Continuity at a Point

You have known what continuity means since you were about six years old. Put a pen on paper and trace a curve: if you can draw the whole thing without lifting the pen, the curve is continuous. Lift the pen — to hop over a gap, to jump to a different height — and it isn't. Rollercoaster tracks and the temperature through an afternoon are pen-never-lifts curves; a staircase, or a stock price at the moment bad news lands, is not.

Calculus keeps the childhood picture but sharpens it into something you can actually check. Why bother? Because the pen picture misleads for exactly the functions where it matters. A graph can look unbroken on a calculator screen and hide a missing point one pixel wide. A function can be given by a formula too wild to draw at all — try sketching \sin(1/x) near zero and your pen gives up long before the mathematics does. And "I drew it in one stroke" is not an argument you can write in a proof. We need a test that works on formulas, not drawings — and it turns out to be a checklist of exactly three conditions, all built from the limits you already own.

A function f is continuous at x = c when all three of the following hold:

Read the three conditions as a tiny story. Condition 1 says the point is there. Condition 2 says the curve knows where it's going. Condition 3 is the punchline: the curve is going exactly where the point is. Where the function is heading and where it actually lands are the same place — no surprise on arrival. That third line is also why direct substitution works: "just plug in c" computes f(c), and it gives the limit precisely when the two are equal — which is the definition of continuity. Substitution isn't a lucky trick; it is continuity in disguise.

The three-part test, live

Here is the checklist in action. All four functions below use the same left half — the line y = x approaching c = 2 — but each treats the point c differently. Flip between the cases and, before reading the verdict, run the checklist yourself: is there a dot at c? Do the two sides of the curve aim at the same height? Does the dot sit at that height?

Notice how each broken case fails at a different line of the checklist. The hole has a curve that aims perfectly at height 2 — condition 2 passes — but there's no dot to arrive at: condition 1 fails. The jump has a dot, but the two halves of the curve can't agree on a destination, so the limit doesn't exist: condition 2 fails. The infinite case fails even harder — no value and no limit. One picture, three distinct diseases; the checklist is the diagnosis.

Worked example 1: a polynomial passes with full marks

Is f(x) = x^2 + 1 continuous at x = 2? Run the checklist in order — each line is a genuine, separate check:

Verdict: continuous at 2 — the pen never lifts. And nothing about the point 2 was special: the same three lines go through for any polynomial at any point, because the limit laws push a limit through sums, products and powers of x. That is the real content of the slogan "polynomials are continuous everywhere". The interesting functions, then, are the ones built with seams — piecewise definitions, fractions whose denominators can vanish — and that's where the checklist earns its keep.

Three ways to break

When a function is not continuous at c, we call c a discontinuity — and the type of discontinuity is simply a record of which condition failed, building directly on when a limit doesn't exist. Learn to name them on sight:

The chart below shows the crudest of the three — a jump at x = 0. Trace it with an imaginary pen: coming in from the left you arrive at height 2; the curve resumes at height 0.5. The pen must lift, and the size of the lift — 1.5 here — is exactly the gap between the one-sided limits. (Flip back to the live figure above to compare it against the hole and the blow-up.)

Worked example 2: diagnosing three patients

Three piecewise functions, each unwell at one point. For each, run the checklist and name the disease.

Patient (a): h(x) = \begin{cases} x^2 & x \neq 1 \\ 5 & x = 1 \end{cases} Check at c = 1. Condition 1: h(1) = 5 — defined. ✓ Condition 2: for every x near 1 (but not equal to it) the rule is x^2, so \lim_{x \to 1} h(x) = 1 — exists. ✓ Condition 3: 1 \neq 5. ✗ The curve aims at height 1 but the dot has been relocated up to 5. Diagnosis: removable discontinuity — two conditions pass, only the match fails.

Patient (b): g(x) = \begin{cases} x + 1 & x < 2 \\ x + 3 & x \ge 2 \end{cases} Check at c = 2. Condition 1: g(2) = 5 — defined. ✓ Condition 2: from the left, \lim_{x \to 2^-} (x+1) = 3; from the right, \lim_{x \to 2^+} (x+3) = 5. The sides disagree (3 \neq 5), so the limit does not exist. ✗ Stop — once condition 2 fails, condition 3 is meaningless (there is no limit to compare against). Diagnosis: jump discontinuity, of size 2.

Patient (c): k(x) = \frac{1}{(x-4)^2} Check at c = 4. Condition 1: k(4) = \tfrac{1}{0} — undefined. ✗ Condition 2: as x \to 4 from either side, (x-4)^2 is a tiny positive number, so k(x) \to +\infty — the limit does not exist either. ✗ Diagnosis: infinite discontinuity. Note the severity ordering: patient (a) failed one condition, (b) effectively one, and (c) failed everything. That ordering matters for the next question — which patients can be cured?

Worked example 3: repairing a removable discontinuity

Here is a function you will meet constantly in calculus — a fraction whose top and bottom both vanish at the same point:

f(x) = \frac{x^2 - 9}{x - 3}.

Step 1 — find the sore point. At x = 3 the denominator is 0, so f(3) is undefined: condition 1 fails, and f is not continuous at 3. But how badly is it broken?

Step 2 — take the limit anyway. For every x \neq 3 we may factor and cancel:

\frac{x^2 - 9}{x - 3} = \frac{(x-3)(x+3)}{x-3} = x + 3,

so \lim_{x \to 3} f(x) = 3 + 3 = 6. The limit exists! Away from the sore point, f is the straight line y = x + 3 — a perfectly healthy function with one point punched out. That is the signature of a removable discontinuity: condition 2 passes, only the value is missing.

Step 3 — prescribe the cure. The curve is aiming at height 6; all that's missing is a dot there. So define one. Build the repaired function

\tilde f(x) = \begin{cases} \dfrac{x^2-9}{x-3} & x \neq 3 \\[2pt] 6 & x = 3 \end{cases}

and re-run the checklist at 3: \tilde f(3) = 6 exists ✓, the limit is still 6 ✓ (limits never care about the value at the point, only near it), and 6 = 6 ✓. Healed. This is why "removable" is exactly the right word: the whole defect was one point's worth of missing (or misplaced) data, and redefining that single point removes it. The recipe is always the same — compute L = \lim_{x \to c} f(x), then declare f(c) := L. Note what it cannot fix: a jump has no single L to aim for, and a blow-up has none at all. Only the mildest disease is curable.

Watch Sal define continuity

Sal Khan walks through the same three-part definition with the epsilon-free pictures used here — a good second pass, especially on telling the hole and the relocated dot apart.

Here is a claim about your own life that mathematics can prove: at some instant in your childhood, your height was exactly one metre. Not roughly — exactly, to infinitely many decimal places. How can anyone know that without a time machine and a very fussy ruler?

Because height is continuous in time — you grow, but you never teleport from 99 cm to 101 cm without passing through every height in between. Once a function is continuous on an interval, it inherits a superpower called the Intermediate Value Theorem: a continuous function that starts below a value and ends above it must hit that value somewhere along the way. You were 50 cm at birth and are more than a metre now; one metre lies in between; therefore some instant caught you at precisely one metre. The pen picture makes it obvious — a pen travelling from below a line to above it must cross the line — but the theorem only holds because all three checklist conditions hold at every point. Give the function a single jump and the guarantee dies: a staircase can step over any value it likes.

The same superpower pulls off a party trick with the weather: at every moment there are two points on exactly opposite sides of the Earth with precisely the same temperature (compare each equator point with its opposite; the difference starts positive, ends negative, so somewhere it is exactly zero). And it is why engineers trust root-finding: a continuous function that is negative here and positive there is guaranteed to solve f(x) = 0 in between. Continuity is the fine print on that guarantee — which is why mathematicians fuss so much about checking it.

The checklist, one last time

Faced with "is f continuous at c?", work in order and stop at the first failure — each later condition only means anything if the earlier ones passed:

Everything on this page compresses into that triage — and it feeds directly into what comes next: continuity at a point is the entry requirement for having a derivative there, where a fourth, stricter condition (no corners!) joins the list.