A removable discontinuity is a point where the limit of a function exists from both sides, but the function is either undefined there or defined with a different value. On the graph it looks like a hole, and you can "remove" it by redefining the function at that single point.
A removable discontinuity is the friendliest kind of discontinuity. The function's limit exists at the point (the left-hand and right-hand limits agree), but the function itself either isn't defined there or is defined at the wrong value. Graphically, it's a single hole, sometimes with a stray dot floating above or below it.
The classic source is a rational function with a common factor on top and bottom. Take f(x) = (x² - 4)/(x - 2). At x = 2 you get 0/0, so the function is undefined. But factoring gives (x - 2)(x + 2)/(x - 2), which simplifies to x + 2 everywhere except x = 2. So the limit as x approaches 2 is 4, and the graph is the line y = x + 2 with a hole at (2, 4). It's called "removable" because you can patch the hole by defining f(2) = 4, and suddenly the function is continuous. Compare that to a jump or a vertical asymptote, where no single-point patch can save you.
Removable discontinuities live in Unit 1 of AP Calculus (Limits and Continuity), right at the heart of the formal definition of continuity. The CED's continuity definition has three conditions: f(a) exists, the limit as x approaches a exists, and the two are equal. A removable discontinuity is exactly what happens when condition two passes but condition one or three fails. That makes it the perfect test of whether you actually understand the definition or just memorized "continuous means no breaks."
It also matters because the 0/0 form you get at a removable discontinuity is the same indeterminate form that drives limit evaluation by algebraic manipulation in Unit 1 and L'Hôpital's Rule in Unit 4. Recognizing "this is a hole, not an asymptote" tells you the limit is findable, you just have to do the algebra.
Limits from Both Sides (Unit 1)
A removable discontinuity is defined by its limit behavior. The left-hand and right-hand limits exist and match, which is what separates a hole from a jump discontinuity, where the one-sided limits disagree.
Asymptote (Unit 1)
Both holes and vertical asymptotes come from zeros in the denominator of rational functions. The difference is whether the factor cancels. If it cancels, you get a hole; if it doesn't, the function blows up to a vertical asymptote and the discontinuity is not removable.
Piecewise Functions (Unit 1)
Piecewise functions are the AP exam's favorite way to manufacture a removable discontinuity on purpose. A classic problem defines f(x) one way everywhere except at x = a, then asks what value of f(a) (or a constant k) makes the function continuous. The answer is always the limit.
L'Hôpital's Rule (Unit 4)
The 0/0 you get when plugging into a removable discontinuity is an indeterminate form. In Unit 1 you resolve it by factoring and canceling, but L'Hôpital's Rule gives you a derivative-based shortcut for the same situation later in the course.
Removable discontinuities show up most often in multiple-choice questions in two flavors. First, limit evaluation: you plug in, get 0/0, and have to factor, cancel, and then substitute to find the limit at the hole. Second, classification: given a function (as an equation, graph, or table), identify where it's discontinuous and what type of discontinuity it is. Graphical and analytical representations of limits are especially useful here, since a hole is easy to spot on a graph but hides inside the algebra of a rational function.
The most common free-response-style move is the "make it continuous" question. You're given a piecewise function with an unknown constant and asked to find the value that removes the discontinuity. Your job is to set the limit equal to the function value and solve. Always justify with the three-part definition of continuity, not just "the pieces meet."
Both come from a zero in the denominator of a rational function, so they look identical in the equation until you factor. If the problem factor cancels with the numerator, the limit exists and you have a removable hole. If it doesn't cancel, the function shoots off to infinity and you have a vertical asymptote, which cannot be removed by redefining one point. Quick check with (x² - 4)/(x - 2): x = 2 cancels, so it's a hole. In 1/(x - 2), nothing cancels, so x = 2 is an asymptote.
A removable discontinuity is a single hole in the graph where the limit exists but the function value is missing or doesn't match the limit.
It's called removable because redefining the function at that one point (setting f(a) equal to the limit) makes the function continuous there.
Removable discontinuities usually come from rational functions where a factor cancels from both the numerator and denominator, producing the 0/0 indeterminate form.
Whether the factor cancels is the dividing line: canceling gives a hole, not canceling gives a vertical asymptote.
On the exam, find the limit at a hole by factoring and canceling first, then substituting; plugging in directly just gives 0/0.
A removable discontinuity fails the continuity definition because f(a) is undefined or f(a) does not equal the limit, even though the limit itself exists.
It's a point where a function's two-sided limit exists, but the function is undefined there or has a value that doesn't match the limit. On a graph it appears as a hole, and defining the function to equal the limit at that point makes it continuous.
Yes. That's exactly what makes it removable. The left and right limits agree, so the limit exists; the function just fails to be defined correctly at that one point. This is the big difference from jump and infinite discontinuities, where the two-sided limit doesn't exist.
Both come from a zero in the denominator, but a hole appears when the problem factor cancels (like x = 2 in (x² - 4)/(x - 2), where the limit is 4), while an asymptote appears when it doesn't cancel (like x = 2 in 1/(x - 2), where the function goes to infinity). Factor first, then decide.
No, not at that point. Continuity requires three things: f(a) exists, the limit exists, and they're equal. A removable discontinuity breaks the first or third condition, so the function is discontinuous there even though the limit is perfectly fine.
Find the limit at the point (usually by factoring and canceling the common factor), then redefine the function so its value equals that limit. For example, if f(x) = (x² - 4)/(x - 2), setting f(2) = 4 fills the hole and makes f continuous at x = 2.