Continuity
f(x) = x³ − x − 1 is −1 at x = 1 and 5 at x = 2. Because it has no breaks, it must cross zero somewhere in between.
Three conditions
Informally, a function is continuous if you can draw its graph without lifting the pencil. That picture is useful, but calculus needs something you can check. OpenStax gives three conditions for f to be continuous at a point a. First, f(a) is defined. Second, limx→a f(x) exists. Third, the two are equal: limx→a f(x) = f(a). If any condition fails, f is discontinuous at a. A function is continuous on an interval if it is continuous at every point of it.
Check them in order and stop at the first failure. For (x² − 4)/(x − 2) at 2, the first condition fails because f(2) is undefined. For the function equal to x + 2 except f(2) = 1, the first two conditions hold (f(2) = 1, limit 4) but the third fails. For x² at 1, all three hold: f(1) = 1, the limit is 1, and they match. Polynomials are continuous everywhere, and rational functions are continuous wherever their denominators are not zero; this follows directly from the limit laws.
Kinds of discontinuity
Discontinuities come in three kinds, matching the three ways a limit can behave. A removable discontinuity has a limit, but the function is either undefined there or defined with the wrong value. A single point can fix it: redefine f(a) as the limit. A jump discontinuity has both one-sided limits, but they are different numbers, so the graph steps up or down. An infinite discontinuity has at least one one-sided limit equal to ±∞, which means a vertical asymptote. Rational functions show the first and third kinds; piecewise functions often show jumps.
Piecewise functions raise a common exercise: choose a constant so that the pieces meet. Suppose f(x) = x + k for x < 1 and f(x) = 2x² for x ≥ 1. The right piece gives f(1) = 2 and right limit 2. The left limit is 1 + k. For continuity at 1 they must match, so 1 + k = 2 and k = 1. Any other k leaves a jump at x = 1.
The Intermediate Value Theorem
Continuity has a powerful consequence. The Intermediate Value Theorem says: if f is continuous on a closed interval [a, b] and z is any number between f(a) and f(b), then there is at least one c in [a, b] with f(c) = z. A continuous function cannot skip values. Take f(x) = x³ − x − 1: f(1) = −1 and f(2) = 5, so f takes the value 0 somewhere between 1 and 2. The theorem says a root exists without telling you where. Halving the interval repeatedly finds it: f(1.5) = 0.875 > 0, so the root is in [1, 1.5]; f(1.25) ≈ −0.297 < 0, so it is in [1.25, 1.5], and so on. The root is about 1.3247.
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- Which condition fails for (x² − 4)/(x − 2) at x = 2?
- f(2) is not defined
- The limit does not exist
- The limit is not 4
- None; it is continuous
- What kind of discontinuity does (x² − 4)/(x − 2) have at 2?
- Removable
- Jump
- Infinite
- None
- What kind of discontinuity does |x − 2|/(x − 2) have at 2?
- Jump
- Removable
- Infinite
- None
- What kind of discontinuity does 1/(x − 2)² have at 2?
- Infinite
- Jump
- Removable
- None
- f(x) = x + k for x < 1 and 2x² for x ≥ 1. Which k makes f continuous at 1?
- f(x) = x³ − x − 1. What is f(1.5)?
- f(1) = −1 and f(1.5) = 0.875. Where does the IVT guarantee a root?
- Between 1 and 1.5
- Between 1.5 and 2
- At exactly 1.25
- Nowhere
- f(x) = x² − 2 has a root between 1 and 2. Bisect twice: f(1.5) = 0.25 > 0, f(1.25) < 0. What is the midpoint of the interval left after these two steps?
- Why does 1/x not contradict the IVT on [−1, 1]?
- It is not continuous on [−1, 1]
- It does take the value 0
- The IVT only applies to polynomials
- f(−1) and f(1) are equal
- Where is a rational function continuous?
- Wherever its denominator is not zero
- Everywhere
- Only at integers
- Only where it is positive
- Continuous at a: f(a) defined, the limit exists, and they are equal.
- Removable (hole or misplaced point), jump (one-sided limits differ), infinite (asymptote).
- Polynomials are continuous everywhere; rational functions wherever the denominator is not 0.
- IVT: on [a, b], a continuous f takes every value between f(a) and f(b).
Sources for this course (5)
- OpenStax, Calculus Volume 1 , chapter 2 Limits: 2.1 A Preview of Calculus (tangent problem and area problem; secant slopes through (1, 1) on f(x) = x² with (2, 4) giving 3 and (3/2, 9/4) giving 2.5; rock s(t) = −16t² + 64 with average velocities −15.84 and −16.16 near t = 0.5; three rectangles under x² + 1 on [0, 3] giving 8: openstax.org/books/calculus-volume-1/pages/2-1-a-preview-of-calculus)
- 2.2 The Limit of a Function ((x² − 4)/(x − 2) → 4 at 2, (sin x)/x → 1 at 0 from a table, sin(1/x) has no limit at 0, one-sided limits of |x − 2|/(x − 2), infinite limits and vertical asymptotes, the formal definition appearing only in the late 19th century: .../pages/2-2-the-limit-of-a-function)
- 2.3 The Limit Laws (sum, difference, constant multiple, product, quotient, power and root laws; direct substitution; factoring and rationalizing 0/0 forms; squeeze theorem: .../pages/2-3-the-limit-laws)
- 2.4 Continuity (three conditions; removable, jump and infinite discontinuities; Intermediate Value Theorem: .../pages/2-4-continuity)
- 2.5 The Precise Definition of a Limit (.../pages/2-5-the-precise-definition-of-a-limit).
Checked: 2026-10-17