Difficulty: Introductory | Prerequisites: understanding of limits (see Limits study notes).
Continuity is the bridge between limits and everything else in calculus. A function is continuous at a point when its limit equals its actual value there; when that fails, you get a discontinuity, and the type of failure tells you what kind. The Intermediate Value Theorem (IVT) is the first major theorem that depends on continuity: it guarantees that a continuous function on a closed interval hits every y-value between its endpoints. This section sits right after limits and right before derivatives in the MTH 132 sequence.
A function is continuous at a point when three things hold: the function is defined there, the limit exists, and the limit equals the function value. When any of these fails, you have a discontinuity (removable, jump, or infinite). The IVT says a continuous function on [a, b] must take on every value between f(a) and f(b), which is how you prove a root exists in an interval.
Continuity at a point
A function f is continuous at x = c if and only if: (1) f(c) is defined, (2) lim(x→c) f(x) exists, and (3) lim(x→c) f(x) = f(c).
In simple terms, you can draw the graph through that point without lifting your pen.
Removable discontinuity (hole)
Occurs when the limit exists at x = c but either f(c) is undefined or f(c) does not equal the limit. The graph has a "hole" that could be filled by redefining f(c).
Think of it as: the function almost works at that point, just the value is missing or wrong.
Jump discontinuity
Occurs when both the left-hand and right-hand limits exist at x = c but they are not equal. The graph has a visible jump.
In simple terms, the function approaches different values from the left and the right.
Infinite discontinuity
Occurs when the function grows without bound (to +∞ or −∞) as x approaches c. The graph has a vertical asymptote there.
Intermediate Value Theorem (IVT)
If f is continuous on the closed interval [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 the output is negative at one end and positive at the other, the function must cross zero somewhere in between.
Removable (hole): The limit exists but does not equal f(c). Example: f(x) = (x² − 1)/(x − 1) at x = 1. The function simplifies to x + 1 for x ≠ 1, so the limit is 2, but f(1) is undefined. The "hole" at (1, 2) can be filled by defining f(1) = 2.
Jump: The left-hand and right-hand limits both exist but differ. Example: a piecewise function that equals 1 for x < 0 and equals 3 for x ≥ 0 has a jump at x = 0.
Infinite: The function heads to ±∞ as x approaches the point. Example: f(x) = 1/(x − 1) at x = 1. The graph has a vertical asymptote there.
A function is continuous on an interval if it is continuous at every point in that interval.
f(x) = |x| is continuous everywhere on (−∞, ∞). It has a corner at x = 0, but no break in the graph.
f(x) = tan(x) is not continuous on all of (−∞, ∞) because it has vertical asymptotes at x = π/2 + nπ. However, it is continuous on its domain.
f(x) = √x has domain [0, ∞) and is continuous there, but not defined for x < 0.
f(x) = 1/x is continuous on its domain, (−∞, 0) ∪ (0, ∞), but not at x = 0.
Statement: if f is continuous on [a, b] and N is any number strictly between f(a) and f(b), there is at least one c in (a, b) with f(c) = N.
How to use it on exams:
Confirm f is continuous on [a, b] (state why: polynomials are continuous everywhere, compositions of continuous functions are continuous, etc.).
Compute f(a) and f(b).
Check that N lies between f(a) and f(b).
Conclude that a c exists in (a, b) with f(c) = N.
Example: f(x) = x³ − x − 1 on [1, 2].
f(1) = 1 − 1 − 1 = −1. f(2) = 8 − 2 − 1 = 5. Since f is a polynomial (continuous everywhere), and 0 is between −1 and 5, the IVT guarantees at least one root c in (1, 2).
Example: cos(x) = x³ on (0, 1).
Define g(x) = cos(x) − x³. Then g(0) = 1 > 0 and g(1) = cos(1) − 1 ≈ −0.46 < 0. Since g is continuous (difference of continuous functions), the IVT guarantees a c in (0, 1) where g(c) = 0, meaning cos(c) = c³.
Example: f(x) continuous on [0, 5], f(0) = −2, f(5) = 10.
Since −2 < 0 < 10, the IVT guarantees a value c in (0, 5) with f(c) = 0.
Three conditions for continuity at x = c: (1) f(c) defined, (2) lim(x→c) f(x) exists, (3) lim(x→c) f(x) = f(c)
IVT: f continuous on [a, b], N between f(a) and f(b) → there exists c in (a, b) with f(c) = N
The IVT is why you can be sure a rising thermometer passed through every temperature between its starting and ending readings. Engineers use it to confirm that a continuous system (pressure, voltage, stress) must have crossed a threshold value at some point in a given range.
Students often confuse removable and jump discontinuities. A removable discontinuity has a hole (the limit exists but the value is wrong or missing). A jump has two different one-sided limits.
The IVT does not tell you what c is, only that it exists. Exams test whether you know to state existence, not compute the root.
Students sometimes apply the IVT to functions that are not continuous on the given interval. Always check continuity first.
A function can be continuous at every point of its domain and still not be continuous on all of (−∞, ∞). For example, f(x) = 1/x is continuous on its domain but not defined at 0.
⚠️ "Which type of discontinuity?" is a reliable multiple-choice question. Know the three types cold.
⚠️ The IVT appears as both a multiple-choice conceptual question and a long-answer proof. The long-answer version asks you to state the theorem, verify continuity, compute endpoint values, and conclude.
⚠️ Know which standard functions are continuous everywhere on their domain: |x|, polynomials, sin(x), cos(x). Know which are not continuous on all of (−∞, ∞): tan(x), 1/x, √x.
⚠️ Removable discontinuity at x = c for (x² − 1)/(x − 1) is a favourite short-answer question. Factor, cancel, identify the hole.
True or False: A function with a hole at x = 3 has a jump discontinuity there.
Fill in the blank: The IVT requires that f be ______ on [a, b].
True or False: f(x) = |x| has a discontinuity at x = 0.
Fill in the blank: If lim(x→c⁻) f(x) ≠ lim(x→c⁺) f(x), the discontinuity is called a ______ discontinuity.
True or False: The IVT can tell you the exact value of c where f(c) = 0.
Answers: 1. False (it is a removable discontinuity). 2. continuous. 3. False (|x| is continuous everywhere). 4. jump. 5. False (it guarantees existence, not the value).
Q: Explain why f(x) = (x² − 1)/(x − 1) has a removable discontinuity at x = 1.
A: Factor the numerator: (x+1)(x−1)/(x−1). For x ≠ 1 this simplifies to x + 1, so lim(x→1) f(x) = 2. But f(1) is undefined (0/0), so the limit exists but does not equal the function value. This is the definition of a removable discontinuity.
Q: Use the IVT to show that f(x) = x³ − x − 1 has a root between x = 1 and x = 2.
A: f is a polynomial, so it is continuous everywhere. f(1) = −1 < 0 and f(2) = 5 > 0. Since 0 is between −1 and 5, the IVT guarantees at least one c in (1, 2) with f(c) = 0.
Q: Prove that cos(x) = x³ has at least one solution in (0, 1).
A: Let g(x) = cos(x) − x³. g is continuous (difference of continuous functions). g(0) = 1 > 0. g(1) = cos(1) − 1 ≈ −0.46 < 0. By the IVT, there exists c in (0, 1) with g(c) = 0, i.e. cos(c) = c³.
Q: Which type of discontinuity occurs where the left-hand and right-hand limits exist but are not equal?
A: Jump discontinuity.
Q: Which of the following is continuous on all of (−∞, ∞): |x|, tan(x), √x, 1/x?
A: f(x) = |x|. The others have restricted domains or vertical asymptotes.
Continuity is a prerequisite for differentiability: if a function is differentiable at a point, it must be continuous there (though the reverse is not always true). The IVT connects to root-finding algorithms like the bisection method, which you may see in numerical methods courses. The ideas in this section also underpin the Mean Value Theorem, which comes later in Calculus I.
continuity, continuous function, removable discontinuity, hole, jump discontinuity, infinite discontinuity, vertical asymptote, Intermediate Value Theorem, IVT, root existence, zero of a function, piecewise function, continuous on a closed interval, MTH 132, Calculus I, Michigan State University