Classifying Discontinuities, and Removing the Removable
- Apply the three-part definition of continuity at a point
- Classify a discontinuity as removable, jump, or infinite
- Choose a parameter value that makes a piecewise function continuous
Continuity is three conditions, and all three are tested
f is continuous at a exactly when: (1) f(a) is defined, (2) lim(x→a) f(x) exists, and (3) the two are equal. Students routinely check only the third and lose points on the first two, so check them in order — and when a question asks you to show continuity or to explain discontinuity, the expected answer names which of the three fails. This is also why continuity is strictly stronger than having a limit: a function can have a perfectly good limit at a point where it is not continuous, which is exactly the removable case below.
The three types, by which condition fails
A removable discontinuity (a hole) occurs when the limit exists but either f(a) is undefined or f(a) ≠ the limit. The name is literal: redefine the single value f(a) to equal the limit and the function becomes continuous. This is what happens to (x² − 9)/(x − 3) at x = 3 — the factor cancels, the limit is 6, and the only defect is that the original expression is undefined there. A jump discontinuity occurs when both one-sided limits exist and are finite but unequal, so nothing you do to f(a) can repair it. An infinite discontinuity occurs where the function grows without bound, giving a vertical asymptote. Only the first is removable, and the exam asks that word specifically.
Rational functions: hole or asymptote?
Given a rational function, factor numerator and denominator and compare. If a factor cancels, the zero of that factor gives a hole — a removable discontinuity, and its height is the limit you get from the reduced expression. If a factor remains in the denominator after canceling, its zero gives a vertical asymptote — an infinite discontinuity. Both can occur in the same function. For (x − 2)(x + 1) / [(x − 2)(x − 5)], the (x − 2) cancels so x = 2 is a hole, while x = 5 remains in the denominator so x = 5 is a vertical asymptote. Do the factoring before saying anything about behavior; guessing from the unfactored denominator is the standard mistake.
Find k so that f is continuous at x = 3, where f(x) = (x² − 2x − 3)/(x − 3) for x ≠ 3, and f(3) = k.
- 1.Factor the numerator: x² − 2x − 3 = (x − 3)(x + 1).
- 2.So for x ≠ 3, f(x) = (x − 3)(x + 1)/(x − 3) = x + 1. The cancellation is legitimate because we are considering x ≠ 3 only.
- 3.Take the limit: lim(x→3) f(x) = lim(x→3) (x + 1) = 4. The limit exists, so the discontinuity is removable.
- 4.Continuity requires f(3) = lim(x→3) f(x), so k must equal 4.
- 5.Check all three conditions with k = 4: f(3) = 4 is defined, the limit is 4 and exists, and they are equal. Continuous.
Canceling (x − 3) does not make the original function equal to x + 1 everywhere. The two agree only for x ≠ 3, and that restriction is the entire content of the problem. Writing "f(x) = x + 1" with no qualifier throws away the discontinuity you were asked about.
For f(x) = (x + 4)(x − 1) / [(x − 1)(x + 6)], which statement is correct?
A function has lim(x→2⁻) f(x) = 5 and lim(x→2⁺) f(x) = 8. The discontinuity at x = 2 is
Answer the 2 checkpoints as you read.
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