Source: MTH 132, Chapter 1 (Sections 8.1–8.4) | Michigan State University
Tags: continuity, continuous function, discontinuity, removable discontinuity, jump discontinuity, infinite discontinuity, Intermediate Value Theorem, IVT, piecewise function, calculus 1, MTH 132, MSU
Difficulty: Intermediate. Prerequisites: limits (Sections 5–6), one-sided limits, vertical asymptotes.
Continuity is the formal version of "you can draw the graph without lifting your pencil." It ties together everything from the limit sections: a function is continuous at a point when its limit there equals its function value. This section also introduces the Intermediate Value Theorem (IVT), a powerful existence result that guarantees a continuous function takes on every value between f(a) and f(b). The IVT is one of the first "big" theorems of the course and appears on nearly every exam.
A function is continuous at x = a when three things hold: f(a) is defined, lim(x→a) f(x) exists, and they are equal. Discontinuities come in three types: removable (a hole you can fill), infinite (a vertical asymptote), and jump (left and right limits disagree). The Intermediate Value Theorem says a continuous function on [a, b] takes every value between f(a) and f(b), which lets you prove that equations have solutions even when you cannot solve them algebraically.
Continuous at a point
A function f is continuous at x = a if and only if: (1) f(a) is defined, (2) lim(x→a) f(x) exists, and (3) lim(x→a) f(x) = f(a).
In simple terms, there is no gap, jump, or blow-up at that point. The function arrives where you expect it to.
Continuous on an interval
A function is continuous on an interval if it is continuous at every point in that interval.
Discontinuity
A point where f is not continuous. If f is defined near a (or at a), and any of the three continuity conditions fails, f has a discontinuity at a.
Removable discontinuity
A discontinuity that can be "fixed" by redefining f at just a single point. This happens when the limit exists but either f(a) is undefined or f(a) does not match the limit. Graphically, it looks like a hole in the curve.
In simple terms, the function has a pinprick missing from an otherwise smooth graph.
Infinite discontinuity
Another name for a vertical asymptote. The function values blow up toward ±∞ near a.
Jump discontinuity
The left-hand and right-hand limits both exist but are not equal. The function "jumps" from one value to another. This typically occurs in piecewise-defined functions.
Intermediate Value Theorem (IVT)
If f is continuous on [a, b] and N is any number between f(a) and f(b) (where f(a) ≠ 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 at 2 and ends at 7, it must pass through every number between 2 and 7 along the way.
Three conditions must all hold at x = a: f(a) exists, the limit exists, they match.
Polynomials, rational functions (in their domain), root functions (in their domain), and trigonometric functions are all continuous wherever they are defined.
Combinations of continuous functions (sums, products, quotients with nonzero denominator) are continuous. Compositions of continuous functions are continuous.
Watch out for: dividing by zero, square roots of negative numbers, and piecewise-defined functions at their boundary points.
Removable: the limit exists, but f(a) is either undefined or has the wrong value. Removable discontinuities typically occur in rational functions where a common factor cancels (the 0/0 case from Section 6). To "remove" one, redefine the function value at that point to equal the limit.
Infinite: a vertical asymptote. The function values blow up. Infinite discontinuities typically occur in rational functions where the denominator is zero but the numerator is not (the N/0 case).
Jump: the left-hand and right-hand limits both exist as finite numbers but disagree. Jump discontinuities typically occur in piecewise-defined functions.
For f(x) = (x² + 5x + 6)/(x² – 3x – 10), factor: (x + 2)(x + 3)/[(x – 5)(x + 2)]. The function is undefined at x = –2 and x = 5.
At x = –2: the (x + 2) cancels. The limit exists (plug –2 into the simplified form). This is a removable discontinuity.
At x = 5: the denominator is zero and the numerator is not. This is an infinite discontinuity (vertical asymptote).
Requires: f continuous on a closed interval [a, b], and N is between f(a) and f(b).
Conclusion: there is at least one c in (a, b) with f(c) = N. The theorem guarantees existence but does not tell you the value of c.
To prove an equation f(x) = 0 has a solution on (a, b): show f is continuous there, compute f(a) and f(b), and confirm they have opposite signs (one positive, one negative). Then N = 0 is between them.
Example: f(x) = x⁴ + x – 3 on (1, 2). f(1) = 1 + 1 – 3 = –1 < 0. f(2) = 16 + 2 – 3 = 15 > 0. Since f is continuous (it is a polynomial) and changes sign, the IVT guarantees a root in (1, 2).
Example: to prove cos(x) = x³ has a solution, rewrite as g(x) = cos(x) – x³. g(0) = 1 > 0, g(π/2) ≈ 0 – (π/2)³ < 0. By the IVT, g has a zero somewhere in (0, π/2).
For f(x) = { cx² + 2x if x < 2; x³ – cx if x ≥ 2 }, set the left-hand limit equal to the right-hand limit at x = 2: c(4) + 4 = 8 – 2c. Solve: 4c + 4 = 8 – 2c → 6c = 4 → c = 2/3.
Continuity of combinations: if f and g are continuous at a and c is a constant, then f + g, f – g, c·f, f·g, and f/g (when g(a) ≠ 0) are all continuous at a.
Continuity of compositions: if g is continuous at a and f is continuous at g(a), then (f ∘ g)(x) = f(g(x)) is continuous at a.
Intermediate Value Theorem: if f is continuous on [a, b] and N is between f(a) and f(b), then there is at least one c in (a, b) with f(c) = N.
The IVT is why your car's speedometer, which reads 0 km/h when parked and 100 km/h on the motorway, must have passed through every speed in between: 30, 50, 72.4, all of them. Engineers use it to prove that a bridge under load has a point where deflection equals a target value, even when the deflection equation cannot be solved by hand.
Students often think continuity means "the function is defined." Being defined is necessary but not sufficient. The limit must also exist and match the function value.
Students confuse removable and jump discontinuities. A removable discontinuity has a limit (both sides agree); a jump discontinuity has two different one-sided limits.
A common error with the IVT: students try to use it on a function that is not continuous on the interval. Continuity on the closed interval is a non-negotiable requirement.
Students sometimes think the IVT tells you the value of c. It does not. It only guarantees that c exists.
⚠️ "Determine where the following function is continuous" is a standard exam question. Know which function families are continuous on their domains.
⚠️ Classifying discontinuities (removable, infinite, jump) from a formula or a piecewise definition appears regularly.
⚠️ IVT proof questions ("show that f(x) = 0 has a root on (a, b)") are a near-certainty on MTH 132 exams. State the three requirements explicitly: f is continuous on [a, b], f(a) and f(b) have opposite signs, therefore by the IVT there exists c with f(c) = 0.
⚠️ "Find c so that the piecewise function is continuous everywhere" is a classic exam problem. Set the left-hand and right-hand limits equal at the boundary and solve for c.
What three conditions must hold for f to be continuous at x = a?
f(a) is defined, lim(x→a) f(x) exists, and lim(x→a) f(x) = f(a).
True or false: a removable discontinuity means the limit does not exist.
False. The limit exists; it just does not match f(a) (or f(a) is undefined).
Fill in the blank: the IVT requires f to be ____ on a ____ interval [a, b].
Continuous, closed.
True or false: the IVT tells you the exact value of c where f(c) = N.
False. It guarantees existence only.
Name the type of discontinuity: the left-hand limit is 3, the right-hand limit is 7.
Jump discontinuity.
Q: Determine and classify the discontinuities of f(x) = (x² + 5x + 6)/(x² – 3x – 10).
A: Factor: (x + 2)(x + 3)/[(x – 5)(x + 2)]. Discontinuities at x = –2 and x = 5. At x = –2 the common factor cancels, so the limit exists: removable discontinuity. At x = 5 the denominator is zero and the numerator is nonzero: infinite discontinuity (vertical asymptote).
Q: Use the IVT to show that f(x) = x⁴ + x – 3 has a root in (1, 2).
A: f is a polynomial, so it is continuous everywhere. f(1) = 1 + 1 – 3 = –1 < 0. f(2) = 16 + 2 – 3 = 15 > 0. Since f is continuous on [1, 2] and f(1) < 0 < f(2), the IVT guarantees a c in (1, 2) with f(c) = 0.
Q: For f(x) = { cx² + 2x if x < 2; x³ – cx if x ≥ 2 }, find c so that f is continuous everywhere.
A: Continuity at x = 2 requires lim(x→2⁻) f(x) = lim(x→2⁺) f(x). Left: c(4) + 4 = 4c + 4. Right: 8 – 2c. Set equal: 4c + 4 = 8 – 2c → 6c = 4 → c = 2/3.
Q: Suppose f is continuous with f(1) = 5, f(3) = 2, f(11) = –1. Which of the following must be true? (a) f(c) = 0 for some c in [1, 3]. (b) f(c) = 0 for some c in [–1, 5]. (c) f(c) = 4 for some c in [1, 3].
A: (a) Not necessarily: f goes from 5 to 2, both positive, so 0 is not between them. (b) This interval notation mixes values with inputs. (c) Yes: 4 is between f(1) = 5 and f(3) = 2, and f is continuous on [1, 3], so the IVT applies.
Q: Where is g(x) = √(1 – x²)/x continuous?
A: The square root requires 1 – x² ≥ 0, so –1 ≤ x ≤ 1. The denominator requires x ≠ 0. Domain: [–1, 0) ∪ (0, 1]. g is continuous on its domain because it is a composition and quotient of continuous functions.
Continuity is the bridge between limits (Sections 5–6) and derivatives (Chapter 2). A function must be continuous at a point to be differentiable there (though continuity alone is not enough for differentiability). The IVT returns in later courses when proving existence of fixed points, in numerical methods like the bisection method for root-finding, and in real analysis as a stepping stone to more advanced theorems.
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