Difficulty: Introductory | Prerequisites: Limits (Sec. 1.4 – 1.6), including one-sided limits and limits at infinity.
Continuity is the formal way of saying a function has no breaks, jumps, or holes. It builds directly on limits: once you know how to evaluate a limit, you can check whether a function behaves "nicely" at a point. This section also introduces the Intermediate Value Theorem (IVT), which lets you prove that an equation has a solution even when you cannot solve it by hand. If you are comfortable with one-sided limits and evaluating limits of rational and trigonometric functions, you have everything you need here.
A function is continuous at a point when the limit equals the function value, with no gaps. Three things can go wrong (hole, jump, vertical asymptote), and each gives a named type of discontinuity. The Intermediate Value Theorem says a continuous function on a closed interval must hit every y-value between its endpoints, which is how you prove roots exist without algebra.
Continuity at a point
A function f is continuous at x = a when three conditions hold: f(a) is defined, lim x→a f(x) exists, and lim x→a f(x) = f(a). In simple terms, you can draw through the point without lifting your pen.
Removable discontinuity
A point where the limit exists but either f(a) is not defined or f(a) does not equal the limit. Think of it as a single hole in the graph that you could "fill in" by redefining f(a).
Jump discontinuity
A point where the left-hand limit and right-hand limit both exist but are not equal. The graph makes a visible jump from one y-value to another.
Infinite discontinuity
A point where the function heads to positive or negative infinity (a vertical asymptote). The limit from at least one side does not exist as a finite number.
Intermediate Value Theorem (IVT)
If f is continuous on [a, b] and N is any number between f(a) and f(b), then there exists at least one c in (a, b) such that f(c) = N. In simple terms, a continuous function cannot skip over a value; if it starts below N and ends above N, it must pass through N somewhere in between.
Piecewise function
A function defined by different formulas on different intervals. To check continuity at a boundary point, you match the left-hand and right-hand pieces so they agree at the join.
For f to be continuous at x = a, check in order:
f(a) is defined (the point exists on the graph).
lim x→a f(x) exists (the left-hand and right-hand limits agree).
lim x→a f(x) = f(a) (the limit matches the function value).
If any one of these fails, f is discontinuous at a. Which condition fails tells you the type of discontinuity.
Removable: The limit exists, but f(a) is missing or wrong. Graphically, a single open circle (hole). Example: f(x) = (x^2 - 1)/(x - 1) at x = 1.
Jump: Left-hand limit ≠ right-hand limit. The graph has a visible step. Example: the floor function at any integer.
Infinite: The function blows up to ±∞ on at least one side (a vertical asymptote). Example: f(x) = 1/x at x = 0.
Consider f(x) = 1/(1 + sin x) on [0, 2π].
The denominator is 1 + sin x. This equals zero when sin x = -1, which happens at x = 3π/2.
At x = 3π/2 the function is undefined, so f is discontinuous there (infinite discontinuity, vertical asymptote).
Everywhere else on [0, 2π] the denominator is positive, and f is a ratio of continuous functions with nonzero denominator, so f is continuous on [0, 3π/2) ∪ (3π/2, 2π].
Given f(x) = { x^2 - 10 for x ≤ c, 8x - 26 for x > c }, find c so f is continuous everywhere.
Each piece is a polynomial, so f is continuous except possibly at x = c.
At x = c, require left-hand value = right-hand limit: c^2 - 10 = 8c - 26.
Solve: c^2 - 8c + 16 = 0, so (c - 4)^2 = 0, giving c = 4.
Statement: If f is continuous on [a, b] and N is any number strictly between f(a) and f(b), then there is at least one c in (a, b) with f(c) = N.
Using IVT to prove a root exists: Show that f is continuous on an interval and that f changes sign (one endpoint positive, the other negative). Then IVT guarantees at least one c where f(c) = 0.
Tree-height example: Let h(t) = height_A(t) - height_B(t). At t = 1, h(1) = 2 - 3 = -1. At t = 2, h(2) = 7 - 5 = 2. Since h is continuous (tree heights change continuously) and h changes sign on [1, 2], IVT gives a time c in (1, 2) where h(c) = 0, meaning the trees are the same height.
Broken-clock example: Let g(t) = (position of stopped clock) - (actual time displayed on a working clock). A stopped clock shows one fixed time. Over a 12-hour cycle the actual time sweeps through every position once. At the moment the stopped clock happens to be correct, g = 0. Twelve hours later the actual time returns to the same position, so g = 0 again. Because the actual time moves continuously and the stopped clock is constant, g is continuous and changes sign between those zeros, guaranteeing two crossings per day.
For f(x) = x^2 - 4/x + 1, test each interval:
(a) [-2, -1]: f(-2) = 4 + 2 + 1 = 7, f(-1) = 1 + 4 + 1 = 6. Both positive, no sign change.
(b) [-1, 1]: f is not continuous here because 4/x is undefined at x = 0. IVT does not apply.
(c) [0, 1]: f is undefined at x = 0. IVT does not apply.
(d) [1, 2]: f(1) = 1 - 4 + 1 = -2. f(2) = 4 - 2 + 1 = 3. Opposite signs and f is continuous on [1, 2]. By IVT, there is a root in [1, 2]. Answer: (d).
Three-part continuity test at x = a:
f(a) is defined.
lim x→a f(x) exists.
lim x→a f(x) = f(a).
Intermediate Value Theorem: If f is continuous on [a, b] and N is between f(a) and f(b), then there exists c ∈ (a, b) with f(c) = N.
Piecewise continuity condition: For f(x) = { g(x) for x ≤ c, h(x) for x > c }, set g(c) = lim x→c+ h(x), i.e. solve g(c) = h(c) for c.
The IVT is the mathematical reason you can be sure that, as you drive from sea level to a mountain pass, your altitude hits every height in between. Engineers rely on IVT-style reasoning when they know a system's output changes continuously and they need to guarantee a specific value is reached (temperature in a heat exchanger, pressure in a pipeline).
Students often think "the limit exists" is enough for continuity. It is not. The function also has to be defined at the point, and the value has to match the limit.
Students confuse removable and jump discontinuities. If the limit exists (both sides agree) but the point is missing or wrong, that is removable. If the two sides disagree, that is a jump.
Students sometimes apply IVT on an interval where the function is not continuous (e.g., across a vertical asymptote). IVT requires continuity on the entire closed interval; if the function is undefined or discontinuous anywhere inside, the theorem does not apply.
When checking piecewise continuity, students forget to verify the boundary point itself. Both pieces must give the same output at x = c, not just near it.
⚠️ Expect at least one problem where you identify the type of discontinuity from a graph or formula. Know all three types cold.
⚠️ Piecewise continuity problems (finding c, or filling blanks to make a function continuous) appear regularly. The technique is always the same: set the two pieces equal at the boundary and solve.
⚠️ IVT questions usually ask you to show a root exists on a given interval. The checklist: confirm continuity on [a, b], evaluate f(a) and f(b), confirm a sign change, cite IVT.
⚠️ A common trap: an interval that contains a point where f is undefined. If the function is not continuous on the entire interval, IVT does not apply. Always check the domain first.
True or false: If lim x→3 f(x) = 5 and f(3) = 5, then f is continuous at x = 3. (True.)
True or false: A removable discontinuity means the left-hand and right-hand limits disagree. (False. That is a jump discontinuity. A removable discontinuity means the limit exists but the function value is missing or does not match.)
Fill in the blank: The IVT requires f to be ____ on the closed interval [a, b]. (Continuous.)
True or false: IVT can be used on f(x) = 1/x over [-1, 1] because f(-1) = -1 and f(1) = 1 have opposite signs. (False. f is not continuous on [-1, 1] because it is undefined at x = 0.)
Fill in the blank: For f(x) = 1/(1 + sin x), the function is discontinuous on [0, 2π] at x = ____. (3π/2.)
Q: Where is f(x) = 1/(1 + sin x) discontinuous on [0, 2π], and what type of discontinuity is it?
A: At x = 3π/2. The denominator is zero there, so f has an infinite discontinuity (vertical asymptote).
Q: Find c so that f(x) = { x^2 - 10 for x ≤ c, 8x - 26 for x > c } is continuous everywhere.
A: Set the pieces equal at x = c: c^2 - 10 = 8c - 26. This gives c^2 - 8c + 16 = 0, so (c - 4)^2 = 0 and c = 4.
Q: For f(x) = { x^2 - 17 for x ≤ □, □x - 29 for x > □ }, fill the blanks with digits 1–9 so f is continuous everywhere. How many solutions are there?
A: Let the boundary be c and the coefficient be m. Then c^2 - 17 = mc - 29, so c^2 - mc + 12 = 0, giving c = (m ± √(m^2 - 48)) / 2. Both c and m must be digits 1–9 and c must be a positive integer. Testing: m = 8 gives c^2 - 8c + 12 = 0, so c = 2 or c = 6. Both are digits 1–9. m = 7 gives c^2 - 7c + 12 = 0, so c = 3 or c = 4. Both valid. That yields four solutions: (c = 2, m = 8), (c = 6, m = 8), (c = 3, m = 7), (c = 4, m = 7).
Q: Two trees: at year 1, heights are 2 ft and 3 ft; at year 2, heights are 7 ft and 5 ft. Show they were the same height at some moment.
A: Let h(t) = height_A(t) - height_B(t). h(1) = -1 < 0 and h(2) = 2 > 0. Tree height is a continuous function of time, so h is continuous. By IVT, there exists c in (1, 2) with h(c) = 0.
Q: On which interval does IVT guarantee a root of f(x) = x^2 - 4/x + 1? Options: [-2,-1], [-1,1], [0,1], [1,2].
A: [1, 2]. f(1) = -2 and f(2) = 3; f is continuous on [1, 2] and changes sign, so IVT guarantees a root there. The other intervals either contain x = 0 (where f is undefined) or have no sign change.
Continuity is a prerequisite for the derivative (Sec. 2.1). A function must be continuous at a point to be differentiable there, though the converse is not true (continuous does not guarantee differentiable). The IVT will reappear when you study the Mean Value Theorem and later in integral calculus, where continuous functions are guaranteed to be integrable.
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