Continuity & Discontinuities

You’ll be able to

State the three-part definition of continuity at a point
Classify discontinuities as removable, jump, or infinite
Determine parameter values that make a piecewise function continuous

Continuity in three conditions

A function f is continuous at x = a when three things all hold: (1) f(a) is defined, (2) lim(x→a) f(x) exists, and (3) that limit equals the value, lim(x→a) f(x) = f(a). Intuitively, you can draw through x = a without lifting your pencil. If any one condition fails, f is discontinuous there. A function continuous at every point of an interval is continuous on that interval.

Three flavors of discontinuity

A removable discontinuity is a hole: the two-sided limit exists but does not equal f(a) (or f(a) is missing) — you could "fill in" the point to fix it. A jump discontinuity occurs when the left and right limits both exist but differ, so the graph leaps. An infinite discontinuity occurs when a one-sided limit is ±∞, producing a vertical asymptote. Only the removable type can be patched by redefining a single value.

Continuity at a point

f continuous at a ⟺ f(a) defined and lim(x→a) f(x) exists and lim(x→a) f(x) = f(a)

All three parts are required. The equality in the third part is the one students most often skip.

Making piecewise functions continuous

When a piecewise function changes rule at a boundary x = a, continuity there forces the two pieces to agree in the limit. You set the left-piece value at a equal to the right-piece value at a and solve for the unknown parameter. This "match at the seam" idea is the standard way parameters get pinned down, and it shows up constantly on both multiple-choice and free-response questions.

Worked example

For what value of k is f continuous at x = 2, where f(x) = x² for x ≤ 2 and f(x) = kx + 1 for x > 2?

1.The left piece at x = 2 gives lim(x→2⁻) f(x) = 2² = 4, which also equals f(2).
2.The right piece gives lim(x→2⁺) f(x) = k(2) + 1 = 2k + 1.
3.Continuity requires the two one-sided limits to match: 2k + 1 = 4.
4.Solve: 2k = 3, so k = 3/2.

Answer: k = 3/2. With this value both pieces meet at the point (2, 4), so the function is continuous there.

Tip

To test continuity fast, compute three things separately: the left limit, the right limit, and the actual value f(a). If all three are the same finite number, the function is continuous. Any mismatch tells you which type of discontinuity you have.

Checkpoint

The function g(x) = (x² − 4)/(x − 2) is undefined at x = 2, but simplifies to x + 2 elsewhere. What kind of discontinuity is at x = 2?

Checkpoint

A piecewise function has lim(x→1⁻) f(x) = 3, lim(x→1⁺) f(x) = 7, and f(1) = 3. Which statement is correct?

On the exam

When justifying continuity on the AP exam, cite all three conditions by name. Writing "f(a) is defined, the limit exists, and they are equal" earns the justification point that a bare numerical answer does not.

Answer the 2 checkpoints as you read.