Removing Discontinuities

Objectives

Explain what a removable discontinuity is and why it matters in calculus.
Identify when a function has a removable discontinuity by using limits.
Rewrite a function to remove a discontinuity when possible.
Connect removable discontinuities to continuity, factoring, and graph behavior.
Use examples to show how AP Calculus AB tests this idea.

Have you ever watched a video freeze for one frame and then keep playing? 📺 The story is still there, but there is a tiny gap. In calculus, a removable discontinuity is like that one-frame glitch: the graph has a hole, but the overall pattern still makes sense. students, this lesson will help you see when a function has a hole, how to find the limit at that point, and when the hole can be “fixed” by redefining the function.

What Is a Removable Discontinuity?

A function is continuous at $x=a$ if three things are true: $f(a)$ is defined, $\lim_{x\to a} f(x)$ exists, and $\lim_{x\to a} f(x)=f(a)$. If any one of these fails, the function is discontinuous at $x=a$.

A removable discontinuity happens when the limit exists, but the function is either not defined at that point or is defined with the wrong value. In other words, the graph has a hole that can be removed by changing one value.

This is important because calculus often cares more about the behavior near a point than the value at the point itself. For example, if a bridge design has a tiny gap in a model but the shape on both sides matches perfectly, engineers want to know whether that gap is just a missing point or a deeper problem. In calculus, a removable discontinuity is the “missing point” case.

A classic sign of a removable discontinuity is a rational function that simplifies after factoring. For example, consider

$$f(x)=\frac{x^2-9}{x-3}.$$

The numerator factors as $x^2-9=(x-3)(x+3)$, so for $x\ne 3$,

$$f(x)=x+3.$$

But at $x=3$, the original function is undefined because the denominator is $0$. The graph matches the line $y=x+3$ everywhere except for a hole at $x=3$. The limit exists:

$$\lim_{x\to 3} \frac{x^2-9}{x-3}=\lim_{x\to 3}(x+3)=6.$$

So the function has a removable discontinuity at $x=3$.

How to Recognize a Removable Discontinuity

The easiest way to spot one is to check whether the function can be simplified around the problematic point. A removable discontinuity usually appears in one of these ways:

A factor cancels from the numerator and denominator.
The function is undefined at a single point because of a zero denominator or a restricted expression.
The limit from both sides matches, but the function value is missing or incorrect.

Let’s look at a second example.

$$g(x)=\frac{x^2-4x+4}{x-2}.$$

Factor the numerator:

$$x^2-4x+4=(x-2)^2.$$

Then for $x\ne 2$,

$$g(x)=\frac{(x-2)^2}{x-2}=x-2.$$

The limit at $x=2$ is

$$\lim_{x\to 2} g(x)=\lim_{x\to 2}(x-2)=0.$$

But $g(2)$ is undefined in the original form. That means there is a hole at $(2,0)$. If we define a new function by setting $g(2)=0$, then the discontinuity is removed and the function becomes continuous at $x=2$.

This is a key AP Calculus AB idea: the limit tells us the value that would fill the hole. The hole can be removed if the limit exists.

How to Remove the Discontinuity

To remove a discontinuity, the goal is to define the function value at the hole so that it matches the limit. The process is usually:

Find the point where the function is discontinuous.
Simplify the expression, often by factoring.
Compute the limit at that point.
Assign the function that limit value at the missing point.

Suppose

$$h(x)=\frac{x^2-1}{x-1}$$

for $x\ne 1$.

Factor the numerator:

$$x^2-1=(x-1)(x+1).$$

So for $x\ne 1$,

$$h(x)=x+1.$$

Now compute the limit:

$$\lim_{x\to 1} h(x)=\lim_{x\to 1}(x+1)=2.$$

If the original definition leaves $h(1)$ undefined, then the discontinuity is removable. To remove it, define

$$h(1)=2.$$

Now the new function is continuous at $x=1$.

A removable discontinuity is very different from a jump discontinuity or an infinite discontinuity. In a jump discontinuity, the left-hand and right-hand limits exist but are not equal. In an infinite discontinuity, the function grows without bound near the point, often creating a vertical asymptote. Those are not fixed by changing one value at one point. A removable discontinuity is special because the graph is already behaving nicely on both sides.

Continuity, Holes, and Real-World Meaning

Why does this matter outside math class? Imagine a temperature sensor in a lab that skips one reading because of a glitch. If the readings just before and just after the missing value are almost the same, the missing value may be recovered by looking at the pattern. That is like a removable discontinuity. The system’s behavior is still predictable at that moment.

On graphs, a removable discontinuity looks like an open circle. The curve approaches the same height from both sides, but the actual point is missing or wrong. For AP Calculus AB, you should connect this visual idea to the formal limit definition:

$$\lim_{x\to a} f(x)=L$$

means that as $x$ gets close to $a$, the values of $f(x)$ get close to $L$, even if $f(a)$ is not equal to $L$ or not defined at all.

Here is a common exam-style example.

Let

$$p(x)=\begin{cases}

$\frac{x^2-16}{x-4}, & x\ne 4 \\$

$7, & x=4$

$\end{cases}$$$

For $x\ne 4$,

$$p(x)=\frac{(x-4)(x+4)}{x-4}=x+4.$$

Then

$$\lim_{x\to 4} p(x)=8.$$

But $p(4)=7$, so the function is not continuous at $x=4$. The discontinuity is removable because the limit exists. To make it continuous, redefine the function so that

$$p(4)=8.$$

This kind of question may ask you to determine whether a function is continuous, find the missing value that makes it continuous, or identify the graph feature at the discontinuity.

AP Calculus AB Thinking and Common Test Moves

When AP Calculus AB asks about removable discontinuities, the question often checks whether you can connect algebra and limits. You may need to:

factor and simplify an expression,
evaluate a limit,
compare the limit with the function value,
identify a hole on a graph,
or rewrite a piecewise function to make it continuous.

A useful strategy is to remember this chain:

$$\text{factor} \rightarrow \text{simplify} \rightarrow \text{limit} \rightarrow \text{compare with } f(a).$$

For example, if you are given

$$r(x)=\frac{x^2-5x+6}{x-2},$$

factor the numerator:

$$x^2-5x+6=(x-2)(x-3).$$

Then for $x\ne 2$,

$$r(x)=x-3.$$

$$\lim_{x\to 2} r(x)=-1.$$

If the function is undefined at $x=2$, the removable discontinuity can be removed by setting $r(2)=-1$.

A very important detail: a limit can exist even if the function value does not. That is why limits are so powerful. They describe the intended behavior of the function near the point, not just the value written in the rule.

Another common AP task is checking continuity on an interval. If a function has a removable discontinuity inside an interval, it is not continuous there unless the hole is filled. Continuity matters in theorems like the Intermediate Value Theorem, which requires continuity on a closed interval. If there is a hole, the theorem cannot be used until the discontinuity is removed.

Conclusion

Removable discontinuities show how calculus uses limits to understand behavior near a point. students, the big idea is simple: if the graph has a hole but the left- and right-hand behavior agree, then the limit exists and the hole may be fixed by redefining the function value. This connects directly to continuity, because a function is continuous only when its value matches its limiting behavior. On AP Calculus AB, you should be ready to factor expressions, find limits, identify holes, and decide whether a discontinuity can be removed. These skills build a strong foundation for later topics like theorems, asymptotes, and deeper limit reasoning.

Study Notes

A removable discontinuity is a hole in a graph that can be fixed by changing or defining one function value.
A function is continuous at $x=a$ if $f(a)$ is defined, $\lim_{x\to a} f(x)$ exists, and $\lim_{x\to a} f(x)=f(a)$.
If the limit exists but the function value is missing or wrong, the discontinuity may be removable.
Factoring and canceling common factors is the most common way to detect a removable discontinuity.
The limit gives the value needed to fill the hole and make the function continuous.
A removable discontinuity is different from a jump discontinuity or infinite discontinuity.
On graphs, removable discontinuities usually appear as open circles.
AP Calculus AB may ask you to find the missing value that makes a function continuous.
Continuity matters for theorems such as the Intermediate Value Theorem.
Limits describe the behavior near a point, which is why they are essential for understanding discontinuities.