Removing Discontinuities

students, have you ever watched a video freeze for a second and then load normally again? 📱 In calculus, a function can sometimes behave in a similar way: it may look broken at one point, but the “break” can be fixed. This lesson explains removing discontinuities, also called removable discontinuities. By the end, you should be able to tell when a function has a hole, why the limit may still exist, and how rewriting the function can make it continuous.

What You Will Learn

In this lesson, students, you will learn to:

• explain what a removable discontinuity is,
• identify when a hole can be fixed,
• use algebra to simplify expressions and find limits,
• connect removable discontinuities to continuity at a point,
• use examples to show how AP Calculus BC treats these situations.

The big idea is that a function can fail to be defined at one point, yet still have a perfectly good limit there. When that happens, the discontinuity may be removable. This idea is important because limits focus on what a function is doing near a point, not just at the point.

What Is a Removable Discontinuity?

A function has a removable discontinuity when the graph has a hole that can be filled in by redefining the function at one point. This usually happens when a factor causes both the numerator and denominator to be zero, creating an expression like $\frac{0}{0}$ that is undefined at a specific input.

For example, consider

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

If you plug in $x=1$, you get

$$f(1)=\frac{1^2-1}{1-1}=\frac{0}{0},$$

which is undefined. But if you factor the numerator,

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

so for $x\neq 1$,

$$f(x)=\frac{(x-1)(x+1)}{x-1}=x+1.$$

That means the graph follows the line $y=x+1$ everywhere except at $x=1$. The limit still exists:

$$\lim_{x\to 1}\frac{x^2-1}{x-1}=\lim_{x\to 1}(x+1)=2.$$

So the function has a hole at the point $(1,2)$, and the discontinuity is removable. To “remove” it, we could define a new function $g$ by setting $g(1)=2$. Then the graph becomes continuous at $x=1$. ✅

Why Limits Matter More Than the Original Formula

A limit asks, “What value does the function approach as $x$ gets close to a number?” It does not ask, “What is the function value exactly at that number?” This difference is the key to removable discontinuities.

Suppose

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

but $f(a)$ is missing or is not equal to $L$. Then the function is not continuous at $x=a$. If the only problem is the value at one point, and the limit exists, the discontinuity may be removable.

A function is continuous at $x=a$ when all three conditions are true:

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

For a removable discontinuity, condition 2 is often true, but condition 1 or 3 fails. That is why the point can be repaired by redefining the function value.

Think of a road with a tiny missing brick. If the rest of the road lines up perfectly, the gap can be fixed. But if the road suddenly changes direction or splits, that is not a removable problem. 🚧

Common Ways Removable Discontinuities Appear

The most common AP Calculus BC situations involve algebraic expressions that create $\frac{0}{0}$ when evaluated directly. Here are a few patterns to recognize.

1. Factoring and Canceling

Example:

$$h(x)=\frac{x^2+5x+6}{x+2}.$$

If we substitute $x=-2$, the denominator becomes $0$. Factor the numerator:

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

So for $x\neq -2$,

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

Then

$$\lim_{x\to -2}h(x)=\lim_{x\to -2}(x+3)=1.$$

The discontinuity is removable because the graph has a hole at $x=-2$ and the nearby values approach $1$.

2. Rationalizing Expressions

Sometimes radicals create a removable discontinuity.

Example:

$$k(x)=\frac{\sqrt{x+4}-2}{x}.$$

At $x=0$, direct substitution gives $\frac{0}{0}$. Multiply by the conjugate:

$$k(x)=\frac{\sqrt{x+4}-2}{x}\cdot\frac{\sqrt{x+4}+2}{\sqrt{x+4}+2}.$$

This gives

$$k(x)=\frac{(x+4)-4}{x(\sqrt{x+4}+2)}=\frac{x}{x(\sqrt{x+4}+2)}.$$

For $x\neq 0$,

$$k(x)=\frac{1}{\sqrt{x+4}+2}.$$

Now the limit is easy:

$$\lim_{x\to 0}k(x)=\frac{1}{\sqrt{4}+2}=\frac{1}{4}.$$

Again, the discontinuity can be removed by defining $k(0)=\frac14$.

How to Decide Whether a Discontinuity Is Removable

When students sees a function with a possible break, use a step-by-step strategy:

1. Try direct substitution. If you get a real number, the function may be continuous at that point.
2. If you get $\frac{0}{0}$, simplify algebraically. Factor, cancel, or rationalize.
3. Check whether the simplified form has a limit. If it does, the discontinuity may be removable.
4. Compare the limit to the actual function value. If the value is missing or different, the hole can often be filled by redefining the function.

Example:

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

Factor the numerator:

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

Then for $x\neq 2$,

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

So

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

Because $p(2)$ is undefined, there is a hole at $(2,0)$. Defining $p(2)=0$ removes the discontinuity.

Connection to Continuity at a Point and Over a Domain

Removing discontinuities is closely tied to continuity. If a function can be repaired at one point, then it can become continuous there. That matters because continuity lets us use many powerful theorems in calculus.

For a function to be continuous on an interval, it must have no breaks, jumps, or holes in that interval. A removable discontinuity is special because it is the one kind of break that can often be fixed without changing the rest of the graph.

This is different from other discontinuities:

• a jump discontinuity has a sudden change in value,
• an infinite discontinuity involves values growing without bound near a vertical asymptote,
• a removable discontinuity is just a hole.

Only the last one can be repaired by changing the function at one point. For AP Calculus BC, this distinction is important because continuity supports the use of the Intermediate Value Theorem and helps with limit reasoning.

For example, if a function is continuous everywhere except a removable discontinuity, you may be able to redefine the function to make it continuous on a whole interval. Then you can apply continuity-based ideas more confidently.

Real-World Meaning of a Hole

A removable discontinuity can model a situation where a formula works almost everywhere, but one input value is excluded because of how the formula was built. For instance, a rate formula might use division by time, and time $t=0$ may not make sense directly. But the limit as $t\to 0$ can still describe the behavior near the start. In science and engineering, this helps analysts understand what the model predicts even when the original formula fails at one point. 🔍

This is one reason limits are so useful: they let us study behavior near a point even when the function is not defined there.

Conclusion

students, removable discontinuities show how limits and continuity work together. A function can have a hole at one point, yet still approach a single value nearby. By simplifying algebraically, you can often find that limit and determine whether the discontinuity can be removed by redefining the function.

On the AP Calculus BC exam, this skill helps you interpret graphs, simplify expressions, and reason about continuity. Remember the main idea: if the only problem is one missing or incorrect function value, and the nearby behavior settles to a finite limit, the discontinuity is likely removable. That is the power of limits in calculus. ✨

Study Notes

• A removable discontinuity is a hole in a graph that can be fixed by redefining a function value.
• A removable discontinuity often appears when direct substitution gives $\frac{0}{0}$.
• To analyze one, simplify the expression by factoring, canceling, or rationalizing.
• If $\lim_{x\to a}f(x)$ exists and is finite, but $f(a)$ is undefined or different from that limit, the discontinuity may be removable.
• A function is continuous at $x=a$ when $f(a)$ is defined, $\lim_{x\to a}f(x)$ exists, and $\lim_{x\to a}f(x)=f(a)$.
• Removing a discontinuity means redefining the function at one point so the graph becomes continuous there.
• Removable discontinuities are different from jump discontinuities and infinite discontinuities.
• These ideas are important for limit questions, continuity questions, and theorem-based reasoning in AP Calculus BC.