Removing Discontinuities — AP Calculus BC
1. What Is a Removable Discontinuity? ★★☆☆☆ ⏱ 3 min
A removable discontinuity is a point $x=a$ where a function $f(x)$ is discontinuous, but the two-sided limit $\lim_{x \to a} f(x)$ exists and is finite. The discontinuity arises only because $f(a)$ is undefined, or $f(a)$ is defined but not equal to this limit. 'Removing' the discontinuity means redefining $f(a)$ to equal the existing limit, resulting in a function that is continuous at $x=a$.
According to the AP Calculus CED, this topic falls within Unit 1: Limits and Continuity, which accounts for 10-12% of the total AP exam score. Removable discontinuities are most commonly tested in multiple-choice questions, but can also appear as a component of free-response questions testing continuity or piecewise function definitions.
The key property that makes a discontinuity removable is the existence of a finite two-sided limit, which differentiates it from non-removable types (jump, infinite, oscillating) where the two-sided limit does not exist.
2. Identifying Removable Discontinuities ★★☆☆☆ ⏱ 4 min
The formal test for a removable discontinuity at $x=a$ has two requirements: 1) $f(x)$ is discontinuous at $x=a$, and 2) $\lim_{x \to a} f(x)$ exists and is finite. For rational functions (the most common context on the AP exam), removable discontinuities occur when there is a common shared factor between the numerator and denominator. The root of this common factor is the location of the removable discontinuity.
To identify removable discontinuities for a rational function $f(x) = \frac{N(x)}{D(x)}$, follow two core steps: first, find all points where $D(x)=0$ (these are all points of discontinuity). Second, for each such point, check if the two-sided limit exists and is finite. If yes, it is removable; if not, it is non-removable (usually an asymptote). It is critical to check every point of discontinuity, not just assume all are removable.
3. Evaluating Limits at Removable Discontinuities ★★★☆☆ ⏱ 4 min
The most common AP exam skill tested on this topic is evaluating the limit at a removable discontinuity, a core Unit 1 skill. When you have a $0/0$ indeterminate form (the signature of a removable discontinuity), you use algebraic manipulation to eliminate the common factor causing the zero in the denominator, then evaluate the resulting limit by substitution.
The intuition behind this technique is that the limit as $x \to a$ only depends on the behavior of $f(x)$ near $x=a$, not at $x=a$. Canceling the common $(x-a)$ term does not change the value of $f(x)$ for any $x \neq a$, so the limit remains unchanged. Common algebraic techniques are factoring (for polynomials), multiplying by the conjugate (for functions with square roots), and simplifying complex fractions.
4. Removing Discontinuities by Redefining the Function ★★★☆☆ ⏱ 3 min
Once you have confirmed that $x=a$ is a removable discontinuity and calculated $\lim_{x \to a}f(x) = L$, removing the discontinuity means redefining $f(a)$ to equal $L$, which makes the function continuous at $x=a$. This is the origin of the term 'removing' the discontinuity: we eliminate the discontinuity by fixing the function's value at that single point.
A very common AP exam question asks: 'For what value of $c$ is the function continuous at $x=a$?' where the function is defined as a piecewise function with a constant at $x=a$. This is exactly a removing discontinuities question: the answer is $c$ equal to the limit of the function as $x \to a$. The entire process only changes the value of the function at one point, leaving all other values unchanged.
Common Pitfalls
Why: Students forget that canceling removes the restriction $x \neq a$, which was required for the original function to be defined.
Why: All roots of the denominator produce discontinuities, but not all are asymptotes, and students assume any zero denominator means an asymptote.
Why: The difference of squares gives $a^2 - x = -(x-a^2)$, which is easy to miss when expanding quickly.
Why: Students associate 0/0 with removable discontinuities, but 0/0 forms can simplify to functions with infinite limits (e.g., $\lim_{x \to 0} \frac{x}{x^3}$).
Why: Students think the entire function changes after canceling the common factor.