Lecture 5 · Barnett ch. 3 · about 35 min

Continuity

The formal test for an unbroken curve, how to find the constant that patches a piecewise function together, and why continuity guarantees a break-even point exists somewhere in between.

By the end you can

State and apply the three-part definition of continuity at a point
Classify a discontinuity as removable, jump, or infinite
Find a parameter that makes a piecewise function continuous
Use the Intermediate Value Theorem to argue that a solution exists in an interval

The three-part definition

A function is continuous at $x = a$ when you can draw its graph through that point without lifting your pen. That informal picture hides three separate conditions, and an exam question that says “show $f$ is continuous” wants all three checked.

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 actual value at a
$the limit$: what f approaches near a

Use it when you must verify continuity rigorously, not just glance at a picture, especially at the seam of a piecewise function.

Each of last lesson’s broken cases fails exactly one condition:

Type of discontinuity	Which condition fails
Removable (hole), e.g. $\frac{x^2-4}{x-2}$ at $x=2$	Condition 1: $f(a)$ is undefined
Jump	Condition 2: the limit does not exist (one-sided limits disagree)
Infinite (vertical asymptote)	Condition 2: the limit does not exist (grows without bound)
Hole with a relocated point, e.g. $f(2)$ redefined to 9 when the limit is 4	Condition 3: the limit exists and $f(a)$ is defined, but they disagree

Patching a piecewise function

The most common exam task built on continuity is finding an unknown constant that makes two pieces of a function meet without a jump.

Where functions are continuous

Polynomials are continuous everywhere. Rational functions are continuous everywhere except where the denominator is zero. Piecewise functions are continuous everywhere each piece is continuous, plus wherever the pieces have been checked to meet.

Continuous everywhereFormula sheet →

\text{Polynomial: continuous on } (-\infty, \infty) \qquad \frac{p(x)}{q(x)}: \text{ continuous wherever } q(x) \neq 0

$p(x), q(x)$: polynomials

Use it when you need to state the interval(s) on which a function is continuous, as a first step before applying the derivative rules of later lessons, which require continuity.

The Intermediate Value Theorem

Continuity has a powerful consequence: a continuous function cannot skip over a value. If it is negative at one point and positive at another, it must cross zero somewhere in between. This is exactly the logic behind guaranteeing that a break-even point or a market-clearing price exists, even before you solve for it exactly.

Intermediate Value TheoremTheoremFormula sheet →

f \text{ continuous on } [a,b],\ f(a) \text{ and } f(b) \text{ have opposite signs} \;\Rightarrow\; \exists\, c \in (a,b): f(c) = 0

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

Use it when a question asks you to argue a root or an equilibrium exists, without necessarily finding its exact value.

Exam practice

Summary and review

Continuity at $a$ : $f(a)$ defined, $\lim_{x\to a} f(x)$ exists, and they are equal.
Removable discontinuity fails condition 1 or 3; jump and infinite discontinuities fail condition 2.
To patch a piecewise function, set the two pieces equal at the seam and solve for the unknown constant.
Polynomials are continuous everywhere; rational functions fail only where the denominator is zero.
The Intermediate Value Theorem: continuous on $[a,b]$ with opposite-sign endpoints guarantees a root somewhere in between.