Removing discontinuities — AP Calculus AB Study Guide

For: AP Calculus AB candidates sitting AP Calculus AB.

Covers: Definition of removable discontinuities (holes), identifying removable discontinuities, factoring to cancel common factors, rationalizing to remove discontinuities, and finding the constant value that removes a discontinuity, connected to limit existence and continuity.

You should already know: How to evaluate two-sided limits algebraically, How to factor polynomials and difference of squares, The three conditions for 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 AB 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?

Removing a discontinuity (also called filling a hole, or eliminating a removable discontinuity) 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. According to the AP Calculus AB 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. Questions on removing discontinuities appear in both multiple-choice (MCQ) and free-response (FRQ) sections, most often as part of a larger question about continuity, limits, or derivatives.

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 just the value of $f(a)$ to match the existing limit, hence the name "removing" the discontinuity.

2. Identifying Removable Discontinuities

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 $x=a$:

1. $f(a)$ is defined
2. ${\lim }_{x\to a}f(x)$ exists
3. ${\lim }_{x\to a}f(x)=f(a)$

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 limit does not exist), the discontinuity is non-removable and cannot be fixed by changing $f(a)$. Most often, removable discontinuities arise in rational functions when the numerator and denominator share a common linear factor $(x-a)$, which creates a $0/0$ indeterminate form at $x=a$, meaning the limit can still exist after canceling the common factor. Other contexts include functions with radicals that can be rationalized to reveal a common factor, and piecewise functions where the limit of the rule for $x\ne a$ exists but does not match the defined value at $x=a$.

Worked Example

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

1. Step 1: Find all points where $f(x)$ is undefined. Factor the numerator and denominator: $$x^2-2x-8=(x-4)(x+2)\text{and}{x}^2-16=(x-4)(x+4)$$ The denominator equals zero at $x=4$ and $x=-4$, so $f(x)$ is undefined (and discontinuous) at both points.
2. 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.
3. Step 3: Check if the limit exists at each discontinuity. For $x=4$: $$\underset{x\to 4}{\lim }f(x)=\underset{x\to 4}{\lim }\frac{(x-4)(x+2)}{(x-4)(x+4)}=\underset{x\to 4}{\lim }\frac{x+2}{x+4}=\frac{6}{8}=\frac{3}{4}$$ The limit exists, so the discontinuity at $x=4$ is removable. For $x=-4$, ${\lim }_{x\to -4^{-}}f(x)=-\infty$ and ${\lim }_{x\to -4^{+}}f(x)=+\infty$, so the limit does not exist, meaning the discontinuity at $x=-4$ is non-removable.
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

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.

A key point to remember: canceling a common factor $(x-a)$ is only valid for $x\ne a$, but this is perfectly fine for evaluating the limit as $x\to a$, because the limit depends only on values of $x$ near $a$, not at $a$ itself. After canceling the common factor, you can evaluate the simplified function at $x=a$ to get the limit, which is exactly the value you need to assign to $f(a)$ to remove the discontinuity.

Worked Example

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

1. Step 1: Confirm the discontinuity is removable. From the previous worked example, we already know ${\lim }_{x\to 4}f(x)=\frac{3}{4}$, so the discontinuity is removable.
2. 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.
3. Step 3: Assign $f(4)$ equal to the limit, so $f(4)=\frac{3}{4}$.
4. 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

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

The conjugate of an expression of the form $\sqrt{x+b}-c$ is $\sqrt{x+b}+c$, because their product is $(\sqrt{x+b}{)}^2-c^2=(x+b)-c^2$, which eliminates the radical and produces a linear term that will often match the factor in the denominator. After simplifying, you cancel the common factor and evaluate the limit as before, then assign that limit as the value of $f(a)$ to remove the discontinuity.

Worked Example

Find the value of $k$ that makes $h(x)=\{\begin{array}{ll}\frac{\sqrt{x+5}-3}{x-4}& x\ne 4\\ k& x=4\end{array}$ continuous at $x=4$.

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

• 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 $x\ne a$, 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. Practice Questions (AP Calculus AB Style)

Question 1 (Multiple Choice)

Let $f(x)=\frac{x^3-5x^2+6x}{x^2-4}$. For how many values of $x$ does $f(x)$ have a removable discontinuity? A) 0 B) 1 C) 2 D) 3

Worked Solution: First, factor the numerator and denominator: $x^3-5x^2+6x=x(x-2)(x-3)$, and $x^2-4=(x-2)(x+2)$. $f(x)$ is undefined at $x=2$ and $x=-2$. The only common factor is $(x-2)$, so only the discontinuity at $x=2$ is removable. At $x=-2$, the numerator is $(-2)(-4)(-5)=-40\ne 0$, so the limit does not exist, and the discontinuity is non-removable. There are no other discontinuities, so there is exactly 1 removable discontinuity. The correct answer is B.

Question 2 (Free Response)

Let $f(x)=\frac{3x^2-10x+8}{x^2-4x+3}$. (a) Find all $x$-values where $f(x)$ has discontinuities, and classify each as removable or non-removable. (b) Find the value that should be assigned to $f(x)$ at each removable discontinuity to remove the discontinuity. (c) After removing all removable discontinuities, is the resulting function continuous at the removable discontinuity point? Justify your answer.

Worked Solution: (a) Factor numerator and denominator: $3x^2-10x+8=(3x-4)(x-2)$, and $x^2-4x+3=(x-1)(x-3)$. Discontinuities occur at $x=1$ and $x=3$, where the denominator is zero. At both points, the numerator is non-zero: $f(1)=(3-4)(1-2)=(-1)(-1)=1\ne 0$, $f(3)=(9-4)(3-2)=(5)(1)=5\ne 0$, so there are no common factors, meaning there are no removable discontinuities. All discontinuities (at $x=1$ and $x=3$) are non-removable. (b) There are no removable discontinuities, so no values need to be assigned. (c) Since there are no removable discontinuities, there is no change to the function, and the function remains discontinuous at $x=1$ and $x=3$. If there were a removable discontinuity, after assigning $f(a)={\lim }_{x\to a}f(x)$, the function would be continuous at $x=a$, but no such point exists here.

Question 3 (Application / Real-World Style)

The average rate of change of the volume of a balloon as it is inflated, from time $t=1$ minute to time $t=x$ minutes, is given by $r(x)=\frac{x^2+x-2}{x-1}$ cubic feet per minute, for $x\ne 1$. What instantaneous rate of change of volume at $t=1$ minute removes the discontinuity at $x=1$? Give units and interpret your answer in context.

Worked Solution: To remove the discontinuity at $x=1$, we calculate $r(1)={\lim }_{x\to 1}\frac{x^2+x-2}{x-1}$. Factor the numerator: $x^2+x-2=(x-1)(x+2)$, so cancel the common factor for $x\ne 1$: $$\underset{x\to 1}{\lim }(x+2)=1+2=3$$ The value that removes the discontinuity is 3 cubic feet per minute. In context, this means the instantaneous rate at which the volume of the balloon is increasing at $t=1$ minute is 3 cubic feet per minute, matching the limit of the average rate of change as the time interval shrinks to zero around $t=1$.

7. Quick Reference Cheatsheet

8. 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 the process of removing discontinuities, you will struggle to compute limit-definition derivatives, the first core skill of differential calculus. This topic also lays the groundwork for analyzing continuity of piecewise and derivative functions in FRQ questions across the rest of the course.

• Limits and Continuity Overview
• Continuity at a Point
• Evaluating Limits Algebraically
• Limit Definition of the Derivative