Removing Discontinuities

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

1. What is Removing a Discontinuity?★☆☆☆☆⏱ 2 min

Removing a discontinuity (also called filling a hole) is the process of redefining a function at a single point to make it continuous at that point, with no changes to the function anywhere else. This topic falls in Unit 1, which accounts for 10-12% of your total AP Calculus AB exam score, and appears in both multiple-choice and free-response questions.

📘 Definition

Removable Discontinuity

A discontinuity at $x = a$ is removable if and only if the two-sided limit of $f (x)$ as $x \to a$ exists, even if $f (a)$ is undefined or $f (a)$ does not equal the limit. Unlike non-removable discontinuities (jump or infinite), removable discontinuities can be fixed by changing only the value of $f (a)$ to match the existing limit.

2. Identifying Removable Discontinuities★★☆☆☆⏱ 4 min

Before you can remove a discontinuity, you must first correctly identify it as removable, rather than non-removable. Recall the three required conditions for continuity at a point $x = a$ :

$f (a)$ is defined
$lim_{x \to a} f (x)$ exists
$lim_{x \to a} f (x) = f (a)$

A discontinuity is removable if only conditions 1 or 3 fail, but condition 2 still holds (the two-sided limit exists). If condition 2 fails, the discontinuity is non-removable and cannot be fixed by changing $f (a)$ . Most removable discontinuities arise in rational functions when the numerator and denominator share a common linear factor $(x - a)$ , creating a $0/0$ indeterminate form.

📐 Worked Example

Identify all removable discontinuities of $f (x) = \frac{x ^{2} - 2 x - 8}{x ^{2} - 16}$ .

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Step 1: Find all points where $f (x)$ is undefined. Factor the numerator and denominator:
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$x^2 - 2x - 8 = (x-4)(x+2) \quad \text{and} \quad x^2 - 16 = (x-4)(x+4)$
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The denominator equals zero at $x = 4$ and $x = - 4$ , so $f (x)$ is discontinuous at both points.
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Step 2: Check for common factors. The numerator and denominator share a common factor of $(x - 4)$ , so the discontinuity at $x = 4$ is a candidate for being removable.
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Step 3: Check if the limit exists at each discontinuity. For $x = 4$ :
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$\lim_{x \to 4} f(x) = \lim_{x \to 4} \frac{(x-4)(x+2)}{(x-4)(x+4)} = \lim_{x \to 4} \frac{x+2}{x+4} = \frac{6}{8} = \frac{3}{4}$
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The limit exists, so the discontinuity at $x = 4$ is removable. For $x = - 4$ , the left and right limits approach $\pm \infty$ , so the limit does not exist, meaning the discontinuity is non-removable.
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Step 4: Conclusion: The only removable discontinuity is at $x = 4$ .

Exam tip:

Always check every root of the denominator, not just the one that cancels. AP exam questions almost always include one non-removable discontinuity alongside a removable one to test if you check all discontinuities.

3. Removing Discontinuities by Factoring★★☆☆☆⏱ 4 min

Once you have identified a removable discontinuity at $x = a$ , removing it requires you to define or redefine $f (a)$ to equal the value of $lim_{x \to a} f (x)$ , which satisfies all three continuity conditions at $x = a$ . For polynomial rational functions, the standard method to find this limit is factoring and canceling common factors.

📐 Worked Example

What value of $f (4)$ makes $f (x) = \frac{x ^{2} - 2 x - 8}{x ^{2} - 16}$ continuous at $x = 4$ ?

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Step 1: Confirm the discontinuity is removable. We already know $lim_{x \to 4} f (x) = \frac{3}{4}$ , so the discontinuity is removable.
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Step 2: For $f (x)$ to be continuous at $x = 4$ , we need $lim_{x \to 4} f (x) = f (4)$ by the definition of continuity.
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Step 3: Assign $f (4)$ equal to the limit, so $f (4) = \frac{3}{4}$ .
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Step 4: Verify all continuity conditions are satisfied: $f (4)$ is defined, $lim_{x \to 4} f (x)$ exists, and the two values are equal, so $f (x)$ is now continuous at $x = 4$ .

Exam tip:

Never skip explicitly stating that $f (a)$ equals the limit. AP graders will deduct points if you only simplify the function and do not explicitly define the new value of $f (a)$ to remove the discontinuity.

4. Removing Discontinuities by Rationalizing★★★☆☆⏱ 4 min

Not all removable discontinuities come from factoring polynomial rational functions. Another common case is functions with radicals, where the function is undefined at $x = a$ (giving a $0/0$ indeterminate form) but the limit still exists. To evaluate this limit, we multiply both the numerator and denominator by the conjugate of the radical expression to eliminate the radical, which reveals a common $(x - a)$ factor that can be canceled.

📐 Worked Example

Find the value of $k$ that makes $h (x) = {\frac{x + 5 - 3}{x - 4} k x \neq = 4 x = 4$ continuous at $x = 4$ .

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Step 1: Check for indeterminate form at $x = 4$ . At $x = 4$ , the numerator is $4 + 5 - 3 = 0$ , and the denominator is $4 - 4 = 0$ , so we have a $0/0$ indeterminate form, indicating a possible removable discontinuity.
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Step 2: Multiply numerator and denominator by the conjugate of the numerator, $x + 5 + 3$ :
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$\lim_{x \to 4} \frac{(\sqrt{x+5} - 3)(\sqrt{x+5} + 3)}{(x-4)(\sqrt{x+5} + 3)} = \lim_{x \to 4} \frac{(x+5) - 9}{(x-4)(\sqrt{x+5} + 3)} = \lim_{x \to 4} \frac{x-4}{(x-4)(\sqrt{x+5} + 3)}$
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Step 3: Cancel the common $(x - 4)$ factor and evaluate the limit:
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$\lim_{x \to 4} \frac{1}{\sqrt{x+5} + 3} = \frac{1}{\sqrt{9} + 3} = \frac{1}{6}$
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Step 4: For continuity at $x = 4$ , set $k = lim_{x \to 4} h (x) = \frac{1}{6}$ , so $k = \frac{1}{6}$ removes the discontinuity.

Exam tip:

Always multiply both the numerator and denominator by the conjugate. If you only multiply the numerator, you change the value of the expression, leading to an incorrect limit.

5. Common Pitfalls

Wrong move:

After canceling $(x - a)$ from numerator and denominator, you conclude the original function is automatically continuous at $x = a$ without redefining $f (a)$ .

Why:

Students confuse the simplified function (defined at $a$ ) with the original function, which remains undefined at $a$ unless explicitly changed.

Correct move:

Always explicitly state that $f (a)$ must be set equal to $lim_{x \to a} f (x)$ to remove the discontinuity, even after simplifying.

Wrong move:

You assume any $0/0$ indeterminate form means the discontinuity is removable, without checking that the two-sided limit exists.

Why:

Students are taught $0/0$ usually means removable, but rare cases can still have non-existent limits.

Correct move:

Always evaluate the two-sided limit after simplifying to confirm it exists before calling the discontinuity removable.

Wrong move:

You multiply only the numerator by the conjugate when rationalizing, leaving the denominator unchanged.

Why:

Students rush to eliminate the radical and forget the multiplication rule for equality.

Correct move:

Whenever you multiply by the conjugate, always multiply both numerator and denominator by the same term to keep the expression equivalent to the original.

Wrong move:

You cancel $(x - a)$ from numerator and denominator when the numerator is non-zero at $x = a$ , leading to a false claim of a removable discontinuity.

Why:

Students develop a habit of canceling any matching factor, even when the numerator is not zero.

Correct move:

Always check that the numerator is also zero at $x = a$ before looking for common factors to cancel.

Wrong move:

For a piecewise function with $f (x) = g (x)$ for $x \neq = a$ , you set $f (a) = g (a)$ instead of $lim_{x \to a} g (x)$ .

Why:

$g (x)$ is often undefined at $a$ in this setup, so students incorrectly plug in $a$ to the simplified $g (x)$ .

Correct move:

Always evaluate the limit of $g (x)$ as $x \to a$ to get the correct value for $f (a)$ .

6. Quick Reference Cheatsheet

Category	Rule/Value	Notes
Removable Discontinuity Definition	Discontinuity at $x = a$ is removable iff $lim_{x \to a} f (x)$ exists	Occurs when $f (a)$ is undefined or $f (a) \neq = lim_{x \to a} f (x)$
Value to Remove Discontinuity	$f (a) = lim_{x \to a} f (x)$	Only changes the function at $x = a$ , no changes for $x \neq = a$
Method 1: Polynomial Rational Functions	Factor numerator/denominator, cancel common $(x - a)$	Cancellation is only valid for $x \neq = a$ , which is fine for limit evaluation
Method 2: Functions with Radicals	Multiply numerator and denominator by the conjugate of the radical	Eliminates the radical to reveal a common $(x - a)$ factor
Non-Removable Discontinuity Check	Discontinuity is non-removable if $lim_{x \to a} f (x)$ does not exist	Common cases: infinite discontinuity, jump discontinuity
Continuity After Removal	All 3 continuity conditions are satisfied	Only removes the discontinuity at $x = a$ , other discontinuities remain

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
Find k for continuous piecewise function
2019 · FRQ
Classify discontinuities as removable

What's Next

Removing discontinuities is a foundational skill for almost all subsequent topics in AP Calculus AB, because it reinforces the core idea that the limit of a function as $x \to a$ depends only on values near $a$ , not at $a$ itself. Immediately after this topic, you will apply this skill to computing derivatives using the limit definition, where you will almost always encounter a $0/0$ indeterminate form that requires canceling a common factor or rationalizing to evaluate the derivative. Without mastering this process, you will struggle to compute limit-definition derivatives, the first core skill of differential calculus, and analyze continuity in FRQ questions across the course.