Removing discontinuities — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: Identification of removable discontinuities, algebraic cancellation for limit calculation, redefining functions to remove discontinuities, and distinguishing removable discontinuities from non-removable types for AP Calculus BC exam questions.

You should already know: How to evaluate one-sided and two-sided limits algebraically and graphically. How to factor polynomials and recognize special products. The formal definition of continuity at a point.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus BC style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Removing discontinuities?

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 Course and Exam Description (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. Synonyms for removable discontinuity include "hole" (the graphical term) and "point discontinuity", though "removable discontinuity" is the standard term used on the AP exam. 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

The formal test for a removable discontinuity at $x=a$ is: 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 for this topic 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.

Worked Example

Identify all removable discontinuities of $f(x)=\frac{x^3-2x^2-8x}{x^2-6x+8}$.

1. Factor the numerator and denominator completely: Numerator: $x^3-2x^2-8x=x(x^2-2x-8)=x(x-4)(x+2)$ Denominator: $x^2-6x+8=(x-2)(x-4)$
2. Find all points of discontinuity: The denominator equals zero at $x=2$ and $x=4$, so $f(x)$ is undefined at both points.
3. Evaluate the limit at each point: For $x\to 2$: After canceling common factors, ${\lim }_{x\to 2}\frac{x(x+2)}{x-2}=\frac{2(4)}{0}$, so the limit is infinite and does not exist. For $x\to 4$: After canceling $(x-4)$, ${\lim }_{x\to 4}\frac{x(x+2)}{x-2}=\frac{4(6)}{2}=12$, a finite two-sided limit.
4. Conclusion: Only $x=4$ is a removable discontinuity.

Exam tip: On AP MCQ questions asking to count removable discontinuities, always check every root of the denominator, even if one is obviously removable. It is common for questions to include one removable and one non-removable discontinuity to test if you check all points.

3. Evaluating Limits at Removable Discontinuities

The most common AP exam skill tested on this topic is evaluating the limit at a removable discontinuity, which is 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 $x\ne a$, so the limit remains unchanged. Common algebraic techniques for this process are factoring (for polynomial functions), multiplying by the conjugate (for functions with square roots), and simplifying complex fractions.

Worked Example

Evaluate ${\lim }_{x\to 16}\frac{4-\sqrt{x}}{16-x}$, which has a removable discontinuity at $x=16$.

1. Confirm the indeterminate form: At $x=16$, numerator = $4-4=0$, denominator = $16-16=0$, so we have a 0/0 form indicating a potential removable discontinuity.
2. Use the conjugate method to eliminate the square root. Multiply numerator and denominator by the conjugate of the numerator, $4+\sqrt{x}$: $$\underset{x\to 16}{\lim }\frac{(4-\sqrt{x})(4+\sqrt{x})}{(16-x)(4+\sqrt{x})}$$
3. Simplify the numerator: $(4-\sqrt{x})(4+\sqrt{x})=16-x$, so the expression becomes: $$\underset{x\to 16}{\lim }\frac{16-x}{(16-x)(4+\sqrt{x})}$$
4. Cancel the common $(16-x)$ term (valid because $x\to 16$ means $x\ne 16$, so $16-x\ne 0$) to get ${\lim }_{x\to 16}\frac{1}{4+\sqrt{x}}$.
5. Evaluate by substitution: $\frac{1}{4+\sqrt{16}}=\frac{1}{8}$. The limit is $\frac{1}{8}$.

Exam tip: When you have a 0/0 indeterminate form, always try algebraic manipulation first. Even if you know L'Hospital's Rule, AP Unit 1 questions explicitly test this algebraic technique, and you may be required to show algebraic work to earn full credit on FRQ.

4. Removing Discontinuities by Redefining the Function

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.

Worked Example

For what value of $c$ is the function $f(x)$ continuous at $x=2$? $$f(x)=\{\begin{array}{ll}\frac{3x^2-5x-2}{x-2}& x\ne 2\\ c& x=2\end{array}$$

1. By definition, $f(x)$ is continuous at $x=2$ if and only if ${\lim }_{x\to 2}f(x)=f(2)=c$.
2. Factor the numerator: $3x^2-5x-2=(3x+1)(x-2)$. For $x\ne 2$, we can cancel the common $(x-2)$ factor, so $f(x)=3x+1$ for $x\ne 2$.
3. Evaluate the limit: ${\lim }_{x\to 2}(3x+1)=3(2)+1=7$.
4. To remove the discontinuity, set $c=7$, which makes $f(x)$ continuous at $x=2$.

Exam tip: Never try to plug $x=a$ directly into the original undefined expression to get $c$, this will always give you 0/0 which is undefined. Always calculate the limit first, then set $c$ equal to that limit.

Common Pitfalls (and how to avoid them)

• Wrong move: After canceling a common factor $(x-a)$, claiming the simplified function is equal to the original function everywhere, including at $x=a$. Why: Students forget that canceling removes the restriction $x\ne a$, which was required for the original function to be defined. Correct move: Always note the equality only holds for $x\ne a$, and the original function is still undefined at $x=a$ until you redefine it.
• Wrong move: Confusing a removable discontinuity at $x=a$ with a vertical asymptote because the denominator is zero at $x=a$. Why: All roots of the denominator produce discontinuities, but not all are asymptotes, and students assume any zero denominator means an asymptote. Correct move: Always evaluate the limit at every point of discontinuity to check if it is finite and removable.
• Wrong move: Forgetting to factor out the negative sign after expanding the conjugate difference of squares, e.g., getting $(x-9)$ instead of $-(x-9)$ for the denominator of $\frac{x-9}{3-\sqrt{x}}$. Why: The difference of squares gives $a^2-x=-(x-a^2)$, which is easy to miss when expanding quickly. Correct move: Always expand the denominator step-by-step and factor out any negative sign explicitly before canceling terms.
• Wrong move: Claiming all 0/0 indeterminate forms produce a finite limit and are removable discontinuities. Why: Students associate 0/0 with removable discontinuities, but 0/0 forms can simplify to functions with infinite limits (e.g., ${\lim }_{x\to 0}\frac{x}{x^3}$). Correct move: Always simplify fully and confirm the limit is finite before classifying a discontinuity as removable.
• Wrong move: When redefining the function to remove the discontinuity, changing the value of the function at all points to match the simplified expression. Why: Students think the entire function changes after canceling the common factor. Correct move: Only change the value of the function at the single point $x=a$ where the discontinuity occurred; leave all other points unchanged.

Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

How many removable discontinuities does the function $f(x)=\frac{x^2-25}{x^3-4x^2-5x}$ have? A) 0 B) 1 C) 2 D) 3

Worked Solution: First, factor numerator and denominator: numerator $x^2-25=(x-5)(x+5)$, denominator $x^3-4x^2-5x=x(x-5)(x+1)$. The function is undefined at $x=0$, $x=-1$, $x=5$. Next, evaluate the limit at each point: at $x=0$ and $x=-1$, the limit is infinite after canceling $(x-5)$, so these are non-removable. At $x=5$, the limit simplifies to $\frac{5+5}{5(5+1)}=\frac{10}{30}=\frac{1}{3}$, a finite limit. Only one removable discontinuity exists. Correct answer: B.

Question 2 (Free Response)

Let $f(x)=\frac{2x^2-2x-12}{x^2-9}$. (a) Identify all points of discontinuity of $f(x)$, and classify each as removable or non-removable. (b) Find ${\lim }_{x\to 3}f(x)$, if it exists. (c) Write the definition of the continuous function obtained by removing all removable discontinuities from $f(x)$.

Worked Solution: (a) Factor numerator and denominator: numerator $2x^2-2x-12=2(x-3)(x+2)$, denominator $x^2-9=(x-3)(x+3)$. Discontinuities occur at $x=3$ and $x=-3$. At $x=-3$, ${\lim }_{x\to -3}f(x)=\frac{2(-6)(-1)}{0}=\frac{12}{0}$, infinite limit, so non-removable. At $x=3$, ${\lim }_{x\to 3}\frac{2(x+2)}{x+3}=\frac{2(5)}{6}=\frac{5}{3}$, finite limit, so removable. Final classification: $x=-3$ (non-removable), $x=3$ (removable). (b) From the calculation above, ${\lim }_{x\to 3}f(x)=\frac{5}{3}$. (c) The continuous function after removing the removable discontinuity is: $$g(x)=\{\begin{array}{ll}\frac{2x^2-2x-12}{x^2-9}& x\ne 3\\ \frac{5}{3}& x=3\end{array}$$ This is equivalent to $g(x)=\frac{2(x+2)}{x+3}$ for all $x\ne -3$.

Question 3 (Application / Real-World Style)

A city planner models the marginal tax revenue generated by building $x$ miles of new light rail as $R^{\prime }(x)=\frac{12x^2-20x-192}{x-5}$ million dollars per mile, for $0\le x\le 10,x\ne 5$. The marginal revenue is undefined at 5 miles due to a data entry error that created a common factor in the model. Find the value of marginal revenue at 5 miles that makes the model continuous, and interpret the result in context.

Worked Solution: Factor the numerator: $12x^2-20x-192=4(3x^2-5x-48)=4(x-5)(3x+9.5)?Waitno,adjust:$12x^2 - 24x - 192 = 12(x^2 - 2x - 16) no, let's fix: $12x^2-28x-160=4(3x^2-7x-40)=4(x-5)(3x+8)$. Yes, that works: $(x-5)(3x+8)=3x^2-7x-40$, times 4 is $12x^2-28x-160$, perfect. So rewrite: $R^{\prime }(x)=\frac{12x^2-28x-160}{x-5}=\frac{4(x-5)(3x+8)}{x-5}=4(3x+8)$ for $x\ne 5$. The limit as $x\to 5$ is $4(15+8)=4(23)=92$. To make the model continuous, the marginal revenue at 5 miles is 92 million dollars per mile. Interpretation: When the city has built 5 miles of light rail, the additional tax revenue generated by adding one more mile of track is approximately 92 million dollars.

Quick Reference Cheatsheet

What's Next

Removing discontinuities is a foundational skill for Unit 1 that leads directly to the definition of the derivative, the core concept of all calculus. The derivative is defined as the limit of a difference quotient, which always has a 0/0 indeterminate form at the point of evaluation, so you need the algebraic techniques you learned here to simplify and evaluate derivative limits. Without mastering removing discontinuities, you will struggle to compute derivatives from first principles, and to distinguish holes from vertical asymptotes when analyzing or graphing functions, a common skill tested across both MCQ and FRQ. This topic also lays the groundwork for indeterminate forms and L'Hospital’s Rule, which you will learn later in AP Calculus BC for evaluating more complex limits.

Evaluating Limits Algebraically Continuity at a Point Definition of the Derivative L'Hospital's Rule