Continuity and Classifying Discontinuities
A function is continuous at \(x = a\) when you can draw its graph through \(a\) without lifting your pencil — formally, three conditions must hold: \(f(a)\) is defined, the limit exists, and the two agree. When one fails, the way it fails names the break: removable, jump, or infinite.
By the end you'll be able to test continuity at a point with the three-part definition, say exactly which condition fails, and classify any break as a removable, jump, or infinite discontinuity.
Predict: for each of the four cases below, which of the three continuity conditions fails — if any? Select a case and watch the checklist light up.
Every case is examined at \(x = 2\). The orange solid dot is the actual value \(f(2)\) — condition 1. The blue arrows and hollow circle show where the one-sided limits are heading — condition 2. Condition 3 asks whether the dot lands exactly on the circle.
Continuous at x = 2 — all three conditions hold.
Continuity at a point is a three-part test — value, limit, agreement — and each way the test can fail has a name: removable, jump, or infinite discontinuity.
\(f\) is continuous at \(x = a\) when all three conditions hold: (1) \(f(a)\) is defined — \(a\) is in the domain; (2) \(\lim_{x \to a} f(x)\) exists — both one-sided limits agree; and (3) \(\lim_{x \to a} f(x) = f(a)\). Condition 3 is why continuity matters computationally: it is exactly the license for evaluating a limit by direct substitution, the shortcut you used with the limit laws — for a continuous function, plugging in \(a\) IS taking the limit. Continuity on an interval means continuity at every point of the interval (with one-sided continuity at closed endpoints). Polynomials are continuous everywhere; rational, root, trig, exponential, and log functions are continuous on their domains.
The pencil test: continuous at \(a\) means the graph passes through \(x = a\) without you lifting your pencil. Each failure looks different, and the look is the classification. A removable discontinuity is a single missing or displaced point — a hole, possibly with the dot floating elsewhere; the limit exists but doesn't match \(f(a)\) (or \(f(a)\) is undefined), and redefining that one point would repair it. A jump is two pieces arriving at different heights — both one-sided limits exist but disagree, so the two-sided limit fails; typical of piecewise and step functions. An infinite discontinuity is the graph blowing up to \(\pm\infty\) along a vertical asymptote, like \(f(x) = 1/x\) at \(x = 0\) — the same picture you met with infinite limits.
Is \(g(x) = \dfrac{x^2 - 1}{x - 1}\) continuous at \(x = 1\)? Run the three-part test in order. Condition 1: \(g(1) = 0/0\) is undefined — the test already fails, so \(g\) is not continuous at 1. To classify the break, still check the limit: for \(x \neq 1\), \(g(x) = \dfrac{(x-1)(x+1)}{x-1} = x + 1\), so \(\lim_{x \to 1} g(x) = 2\). The limit exists but \(g(1)\) does not — a removable discontinuity: defining \(g(1) = 2\) would repair it.
Let \(h(x) = x + 2\) for \(x < 1\) and \(h(x) = x\) for \(x \geq 1\). Test continuity at \(x = 1\). Condition 1: \(h(1) = 1\), defined. Condition 2: the left-hand limit is \(1 + 2 = 3\); the right-hand limit is ____. Do they agree — and how do you classify the point?
Reveal the answer
The right-hand limit is \(1\). Since \(3 \neq 1\), the one-sided limits disagree, condition 2 fails, and \(h\) has a jump discontinuity at \(x = 1\). Note you can't repair it by changing \(h(1)\) — no single value can meet both pieces. Compare the "Two disagreeing pieces" case in the classifier above.
More info — continuity is the substitution shortcut, made official
In the limit-laws lesson you evaluated limits of polynomials by just plugging in \(a\). Condition 3 is that shortcut written as a definition: \(\lim_{x \to a} f(x) = f(a)\) says "the limit is whatever substitution gives." So the logic runs both ways — knowing a function is continuous (every polynomial, and the other standard families on their domains) licenses substitution, and verifying the three conditions is how you certify that license at a particular point. When substitution produces \(0/0\) instead, you're usually staring at a removable hole like the worked example above. The OpenStax section in Dive deeper walks the same definition with more examples.
Check your understanding
A piecewise function has \(f(x) = x + 1\) for \(x \neq 2\) and \(f(2) = 5\). Classify the behavior at \(x = 2\).
Which list states exactly the three conditions for \(f\) to be continuous at \(x = a\)?
\(f(x) = x^2 + 3x\) is a polynomial, so it is continuous everywhere. What is \(\lim_{x \to 2} f(x)\), and which continuity condition justifies the shortcut you used?
At \(x = a\), both one-sided limits of \(f\) exist but are unequal, and \(f(a)\) is defined. Which condition fails, and what type of discontinuity is this?
Recap
- Continuous at \(x = a\) means three conditions hold: \(f(a)\) is defined, \(\lim_{x \to a} f(x)\) exists (one-sided limits agree), and \(\lim_{x \to a} f(x) = f(a)\).
- Condition 3 is the substitution shortcut: for a continuous function, the limit is found by plugging in \(a\).
- Continuity on an interval = continuity at every point; polynomials are continuous everywhere, and rational, root, trig, exponential, and log functions on their domains.
- Classify failures by which condition breaks: removable — limit exists but ≠ \(f(a)\) (or \(f(a)\) undefined), fixable at one point; jump — one-sided limits exist but disagree; infinite — \(f(x) \to \pm\infty\) at a vertical asymptote.
Dive deeper
- OpenStax Calculus Volume 1 — 2.4 Continuity Study the three-part definition and classify removable, jump, and infinite discontinuities
- Paul's Online Notes — Continuity Practice testing continuity at a point and reading discontinuities from graphs
Sources
- Continuity and Classifying Discontinuities