Continuity: Proofs, Jump Discontinuities and the Intermediate Value Theorem
A function f is continuous at a point a if f(a) is defined, limx→a f(x) exists and limx→a f(x) = f(a); the intermediate value theorem says that a continuous function on [a, b] takes every value between f(a) and f(b). In brief: polynomials, sine, cosine and ex are continuous everywhere, 1/x is continuous wherever it is defined, and a function like ⌊x⌋ has jump discontinuities at the integers. Since f(x) = x3 + x − 1 is continuous with f(0) = −1 < 0 < f(1) = 1, it has a root between 0 and 1.
Continuity captures the idea of a graph that can be drawn without lifting the pen. It guarantees that small changes in the input cause small changes in the output, and it is the condition for several powerful theorems: the intermediate value theorem, the extreme value theorem and the existence of integrals. This guide gives the definition, how to prove continuity, the types of discontinuities (removable, jump, infinite), the intermediate value theorem with the bisection method and applications, and a table of examples.
What is continuity?
A function f is continuous at a if three conditions hold: (1) f(a) is defined; (2) limx→a f(x) exists; (3) limx→a f(x) = f(a). Equivalently, for every ε > 0 there is δ > 0 such that |x − a| < δ implies |f(x) − f(a)| < ε. The function is continuous on an interval if it is continuous at every point of it (one-sided at the endpoints).
Basic results: sums, differences, products, quotients (where the denominator is nonzero) and compositions of continuous functions are continuous. Polynomials, rational functions (on their domains), roots, trigonometric functions, exponentials and logarithms (on their domains) are continuous. Therefore, most elementary functions are continuous wherever they are defined.
How do you prove that a function is continuous?
Example 1: from the definition
Show that f(x) = 3x − 2 is continuous at 1. |f(x) − f(1)| = |3x − 3| = 3|x − 1|. Given ε, choose δ = ε/3 ✓.
Example 2: using limit laws
f(x) = (x2 − 1)/(x + 2) at a = 3: the denominator at 3 is 5 ≠ 0, so lim f = (9 − 1)/5 = 8/5 = f(3). Continuous. At a = −2 the function is not defined.
Example 3: a piecewise function
f(x) = (x2 − 4)/(x − 2) for x ≠ 2 and f(2) = 3. The limit at 2 is 4 (see the factoring), but f(2) = 3 ≠ 4, so f is discontinuous at 2. If we define f(2) = 4, the function becomes continuous: this is a removable discontinuity.
Example 4: a composition
f(x) = sin(1/x) is continuous for x ≠ 0 as a composition of continuous functions, but it has no limit at 0 (oscillation), so no value f(0) makes it continuous. In contrast, g(x) = x sin(1/x) for x ≠ 0 and g(0) = 0 is continuous at 0.
What types of discontinuities are there?
- Removable: the limit exists but differs from f(a) or f(a) is undefined, e.g., (x2 − 1)/(x − 1) at x = 1 (a hole).
- Jump: both one-sided limits exist but differ, e.g., the sign function sgn x at 0 (limits −1 and 1), the floor function at integers.
- Infinite: at least one one-sided limit is infinite, e.g., 1/x at 0, tan x at π/2.
- Essential (oscillating): a one-sided limit does not exist, e.g., sin(1/x) at 0.
Jump and removable discontinuities are called discontinuities of the first kind, since the one-sided limits exist. A function with finitely many such discontinuities on [a, b] is piecewise continuous, and it is still integrable.
How do you choose parameters for continuity?
Equate the one-sided limits (and the function value) at the point where the formula changes.
Example 5
f(x) = kx + 1 for x ≤ 2, and f(x) = x2 − k for x > 2. At x = 2: left side 2k + 1, right limit 4 − k. Continuity requires 2k + 1 = 4 − k, so k = 1. Then f(2) = 3, and both sides give 3.
Example 6: two parameters
f(x) = x + 2 for x < −1; f(x) = ax + b for −1 ≤ x ≤ 1; f(x) = 3 − x for x > 1. At x = −1: −a + b = 1. At x = 1: a + b = 2. Adding: 2b = 3, so b = 3/2 and a = 1/2.
What is the intermediate value theorem?
If f is continuous on [a, b] and c lies between f(a) and f(b), then there is at least one x in [a, b] with f(x) = c. In particular, if f(a) and f(b) have opposite signs, f has a root in (a, b). The theorem fails for functions with jumps: f(x) = −1 for x < 0 and 1 for x ≥ 0 takes no value between −1 and 1 except at the ends.
Example 7: existence of a root
f(x) = x3 + x − 1: f(0) = −1 and f(1) = 1. By the theorem, there is a root in (0, 1). Since f′(x) = 3x2 + 1 > 0, f is increasing, so the root is unique.
Example 8: the bisection method
Narrow down the root by repeatedly halving the interval and keeping the half where the sign changes. f(0.5) = −0.375 < 0, so the root is in (0.5, 1). f(0.75) = 0.172 > 0, so (0.5, 0.75). f(0.625) = −0.131 < 0, so (0.625, 0.75). f(0.6875) = 0.0125 > 0, so (0.625, 0.6875). After 4 steps the interval has length 0.0625, and the root is about 0.682. The exact root is 0.6823…. Each step halves the error.
Example 9: a fixed point
Show that cos x = x has a solution. Let g(x) = cos x − x. g(0) = 1 > 0 and g(π/2) = −π/2 < 0. So there is a root in (0, π/2). Numerically x ≈ 0.7391. Any continuous function mapping [0, 1] into itself has a fixed point (apply the theorem to f(x) − x).
Example 10: a surprising consequence
At any moment, there are two opposite points on the equator with the same temperature (if temperature is continuous): let T(θ) be the temperature at angle θ and define d(θ) = T(θ) − T(θ + π). Then d(0) = −d(π), so d changes sign, and has a zero.
What is the extreme value theorem?
A function continuous on a closed bounded interval [a, b] attains a maximum and a minimum there. On open or unbounded intervals, this can fail: f(x) = 1/x on (0, 1] has no maximum, and f(x) = x on (0, 1) has neither maximum nor minimum. The theorem is the basis of optimization on closed intervals.
Table of discontinuity types
| Function | Point | Left limit | Right limit | Type |
|---|---|---|---|---|
| (x2 − 1)/(x − 1) | 1 | 2 | 2 | removable |
| |x|/x | 0 | −1 | 1 | jump |
| ⌊x⌋ | 3 | 2 | 3 | jump |
| 1/x | 0 | −∞ | +∞ | infinite |
| 1/x2 | 0 | +∞ | +∞ | infinite |
| sin(1/x) | 0 | does not exist | does not exist | essential |
| x2 | any | f(a) | f(a) | continuous |
Where do these ideas come from?
Bernard Bolzano proved the intermediate value theorem in 1817 with a definition of continuity very close to the modern one, and Cauchy gave a similar definition in 1821. Karl Weierstrass developed the ε–δ formalism in the 1860s. The extreme value theorem is also associated with Weierstrass. Before these definitions, “continuous” was understood intuitively as a curve drawn without a break. Weierstrass’s example of a function continuous everywhere but differentiable nowhere (1872) shocked mathematicians.
Where are they used?
The intermediate value theorem is the foundation of root-finding algorithms (bisection, used in calculators and software). Continuity ensures that physical quantities, such as temperature and position, do not jump. In economics, it guarantees the existence of equilibrium prices in models; the fixed point theorems of Brouwer generalise the one-dimensional result of Example 9. Continuity is needed for a function to be integrable.
Which mistakes are common?
Forgetting to check that f(a) is defined
(x2 − 1)/(x − 1) has the limit 2 at x = 1, but it is not continuous there, since it is not defined at 1.
Thinking that “continuous” means “defined everywhere”
1/x is continuous on its domain (all x ≠ 0), though it has an infinite discontinuity at 0, which is not in the domain.
Using the intermediate value theorem without continuity
The theorem requires continuity on the whole closed interval. A single jump invalidates the conclusion.
Assuming that a sign change implies exactly one root
The theorem gives at least one root. Use monotonicity (the derivative sign) to show uniqueness.