takes every value in $[- 1, 1]$ on every interval around $0$ , so it maps intervals onto intervals, yet it is discontinuous at $0$ by the example Classifying three discontinuities at the origin. A deeper example is Darboux's theorem: every derivative has the intermediate value property, even when the derivative is discontinuous — the function $x^{2} sin (1/ x)$ of Differentiation being the standard instance. So "takes all intermediate values" is strictly weaker than "continuous".

Exercises for §3.15 →

3.16The Extreme Value Theorem: statement

The second great theorem is about attainment. A continuous function on a closed bounded interval does not merely have a finite supremum; it actually reaches it.

Theorem 3.56 (Boundedness theorem). If $f$ is continuous on $[a, b]$ , then $f$ is bounded on $[a, b]$ .

Proof. Suppose $f$ were unbounded. Then for each $n \in N$ there is $x_{n} \in [a, b]$ with $∣ f (x_{n}) ∣ > n$ . The sequence $(x_{n})$ lies in the bounded set $[a, b]$ , so Bolzano–Weierstrass gives a convergent subsequence $x_{n_{k}} \to c$ , and $c \in [a, b]$ because $[a, b]$ is closed.

By the theorem Sequential characterisation of continuity, $f (x_{n_{k}}) \to f (c)$ . A convergent sequence is bounded, so ${∣ f (x_{n_{k}}) ∣}$ is bounded; but $∣ f (x_{n_{k}}) ∣ > n_{k} \geq k$ for every $k$ , which is unbounded. Contradiction.∎

Theorem 3.57 (Extreme Value Theorem). If $f$ is continuous on the closed bounded interval $[a, b]$ , then there exist $c, d \in [a, b]$ with

f (c) \leq f (x) \leq f (d) for all x \in [a, b] .

The theorem asserts two things that must be kept apart. That $sup f$ and $in f f$ are finite is the Boundedness theorem; that they are attained is the new content, and it is exactly the part that fails on an open interval. Combined with Continuous images of intervals are intervals this sharpens to: $f ([a, b])$ is precisely the closed bounded interval $[min f, max f]$ .

A concrete instance keeps the two claims apart. On $[- 1, 1]$ the function $x^{2}$ is continuous and the interval is closed and bounded, so extrema exist; and indeed $x^{2} \geq 0$ with equality at the interior point $0$ , while $∣ x ∣ \leq 1$ gives $x^{2} \leq 1$ with equality at both endpoints. So $min f = 0$ , $max f = 1$ , and the image is exactly $[0, 1]$ . Note that the maximum lives at the endpoints rather than at an interior critical point: searching only where the derivative vanishes would have found the minimum and missed the maximum entirely.

Exercises for §3.16 →

3.17Proving the Extreme Value Theorem from completeness

Proof. We prove that the supremum is attained; the infimum follows by applying the result to $- f$ , whose supremum is $- in f f$ .

By the theorem Boundedness theorem, the image $f ([a, b])$ is a bounded nonempty set of reals, so by the least-upper-bound property $M = sup f ([a, b])$ exists and is finite.

By the characterisation of the supremum — for each $n$ , the number $M - 1/ n$ is not an upper bound — there is $x_{n} \in [a, b]$ with

M - \frac{1}{n} < f (x_{n}) \leq M .

By Bolzano–Weierstrass the bounded sequence $(x_{n})$ has a convergent subsequence $x_{n_{k}} \to d$ , and $d \in [a, b]$ since $[a, b]$ is closed.

Two facts now collide. On one hand, the Squeeze Theorem applied to $M - 1/ n_{k} < f (x_{n_{k}}) \leq M$ gives $f (x_{n_{k}}) \to M$ . On the other, the theorem Sequential characterisation of continuity at $d$ gives $f (x_{n_{k}}) \to f (d)$ . By uniqueness of sequential limits, $f (d) = M$ , so the supremum is attained at $d \in [a, b]$ .∎

Three ingredients did the work and each is indispensable: the least-upper-bound property manufactured the target $M$ ; Bolzano–Weierstrass, itself a consequence of completeness, stopped the near-optimal points escaping; continuity carried the values along with the points. Remove completeness and even the first step fails — over $Q$ the continuous $f (x) = 1/ (x^{2} - 2)$ on $[0, 2] \cap Q$ is unbounded, its pole sitting at an irrational point the rational interval cannot see.

The closed end is not negotiable. On $(0, 1]$ the function $1/ x$ is continuous, since the denominator never vanishes, but $f (1/ n) = n$ makes it unbounded above, so it has no maximum — while its infimum $1$ is attained at $x = 1$ , showing that attainment is a separate question at each end. The step that fails is the appeal to closedness in Boundedness theorem: the escaping subsequence $1/ n$ converges to $0 \in / (0, 1]$ , so there is no point of the domain at which to invoke continuity. Adding $0$ does not help, because $1/ x$ cannot be defined there at all.

Exercises for §3.17 →

3.18Both hypotheses are necessary

The Extreme Value Theorem needs continuity and a domain that is both closed and bounded, so there are three ways to break it, and all three occur.

Continuity fails. On the compact interval $[0, 1]$ define $f (x) = x$ for $x < 1$ and $f (1) = 0$ . Then $f (x) < 1$ everywhere while values arbitrarily close to $1$ occur, so $sup f = 1$ is never attained. The domain is compact, so the only defective hypothesis is continuity, which fails at $1$ where the left limit is $1 \neq = 0 = f (1)$ .

Closedness fails. On $(0, 1]$ the continuous $f (x) = 1/ x$ has $f (1/ n) = n$ , so no maximum exists — the case discussed just above.

Boundedness fails. On $[0, \infty)$ , which is closed but unbounded, the continuous $f (x) = x$ satisfies $f (x + 1) > f (x)$ for every $x$ , so no value is maximal.

A fourth failure is the surprising one: a function can be continuous and bounded on a non-compact domain and still attain nothing.

Example 3.58 (Two ways to lose an extremum). (a) Show $f (x) = \frac{x}{1 + ∣ x ∣}$ on $R$ is continuous and bounded with $sup f = 1$ , $in f f = - 1$ , neither attained. (b) Show $g (x) = tan x$ on $(- \frac{π}{2}, \frac{π}{2})$ has neither a maximum nor a minimum.

Solution.

(a) $1 + ∣ x ∣ \geq 1 > 0$ everywhere and $∣ x ∣$ is continuous, so $f$ is a quotient of continuous functions with nonvanishing denominator. And $∣ f (x) ∣ = \frac{∣ x ∣}{1 + ∣ x ∣} < 1$ for every $x$ , so no value reaches $\pm 1$ .
The bounds are sharp: $f (n) = n / (1 + n) \to 1$ gives $sup f = 1$ , and by oddness $in f f = - 1$ . Numerically $f (10) \approx 0.909$ and $f (100) \approx 0.990$ , creeping toward $1$ .
Which hypothesis broke. $R$ is closed but unbounded, hence not compact. Boundedness of the function is irrelevant — the hypothesis is about the domain. The optimum has run away to $+ \infty$ , where there is no point to attain it at.
(b) $tan x = sin x / cos x$ with $cos x \neq = 0$ throughout the open interval, so the quotient clause of Algebra of continuous functions gives continuity.
Which hypothesis broke. The domain is bounded but not closed, so compactness fails. As $x \to (π /2)^{-}$ we get $sin x \to 1$ , $cos x \to 0^{+}$ and $tan x \to + \infty$ ; symmetrically $tan x \to - \infty$ at the other end. Numerically $tan (1.5) \approx 14.10$ and $tan (1.57) \approx 1255.77$ .
Unlike a removable case, the interval cannot simply be closed up: $tan$ has no finite limit at $\pm π /2$ , so no continuous extension exists. The defect is intrinsic to the function, not to the notation for its domain.

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Exercises for §3.18 →

3.19Compactness, in the end, is the whole story

Both great theorems used exactly the same two facts about $[a, b]$ : every sequence in it has a convergent subsequence, and the limit stays inside. That is sequential compactness, and restating the theorems in those terms both generalises and explains them.

Theorem 3.59 (Continuous images of compact sets are compact). Let $K \subseteq R$ be closed and bounded and $f : K \to R$ continuous. Then $f (K)$ is closed and bounded.

Proof. Bounded. If not, there are $x_{n} \in K$ with $∣ f (x_{n}) ∣ > n$ . Bolzano–Weierstrass gives $x_{n_{k}} \to c$ with $c \in K$ by closedness, and Sequential characterisation of continuity gives $f (x_{n_{k}}) \to f (c)$ , so that subsequence of images is bounded — contradicting $∣ f (x_{n_{k}}) ∣ > n_{k}$ .

Closed. Let $y$ be a limit of points of $f (K)$ , say $y_{n} = f (x_{n}) \to y$ with $x_{n} \in K$ . Bolzano–Weierstrass and closedness give a subsequence $x_{n_{k}} \to c \in K$ , and continuity gives $f (x_{n_{k}}) \to f (c)$ . But $f (x_{n_{k}})$ is a subsequence of the convergent sequence $(y_{n})$ , hence converges to $y$ . By uniqueness of limits, $y = f (c) \in f (K)$ . So $f (K)$ contains all its limit points.∎

Corollary 3.60 (Extreme Value Theorem on a compact set). A continuous real-valued function on a nonempty closed bounded set $K$ attains a maximum and a minimum on $K$ .

Proof. By the theorem Continuous images of compact sets are compact, $f (K)$ is nonempty, closed and bounded. Being bounded and nonempty, it has a supremum $M$ ; being closed, it contains $M$ , because $M$ is either a member of $f (K)$ already or a limit of members, the characterisation of the supremum supplying points of $f (K)$ within $1/ n$ of $M$ for every $n$ . Hence $M \in f (K)$ , so $M = f (d)$ for some $d \in K$ , and $M$ is a maximum. The minimum is the same argument with the infimum, or an application to $- f$ .∎

This is the form the theorem takes in a general metric space, where by Characterizations of compactness in a metric space the sequential and open-cover definitions agree, and in $R^{n}$ the theorem Heine-Borel identifies the compact sets as exactly the closed bounded ones. The interval structure was never used above — only compactness — which is why the statement applies equally to a Cantor set or a finite union of intervals.

A one-line application shows the gain. The unit circle $S^{1} = {(x, y) : x^{2} + y^{2} = 1}$ is bounded, lying inside the disc of radius $1$ , and closed, being the preimage of ${1}$ under the continuous map $(x, y) \mapsto x^{2} + y^{2}$ ; by Heine-Borel it is compact. The height $h (x, y) = y$ is a coordinate projection, hence continuous, so by the corollary Extreme Value Theorem on a compact set it attains a maximum and a minimum — namely $1$ at $(0, 1)$ and $- 1$ at $(0, - 1)$ , as the parametrisation $h = sin θ$ confirms. Note that $S^{1}$ is not an interval, so only the compactness formulation makes the application legal.

Summary. The $ε$ - $δ$ definition makes "approaches" precise; the two sequential characterisations convert it into statements about sequences, which is how nearly every later proof gets leverage. Continuity is the agreement of limit and value, preserved by the algebra and by composition — which is why explicit $ε$ - $δ$ work is almost never needed in practice. Uniform continuity moves one quantifier, and Heine–Cantor says compactness restores it for free. The two great theorems split by which property of $R$ they use: the Intermediate Value Theorem is the least-upper-bound property, the Extreme Value Theorem is compactness.

Exercises for §3.19 →

Common Mistakes

Choosing a $δ$ that depends on $x$ . The order is $\forall ε \exists δ \forall x$ , so $δ$ is fixed before $x$ is named. An answer of the form $δ = ε /∣ x + 3∣$ has proved nothing; bound the offending factor by a constant on a pre-restricted neighbourhood instead.
Reversing the quantifiers. "There is $δ$ such that for all $ε$ ... " says the function is locally constant, not that it has a limit.
Requiring $f (a)$ to exist or to equal the limit. The definition inspects only the punctured neighbourhood: $lim_{x \to 1} (x^{2} - 1) / (x - 1) = 2$ although the function is undefined there.
Treating "the limit is $+ \infty$ " as a kind of convergence. It is a specific mode of divergence; the limit laws do not apply to it.
Using the limit laws without checking the component limits exist. $lim (f + g)$ can exist when neither $lim f$ nor $lim g$ does, so the laws run one way only.
Assuming continuity at a point says something about a neighbourhood. The function equal to $x$ on $Q$ and $0$ elsewhere is continuous at exactly one point.
Confusing continuity with uniform continuity. Pointwise continuity lets $δ$ shrink as the base point moves; $x^{2}$ on $R$ and $1/ x$ on $(0, 1]$ are continuous and not uniformly so.
Believing uniform continuity requires a bounded derivative. That is the Lipschitz condition, strictly stronger: $x$ on $[0, 1]$ is uniformly continuous with unbounded difference quotients at $0$ .
Applying Heine–Cantor to a bounded but non-closed domain. Both halves of compactness are used in the proof, and $1/ x$ on $(0, 1]$ is the standard counterexample.
Dropping a hypothesis from the two great theorems. A step function skips intermediate values; $x$ on $(0, 1)$ attains no extremum. In each case exactly one hypothesis is missing and the counterexample is one line.
Thinking the intermediate value property implies continuity. It does not: $sin (1/ x)$ with value $0$ at the origin has the property and is discontinuous, and by Darboux's theorem every derivative has it.
Confusing attainment of extrema with the existence of critical points. The theorem says a maximum exists; locating it may require checking endpoints, where the derivative need not vanish, as $x^{2}$ on $[- 1, 1]$ shows.

About the practice questions. They check that you can carry out this chapter's computations correctly, and each one is graded on a single answer. They are not proof exercises: working through them confirms the mechanics, not that you could prove the results yourself. For that, re-read the statements above and try to reconstruct their proofs with the page closed.

Practice this chapter

Introduction

3.1Approaching a point: one-sided limits

3.2The precise ε -δ definition

3.3Infinite limits, limits at infinity, and the laws in outline

3.4The definition in full, and what it forces

3.5Strategies for proving a limit from the definition

3.6Proving the limit laws

3.7Continuity at a point

3.8Continuity on a set, and how continuous functions are built

3.9The sequential characterisation, and a taxonomy of discontinuities

3.10Uniform continuity: moving the quantifier

3.11Heine–Cantor: compactness buys uniformity

3.12The Cauchy criterion and continuous extension

3.13The Intermediate Value Theorem

3.14Consequences of the Intermediate Value Theorem

3.15Why continuity cannot be dropped

3.16The Extreme Value Theorem: statement

3.17Proving the Extreme Value Theorem from completeness

3.18Both hypotheses are necessary

3.19Compactness, in the end, is the whole story

3.2The precise $ε$ - $δ$ definition