The framework defines continuity at a point in one sentence with three parts. A function \(f\) is continuous at \(x = c\) provided that \(f(c)\) exists, that \(\lim_{x \to c} f(x)\) exists, and that

\[\lim_{x \to c} f(x) = f(c).\]

Three conditions, and the useful thing about them is that they fail independently. Every named discontinuity in the course is just a record of which one broke.

Break them one at a time

The buttons below break a single condition each. The checklist reports the three parts separately, and the verdict names the result.

Each button leaves the curve alone and changes only what happens at x = 2. "No value" removes the point; "wrong value" puts it somewhere the neighbourhood does not point to; "sides disagree" and "unbounded" break the limit itself, in the two ways a limit can fail here. The checklist is the definition read one clause at a time, and the verdict is the name that clause pattern earns. Repair works on exactly the two cases where the limit exists, which is the whole content of what "removable" means.

Start at “continuous” and step right. Each button changes only what happens at \(x = 2\); the curve on either side is untouched, which is the point. The three conditions are about three different things, and a function can satisfy any two of them while failing the third.

“No value” removes \(f(2)\). The first condition fails, the other two are fine — the limit is still 4, because a limit never consulted the point anyway.

“Wrong value” puts \(f(2) = 1\) while the neighbourhood still points at 4. Now \(f(2)\) exists and the limit exists, and they disagree, so the third condition is the one that breaks.

Those two cases share a name. The framework calls both a removable discontinuity, and the name is a promise: because the limit exists, you can define or redefine \(f(2)\) to equal it and continuity is restored. Press “repair it” in either case and watch all three ticks turn.

When the limit itself is what fails

The last two buttons break the second condition, and they break it in the two ways a limit can fail at a point where the function is otherwise well behaved.

“Sides disagree” gives one-sided limits of 4 and 1. Both exist; they are not equal; the two-sided limit does not exist. That is a jump discontinuity, and repair is impossible — there is no single value you could assign \(f(2)\) that would agree with both sides at once. The repair button is greyed out, which is the honest interface.

“Unbounded” uses \(1/(x-2)^2\), which grows without bound from both directions. The framework’s third category is a discontinuity due to a vertical asymptote, and again nothing can be assigned at \(x = 2\) to fix it.

Those are the three types the course names: removable, jump, and vertical asymptote. They are not a list to memorise so much as a consequence of there being three conditions and a limit having two ways to fail.

Continuity on an interval, and what you may assume

A function is continuous on an interval when it is continuous at every point in that interval. That sounds like infinitely much work, and it would be, except the framework hands you a shortcut worth knowing by heart: polynomial, rational, power, exponential, logarithmic, and trigonometric functions are continuous at every point in their domains.

The qualifier is the whole sentence. A rational function is continuous on its domain, and its domain excludes the zeros of the denominator, so \(\tfrac{1}{x-2}\) is continuous everywhere it is defined and there is no contradiction with the asymptote at 2. Students lose the point by reporting that a rational function is “continuous everywhere” or, in the other direction, by treating a removable hole as though it disqualified the function from being continuous anywhere.

Piecewise functions are where this gets asked with a parameter attached. For a piecewise function to be continuous at a boundary, the expression on one side must equal the expression on the other side at that boundary, and both must equal the value the function takes there. That is the same three conditions, written for the case where you control the pieces: set the two expressions equal at the boundary and solve.

A self-test with the buttons: predict, before pressing, which of the five cases will let the repair button light up, and why. Then check. The rule you should end up with is not “removable discontinuities are repairable” — that is circular — but “the limit existing is what makes repair possible,” which is a claim about the second condition and says nothing about the first or third.