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Continuity & Its Failures

You’ve been meeting this word since Lesson 1 — the functions where substitution “just works.” Now it gets its full treatment. Continuity is what it means for a graph to flow through a point with no hole, no jump, no break — and there are exactly three things that have to line up for it.
A function is continuous at a point when where it’s heading equals where it is — the limit and the value agree, with no surprises. Informally: you can draw the graph through that point without lifting your pen. When that fails, it fails in one of a few specific, namable ways — a whole zoo of discontinuities, each with its own signature.

Drawing without lifting your pen

Here’s the intuition before any rules. A continuous function is one you can trace in a single unbroken stroke. The moment you have to lift your pen — to skip a hole, leap a gap, or escape to infinity — you’ve hit a discontinuity. Watch the pen try four different functions:

What you’re looking at: the pen traces left to right. If it ever has to jump, the function is not continuous there.

Pick a function

Only the first traces in one stroke. The others each force a lift — and each lift has a name you’ll learn below.

Continuity, written precisely

“Draw without lifting your pen” is the picture. Here it is as a single clean equation — the whole definition of continuity at a point lives in one line:

Read it aloud: “the limit of \(f\) as \(x\) approaches \(a\) equals \(f\) of \(a\)” — where the function is heading is exactly where it is. That single equals sign is the entire idea.
The left side — \(\lim_{x\to a}f(x)\) — is where the function is heading as you approach \(a\) (Lessons 1–3).
The right side — \(f(a)\) — is where the function actually is: its value exactly at \(a\).
The equals sign is the whole claim. For it to even make sense, both sides must exist first — which is exactly why the test below has three parts, not one.

The three-part test

“Draw without lifting” is the feeling; here’s the precise check. For \(f\) to be continuous at a point \(a\), three things must all be true — and if any one fails, continuity fails. Run a function through the gate and watch each box get checked (or not):

The gate
Pick a function

Feel it click into place

The three checks are clearest when you operate them. Below, the curve heads toward height 2 at \(x=1\) — that’s the limit. Drag the point’s value up and down and watch the three boxes respond live. There’s exactly one spot where all three turn green:

drag the value at x = 1: f(1) = 0.0

What you’re looking at: the magenta curve heads to 2 from both sides. The teal dot is the actual value \(f(1)\) — you control its height. Continuity needs it to land exactly where the curve is heading.

The gate, live

The discontinuity zoo

When continuity fails, it fails in one of four ways. Each is a different “creature” with a recognizable shape and a specific broken condition. Meet them all:

Name that break

The exam loves this: show a graph, ask what kind of discontinuity it has — and which part of the three-part test it violates. Diagnose each one:

Graph 1 of 6
Score: 0 / 0

Close the gap yourself

Piecewise functions are continuous only if the two branches meet at the seam. Here the right branch is fixed at \(x^2\); the left is \(kx\). Drag \(k\) and slide the left branch until it meets the right one exactly at \(x=2\) — watch the gap shrink to zero:

drag k: k = 0.80

What you’re looking at: the left branch \(kx\) (its slope changes with \(k\)) and the fixed right branch \(x^2\). The gap at the seam is the vertical distance between where they land at \(x=2\).

The seam at x = 2

Left branch lands at \(2k\); right branch lands at \(4\). They’re equal exactly when \(2k=4\) — the value of \(k\) that makes \(f\) continuous.

Fixing a removable hole

One type of discontinuity is special: the removable one. The limit exists — the function just has the wrong value (or none) at a single point. You can patch it: redefine that one point to equal the limit, and the hole seals. These “find the value that makes \(f\) continuous” problems are an exam staple. Step through:

press step
The work, shown
Pick a problem

The recipe is always the same: compute the limit at the trouble point (using Lesson 2’s tools), then set the function’s value there equal to it. The hole fills.

The lesson on one card

All three must hold: \(f(a)\) exists, \(\lim_{x\to a}f\) exists, and they’re equal
Four failure types: removable (hole), jump (sides disagree), infinite (asymptote), oscillating
Only removable discontinuities can be patched — redefine the point to equal the limit

The ideas everything else builds on

Heading = being

Continuity is exactly limit equals value. It’s why substitution works for friendly functions — they’re continuous, so the limit IS \(f(a)\).

Three independent checks

Value exists, limit exists, they match. Any one can fail on its own — that’s what makes the test diagnostic.

Removable means patchable

If the limit exists, the break is just one bad point — redefine it and the function becomes continuous. Jumps and asymptotes can’t be patched.

Where functions live

Polynomials are continuous everywhere; rational functions everywhere except where the denominator is zero. Knowing the type tells you where to look for trouble.

At a point vs. over an interval

A function is continuous on an interval when it’s continuous at every point inside. At a closed endpoint, only the one side that exists needs to match — e.g. \(\sqrt{x}\) is continuous on \([0,\infty)\) because it’s right-continuous at \(0\).

Your turn

Five problems — the three-part test, classification, and a solve-for-k.

Quick quiz

Eight fast checks across the whole lesson.

Question 1 of 8
Score: 0 / 0

Common mistakes & exam tips

Thinking “limit exists” means continuous

The limit can exist while the function is still discontinuous — if \(f(a)\) is missing or has the wrong value. All THREE conditions are required, not just the limit.

Confusing removable with the others

Removable means the LIMIT EXISTS (both sides agree) — it’s just a hole. Jumps have disagreeing sides; infinite discontinuities blow up. Only removable ones can be patched.

Forgetting to check the value, not just the limit

A function can have a perfect limit and still be discontinuous because \(f(a)\) was defined as something else. Always check all three boxes.

Solving for k using only one side

For piecewise functions, continuity needs the left piece, right piece, AND the defined value to all agree at the seam. Set the one-sided limits equal — both of them.

Assuming all discontinuities are holes

“Discontinuous” isn’t automatically “removable.” Identify the type first — a jump or asymptote can never be patched by redefining one point.

Ignoring domain edges

A rational function is discontinuous wherever its denominator is zero — check those points specifically. Polynomials never have this problem; they’re continuous everywhere.

On the AP exam

You can now spot exactly where and how a function breaks. The last lesson zooms out to the extremes: limits that run off to infinity, the asymptotes that mark them, and a powerful theorem that continuity unlocks — the guarantee that a continuous function hits every value in between. Asymptotes & the IVT is next — or head back to the Unit 1 Guide.