Lecture 5 · Barnett 2-3 · about 35 min

Continuity

The three-condition test for continuity at a point, the three types of discontinuity, finding the constant that makes a piecewise function continuous, solving inequalities with a sign chart, and the Intermediate Value Theorem.

By the end you can

Check continuity at a point with the three conditions and write the conclusion
Classify a discontinuity as removable, jump or infinite, from a formula or a graph
Find one or two constants that make a piecewise function continuous
Solve polynomial and rational inequalities with a sign chart
Use the Intermediate Value Theorem to show a root exists in an interval

Key idea · Exam recipe (memorise this)

Continuity at $a$ : write all three lines. (1) $f(a) = \dots$ (defined?). (2) $\lim_{x\to a^-} f = \dots$ , $\lim_{x\to a^+} f = \dots$ , so the limit is $\dots$ (exists?). (3) limit $= f(a)$ ? Conclude: “continuous at $a$ ” or “not continuous: condition $n$ fails”.
Where is $f$ continuous? Polynomials: everywhere. Rational: everywhere except denominator zeros. $\sqrt{u}$ : where $u \ge 0$ . $\ln u$ : where $u \gt 0$ . Piecewise: each piece on its interval, then check every boundary.
Find $k$ : value of the left piece at the boundary = value of the right piece at the boundary. Solve. (Two constants: two boundaries, two equations.)
Classify: limit exists but $\ne f(a)$ or $f(a)$ undefined → removable; one-sided limits finite and different → jump; a one-sided limit is $\pm\infty$ → infinite.
Inequality $f(x) \gt 0$ : factor, partition numbers (zeros of top and bottom), sign chart, answer in intervals; denominator zeros always open.
IVT: $f$ continuous on $[a, b]$ and $f(a)$ , $f(b)$ of opposite signs → a root in $(a, b)$ .

How the exam asks it: three core exam templates

In midterm exams, continuity questions focus on three canonical structures: writing out the full three-condition proof at a boundary or point, identifying and classifying discontinuities (removable, jump, infinite), and enforcing continuity across piecewise tax/pricing schedules or using sign charts and the Intermediate Value Theorem. Master these three worked solutions, then follow the cheat-sheet and 4-step method below.

Exam template 1: Testing continuity at a point with the three-condition test

Question: Let $f(x)$ be defined by: $f(x) = \begin{cases} x^2 & \text{if } x \lt 1 \\ 2x - 1 & \text{if } x \ge 1 \end{cases}$ Separately, consider $g(x) = \dfrac{x^2 - 9}{x - 3}$ for $x \ne 3$ with $g(3) = 5$ .
a) State the three formal conditions required for a function to be continuous at $x = a$ .
b) For $f(x)$ at $x = 1$ , evaluate condition (1), condition (2), and condition (3), and conclude whether $f(x)$ is continuous at $x = 1$ .
c) For $g(x)$ at $x = 3$ , evaluate condition (1), condition (2), and condition (3). State which condition fails and classify the discontinuity.

Exam template 2: Identifying and classifying discontinuities from algebraic formulas

Question: Find and classify all discontinuities of the rational function $f(x) = \dfrac{x^2 - 1}{x^2 - x}$ .

Exam template 3: Solving rational inequalities using partition numbers and sign charts

Question: Solve the inequality $\dfrac{x^2 - x - 12}{x - 1} \gt 0$ , writing the solution in interval notation.

Cheat-sheet & the 4-step method

· Cheat-sheet: continuity criteria, discontinuity types, and sign chart inequalities

Concept / Task	Mathematical Criteria	When to use it	How to write the exam answer
Continuity at a point	(1) $f(a)$ defined; (2) $\lim_{x\to a} f(x)$ exists; (3) $\lim = f(a)$	Checking continuity of piecewise rule or point	Write all three numbered steps explicitly, then conclude continuous/discontinuous
Removable discontinuity (hole)	Factor cancels; $\lim_{x\to a} f(x)$ exists but $f(a)$ undefined	Denominator zero cancels with numerator factor	“Removable discontinuity at $(a, L)$ ”
Removable discontinuity (moved point)	$\lim_{x\to a} f(x) = L \ne f(a)$	Point defined at different height than limit	“Removable discontinuity: condition 3 fails ( $L \ne f(a)$ )”
Jump discontinuity	$\lim_{x\to a^-} f(x) \ne \lim_{x\to a^+} f(x)$	Piecewise formulas do not meet at boundary	“Jump discontinuity: condition 2 fails (one-sided limits differ)”
Infinite discontinuity	Form $\dfrac{c}{0}$ ( $c \ne 0$ ), one-sided limits are $\pm\infty$	Denominator zero does not cancel	“Infinite discontinuity: vertical asymptote $x = a$ ”
Piecewise parameter $k$	Left piece at boundary = Right piece at boundary	Finding parameter to ensure continuity	Set two formulas equal at $x = a$ , solve for $k$
Sign chart inequality	Factor top & bottom $\implies$ partition numbers	Solving $f(x) \gt 0$ or $f(x) \le 0$	Partition line into intervals, test one point each, write interval union
Intermediate Value Theorem	$f$ continuous on $[a, b]$ , $f(a) \cdot f(b) \lt 0$	Proving existence of root or break-even	State continuity, show opposite signs, conclude root exists in $(a, b)$

Key idea · The 4-step method for any exam question on continuity

Step 1: Check domain restrictions and boundary points.
- Locate any denominator zeros, roots, or piecewise boundary points $x = a$ .
Step 2: Evaluate the three conditions systematically.
- Condition 1: Evaluate $f(a)$ using the piece with $\le$ or $\ge$ .
- Condition 2: Evaluate left-hand limit $\lim_{x\to a^-}$ and right-hand limit $\lim_{x\to a^+}$ .
- Condition 3: Compare $\lim_{x\to a} f(x)$ with $f(a)$ .
Step 3: Classify any failure accurately.
- If limits differ $\implies$ jump.
- If limit is infinite $\implies$ infinite (vertical asymptote).
- If limit exists but $f(a)$ is missing or mismatched $\implies$ removable.
Step 4: State the final bold exam answer.
- For parameter problems: state $k$ with units.
- For inequalities: state interval notation with open/closed brackets matched to $\lt$ , $\le$ , and denominator restrictions.

The three conditions

A function is continuous at $x = a$ when you can draw through that point without lifting the pen. Formally, three conditions:

Continuity at a pointDefinitionFormula sheet →

f \text{ is continuous at } a \iff \begin{cases} 1.\ f(a) \text{ is defined} \\ 2.\ \lim_{x \to a} f(x) \text{ exists} \\ 3.\ \lim_{x \to a} f(x) = f(a) \end{cases}

$f(a)$: the function's value at a
$\lim_{x \to a} f(x)$: what f approaches near a

Use it when you must show (dis)continuity, especially at the boundary of a piecewise function.

Type of discontinuity	What you see	Condition that fails
Removable (hole)	open dot, no value	1: $f(a)$ undefined
Removable (moved point)	open dot plus a closed dot elsewhere	3: limit $\ne f(a)$
Jump	two heights	2: one-sided limits differ
Infinite	vertical asymptote	2 (and usually 1)

Where functions are continuous

Continuity of common functionsFormula sheet →

\text{polynomial: } (-\infty, \infty); \quad \frac{p(x)}{q(x)}: q(x) \neq 0; \quad \sqrt{u(x)}: u(x) \ge 0; \quad \ln u(x): u(x) > 0

$p, q$: polynomials
$u(x)$: the expression inside the root or the log

Use it when you must give the intervals on which a function is continuous.

Classifying discontinuities from a formula

Making a piecewise function continuous

Solving inequalities with a sign chart

A continuous function can only change sign where it is 0 or undefined. So the sign is constant between the partition numbers.

The Intermediate Value Theorem

Intermediate Value Theorem (root form)TheoremFormula sheet →

f \text{ continuous on } [a,b],\ f(a) \cdot f(b) < 0 \;\Rightarrow\; \exists\, c \in (a,b): f(c) = 0

$a, b$: endpoints of a closed interval
$c$: a point strictly between a and b

Use it when you must show a root (break-even point, equilibrium) exists without finding it.

Exam practice

Summary and review

Continuous at $a$ : $f(a)$ defined, limit exists, limit $= f(a)$ . Write all three.
Removable: limit exists (fix by redefining). Jump: one-sided limits differ. Infinite: asymptote.
Cancelled denominator factor → hole; surviving factor → vertical asymptote.
$k$ : set the pieces equal at each boundary.
Sign chart: partition numbers from top and bottom, one sign per interval, denominator zeros open.
IVT: continuous plus a sign change guarantees a root in between.