A machinist is handed a drawing for a piston. In the corner, in a little box, sits the customer's demand: diameter 80 mm ± 0.01 mm. The customer does not care how the machinist achieves it — only that the output lands inside that band. The machinist's job is to translate the demand backwards: how precisely must I control the inputs — the tool position, the temperature of the blank, the feed rate — so that the finished diameter is guaranteed to land within ± 0.01? Every engineering shop on Earth runs on this contract: you name the output tolerance; I must find an input tolerance that secures it. And the contract is only honourable if it can be met for every demand, however fussy — ± 0.01, ± 0.001, ± 10⁻⁹.

That contract is the rigorous definition of continuity. Informally, a function is continuous if you can draw its graph "without lifting the pen". To prove theorems we need that intuition pinned to an inequality, with no appeal to pens, motion, or time. The working definition is that the limit equals the value, and the ε–δ definition of a limit spells that out. Here we put the two together and — more importantly — learn to write the proofs.

The definition, dissected

A function f is continuous at c when: for every output tolerance \varepsilon > 0, there exists an input tolerance \delta > 0 such that

Every symbol earns its keep. |f(x) - f(c)| < \varepsilon is the customer's spec: the output must land within \varepsilon of the target value f(c). |x - c| < \delta is the machinist's input control: hold the input within \delta of c. The implication says the input control guarantees the output spec — every x in the \delta-window, no exceptions. And the quantifiers say the guarantee must be available for every demand:

Read it as a game. An adversary moves first and names \varepsilon — how close to f(c) the outputs must land. You move second and answer with a \delta-window around c so narrow that every input inside it is mapped inside the adversary's band. Continuity at c means you have a winning strategy: whatever \varepsilon is thrown at you, you can produce a working \delta. A proof of continuity is nothing more than writing that strategy down — usually as a formula giving \delta in terms of \varepsilon.

Warm-up proof: f(x) = 3x + 1 is continuous everywhere

Fix any point c and let the adversary name \varepsilon > 0. Before choosing \delta, do the scratch work: relate the output gap to the input gap.

The output gap is always exactly 3 times the input gap — a line of slope 3 stretches every interval by a factor of 3, no more, no less. So to force 3|x - c| < \varepsilon we need |x - c| < \varepsilon / 3, and the strategy writes itself.

Proof. Given \varepsilon > 0, choose \delta = \varepsilon / 3. Then whenever |x - c| < \delta,

Two lines, and it works at every c — the same \delta = \varepsilon/3 everywhere, because a line is equally steep everywhere. The general pattern for f(x) = mx + b with m \ne 0 is \delta = \varepsilon / |m|: the steeper the line, the tighter the input tolerance you must promise. (A slope-m machine amplifies input error by |m| — exactly the "sensitivity" a tolerance engineer computes.)

The first real proof: f(x) = x^2 at c = 2

Now the slope isn't constant, and a new idea is needed — the one that powers most ε–δ proofs you will ever write. Scratch work first: factor the output gap.

The factor |x - 2| is under our control — it's the input gap. But the factor |x + 2| is a nuisance: it varies with x, so we can't just divide \varepsilon by it. The trick: tether x first, then bound the nuisance factor by a constant. Promise in advance that \delta \le 1. Then |x - 2| < 1, so 1 < x < 3, and by the triangle inequality

Inside the tether, the awkward factor never exceeds 5, so |x^2 - 4| < 5\,|x - 2| — the problem is now as easy as the linear one, with "slope" 5.

Proof. Given \varepsilon > 0, choose

Suppose |x - 2| < \delta. Since \delta \le 1, the tether holds and |x + 2| < 5. Since \delta \le \varepsilon / 5,

The \min makes both promises true at once: small enough to keep the bound on |x + 2| honest, and small enough to beat \varepsilon. This tether–bound–divide pattern is the workhorse of analysis; you will use it for x^3, for 1/x, for products of continuous functions — everywhere a non-constant factor gets in the way.

And at a general point a? The identical argument runs with |x^2 - a^2| = |x - a|\,|x + a|: tether with \delta \le 1, bound |x + a| \le |x - a| + 2|a| < 1 + 2|a|, and choose

So x^2 is continuous at every real number. But notice the catch worth remembering: this \delta shrinks as |a| grows — far from the origin the parabola is steep, and the input tolerance must tighten. That dependence is the seed of the next idea.

See the contract in action

Here is the game made visual on y = x^2 at the point a = 1. The horizontal band is the adversary's \varepsilon-tolerance around f(a) = 1; the vertical strip is your answering \delta-window, computed by the proof's own formula \delta = \min(1, \varepsilon/3). Pull the slider down: as \varepsilon shrinks, the strip narrows just enough that the parabola never escapes the band inside it. Winning the game means this picture can be drawn for every \varepsilon, however small — the readout shows the strategy responding live.

Notice the picture's honest asymmetry: the parabola is steeper on the right edge of the strip than on the left, so it is the right edge that touches the band first. The proof's \delta is chosen to survive the worst case — the tolerance engineer's habit of designing for the steepest part of the response.

Proving a DIScontinuity — the other half of the skill

To prove continuity you must beat every \varepsilon. To prove discontinuity you need only exhibit one \varepsilon that defeats every \delta — the roles in the game swap, and now the adversary is you. Take the step function

which jumps by 1 at the origin. Claim: H is not continuous at 0.

Proof. Take \varepsilon = \tfrac{1}{2} — half the jump. Let \delta > 0 be arbitrary: whatever \delta is offered, the point x = -\delta/2 satisfies |x - 0| = \delta/2 < \delta, yet

So no \delta honours the contract for this \varepsilon: every input window around 0, however narrow, contains points from the low side of the jump whose outputs miss the band. Discontinuous at 0. \blacksquare

The logical shape is worth memorising, because it is the exact negation of the definition:

Each quantifier flips (\forall \leftrightarrow \exists) and the final implication becomes a counterexample pair. For a jump of height g, any \varepsilon \le g does the job — \varepsilon = g/2 is the traditional choice because it leaves room to spare.

The sequential characterisation

There is a second, equivalent language for the same idea, and analysts switch between the two constantly:

In words: continuous functions are exactly the ones that preserve limits of sequences — you may push a limit through f, writing \lim f(x_n) = f(\lim x_n). The forward direction is a pleasant ε–δ exercise (given \varepsilon, get \delta from continuity, then get N from x_n \to c). The reverse direction is proved by contraposition and is where the real power lies: to prove discontinuity, one bad sequence suffices. For the step function above, take x_n = -1/n \to 0; then H(x_n) = 0 \to 0 \ne 1 = H(0) — a one-line second proof of the discontinuity, with no quantifier gymnastics at all.

Pointwise versus uniform

In the x^2 proof, \delta depended on both \varepsilon and the point a. That is ordinary (pointwise) continuity. A stronger demand is that one \delta work for all points at once — one input tolerance for the whole production line, not a separate one per part.

f is uniformly continuous on a set S when, for every \varepsilon > 0, there is a single \delta > 0 (depending on \varepsilon only) with

Why x^2 is NOT uniformly continuous on \mathbb{R}. Take two points a fixed gap h apart, x and x + h:

However small the gap h > 0, the output gap h|2x + h| \to \infty as x \to \infty. So no single \delta can keep the outputs within (say) \varepsilon = 1 everywhere: out at large x the parabola is too steep. Uniform continuity fails precisely because the \delta from the proof above had to keep shrinking with |a|.

The rescue is to stay on a closed, bounded interval. There x^2 can't run off to steep regions, and uniform continuity is automatic.