MISSION · Calculus
Continuity and Where It Breaks
The three conditions a point has to pass, and the three ways it fails
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The idea in one sentence
A function is continuous at a point when the value it approaches there and the value it takes there are the same number.
The three conditions
\[f \text{ is continuous at } a \iff \lim_{x \to a} f(x) = f(a)\]That one line hides three separate requirements, and a discontinuity is a failure of any one of them:
- \(f(a)\) exists. There has to be a point there at all.
- \(\lim_{x \to a} f(x)\) exists. The left and right limits must agree on a finite value.
- They’re equal. Approaching and arriving give the same answer.
Continuity on an interval just means continuity at every point of it. Informally: you can draw the graph without lifting the pen — a decent picture, though the real definition is the limit statement, and the picture starts lying for stranger functions.
The three ways it fails
Removable. \(f(x) = \dfrac{x^2 - 9}{x - 3}\) at \(x = 3\). Everywhere except 3 this simplifies to \(x + 3\), so the limit is 6 — but at 3 itself the formula is \(0/0\) and the function has no value. There’s a single-point hole in an otherwise perfect line. Defining \(f(3) = 6\) fixes it, which is why it’s called removable.
Jump. The left-hand and right-hand limits both exist but differ. Postage rates, a step function, a piecewise function whose pieces don’t meet. No single value at the joining point can repair this — the two sides disagree about where to be.
Infinite. \(f(x) = 1/x\) at \(x = 0\). The function runs off to \(\pm\infty\), there’s a vertical asymptote, and no limit exists.
Making piecewise functions continuous
\[f(x) = \begin{cases} x^2 & x \leq 2 \\ ax + 1 & x > 2\end{cases}\]For continuity at \(x = 2\), the pieces must meet: \(4 = 2a + 1\), so \(a = 1.5\). This is the standard exam use of the definition, and it’s the definition doing exactly what it says — the approach from the right must land where the left-hand piece ends.
Why it matters
Continuity is the licence for the theorems that make calculus useful.
Intermediate Value Theorem. A continuous function that is negative somewhere and positive somewhere else must be zero in between. That’s how you know a root exists before you can find it, and it’s the basis of every bisection method.
Extreme Value Theorem. A continuous function on a closed interval attains a maximum and a minimum. Optimisation problems quietly depend on this.
And differentiability requires continuity. A function with a break has no tangent line there — though the reverse fails: \(|x|\) is continuous at 0 and has a corner, so continuity is necessary for differentiability and not sufficient.
Common mix-ups
- Assuming a formula’s domain is the whole line. \(1/x\) isn’t discontinuous at 0 in the sense of a broken graph; 0 simply isn’t in the domain — but the graph still breaks there, which is what you’re asked about.
- Checking only that the limit exists, and forgetting to compare it with \(f(a)\).
- Thinking continuous means smooth. Corners are continuous.
- Applying the IVT to a discontinuous function, which can skip straight over a value.
Check yourself
What kind of discontinuity does \((x^2-9)/(x-3)\) have at \(x = 3\)? And what must the limit as \(x \to 3\) equal for \(f\) to be continuous there?
It’s removable: the function agrees with \(x + 3\) everywhere else, so the limit is 6 and only the single point is missing. For continuity the limit must equal f(3) itself.
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