Limits · Lesson 4 of 5 · ~6 min

Continuity

Predict

A function has a hole at $x = 2$, but $f(2)$ is defined — the point sits at a different height than the hole. Is $f$ continuous at $x = 2$?

Hold that thought. Press Hole in the instrument, park probe a on 1.00, and read which of the three checks fails.

Try

Try itSweep probe a along the Continuous curve: all three checks read PASS at every point. Now press Hole and park probe a at exactly $1.00$. Then press Jump and watch which check flips.

Notice

On Hole at $a = 1$: f(a) defined? reads PASS, since $f(1) = 2.000$ exists. lim exists? reads PASS, since both sides head to $0.833$. equal? reads FAIL, because $2.000 \neq 0.833$. On Jump the failure moves up a line: $f(1) = 3.000$ is defined, but the left side approaches $0.833$ while the right approaches $3.000$, so lim exists? fails.

Explain

Continuity at a point is exactly the checklist the instrument runs: $f(a)$ has a value, the limit exists, and the two numbers match. Fail any one check and $f$ is discontinuous at $a$. So the prediction resolves to no — a value parked at the wrong height passes the first two checks and still fails the third.

The failing check tells you whether the break is repairable. On Hole, only equality fails: redefine $f(1)$ from $2$ to $0.833$ and all three checks pass. That is a removable discontinuity. On Jump, the limit itself fails — the sides disagree by $2.167$ — and no single value of $f(1)$ can agree with both sides at once. That is a jump discontinuity.

f \ \text{continuous at}\ a \iff f(a)\ \text{defined}, \quad \lim_{x \to a} f(x)\ \text{exists}, \quad \lim_{x \to a} f(x) = f(a)

Apply

Checkpoint — prove it to yourself

On the Hole specimen, which check fails at $x = 1$?

Let $g(x) = \dfrac{x^2 - 9}{x - 3}$ for $x \neq 3$, with $g(3) = 5$. What is $\lim_{x \to 3} g(x)$, and is $g$ continuous at $3$?

$h(x) = |x|/x$ jumps at $x = 0$: left limit $-1$, right limit $1$, $h(0)$ undefined. Which choice of $h(0)$ makes $h$ continuous at $0$?

Key idea

Continuity is a three-item checklist: $f(a)$ defined, $\lim_{x \to a} f(x)$ exists, and the two equal — compressed, $\lim_{x \to a} f(x) = f(a)$. A removable discontinuity fails only the last item, so moving one point repairs it. A jump breaks the limit itself, and no single point can fix that.

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