Removing Discontinuities

AP Calculus BC· AP Calculus BC CED — Limits and Continuity· 14 min read

1. What Is a Removable Discontinuity?★★☆☆☆⏱ 3 min

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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.

📘 Definition

Removable Discontinuity

Also called a 'hole' or 'point discontinuity'

A discontinuity at $x = a$ where the two-sided limit $lim_{x \to a} f (x)$ exists and is finite. The discontinuity can be removed by redefining $f (a)$ to equal this limit.

Example:

A rational function with a common factor in numerator and denominator has a removable discontinuity at the root of that common factor.

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

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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.

📐 Worked Example

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

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1. Factor the numerator and denominator completely:
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$x^3 - 2x^2 - 8x = x(x^2 - 2x - 8) = x(x-4)(x+2) \\ x^2 - 6x + 8 = (x-2)(x-4)$
3
1. Find all points of discontinuity: The denominator equals zero at $x = 2$ and $x = 4$ , so $f (x)$ is undefined (discontinuous) at both points.
4
1. Evaluate the limit at each point:
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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.
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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.
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1. Conclusion: Only $x = 4$ is a removable discontinuity.

3. Evaluating Limits at Removable Discontinuities★★★☆☆⏱ 4 min

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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.

📐 Worked Example

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

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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
1. Use the conjugate method to eliminate the square root. Multiply numerator and denominator by the conjugate of the numerator, $4 + x$ :
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$\lim_{x \to 16} \frac{(4 - \sqrt{x})(4 + \sqrt{x})}{(16 - x)(4 + \sqrt{x})}$
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1. Simplify the numerator: $(4 - x) (4 + x) = 16 - x$ , so the expression becomes:
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$\lim_{x \to 16} \frac{16 - x}{(16 - x)(4 + \sqrt{x})}$
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1. Cancel the common $(16 - x)$ term (valid because $x \to 16$ means $x \neq = 16$ , so $16 - x \neq = 0$ ) to get:
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$\lim_{x \to 16} \frac{1}{4 + \sqrt{x}}$
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1. Evaluate by substitution: $\frac{1}{4 + 16} = \frac{1}{8}$ . The limit is $\frac{1}{8}$ .

4. Removing Discontinuities by Redefining the Function★★★☆☆⏱ 3 min

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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 $x = 2$ ? $f (x) = {\frac{3 x ^{2} - 5 x - 2}{x - 2} c x \neq = 2 x = 2$

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

5. Common Pitfalls

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 $x \neq = a$ , which was required for the original function to be defined.

Correct move:

Always note the equality only holds for $x \neq = a$ , 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 $\frac{x - 9}{3 - x}$ .

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.

6. Quick Reference Cheatsheet

Category	Formula / Rule	Notes
Removable Discontinuity Definition	Discontinuity at $x = a$ is removable iff $lim_{x \to a} f (x)$ exists and is finite	$f (a)$ can be undefined or defined incorrectly, this does not change the classification
Non-Removable Discontinuity	Any discontinuity where the two-sided limit does not exist (finite)	Includes jump, infinite, and oscillating discontinuities; cannot be removed by redefining $f (a)$
Factoring Method for Limits	If $f (x) = \frac{( x - a ) g ( x )}{( x - a ) h ( x )}$ , then $lim_{x \to a} f (x) = \frac{g ( a )}{h ( a )}$	Only valid if $h (a) \neq = 0$ ; equality holds for all $x \neq = a$
Conjugate Method for Limits	For $lim_{x \to a} \frac{N ( x )}{c - x}$ , multiply numerator and denominator by $c + x$	Used for functions with square roots where factoring is not possible
Find $c$ for Continuity	If $f (x) = {g (x) c x \neq = a x = a$ , set $c = lim_{x \to a} g (x)$	This is the most common AP question type for this topic
Remove Discontinuity Rule	New continuous function $g (x) = {f (x) lim_{x \to a} f (x) x \neq = a x = a$	Only changes the value of $f (x)$ at the single point $x = a$
Indeterminate Form Clue	A $0/0$ form when evaluating $f (a)$ indicates a potential removable discontinuity	Always confirm the limit is finite after simplification, not all $0/0$ are removable

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · MCQ
Count removable discontinuities
2023 · FRQ
Find c for continuity at a point

What's Next

Removing discontinuities is a foundational skill for AP Calculus 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 multiple-choice and free-response questions. This topic also lays the groundwork for indeterminate forms and L'Hospital’s Rule for evaluating more complex limits later in the course.