Removing Discontinuities

Lesson ~10 min read

In simple terms: In simple terms, removing discontinuities is about finding and fixing "holes" or "breaks" in a graph to make it a single, unbroken line.

Why this matters

Imagine you're driving on a brand-new highway stretching from Dallas to Austin. The road is perfectly smooth, except for one spot. A construction crew forgot to pave a single, tiny section, leaving a jarring pothole. The road almost connects, but that one missing point makes the ride discontinuous. You can see the road on the other side, you know exactly where you should be, but there's a gap.

In calculus, we call that gap a "removable discontinuity." It's a hole in the graph of a function. The good news is, just like a road crew can fill a pothole, we can "fix" these mathematical holes. We can define (or redefine) a single point to make the function continuous, creating a perfectly smooth path. In this lesson, we'll learn how to spot these holes and, more importantly, how to patch them up so our functions are perfectly connected.

Concept overview

flowchart TD
    A[Start: Analyze function f(x) at x=c] --> B{Does lim (x->c) f(x) exist?};
    B -- No --> C[Non-removable discontinuity (Jump or Infinite)];
    B -- Yes --> D{Does f(c) exist?};
    D -- No --> E[Removable Discontinuity (Hole)];
    D -- Yes --> F{Is f(c) == lim (x->c) f(x)?};
    F -- No --> E;
    F -- Yes --> G[Function is continuous at x=c];
    E --> H[To remove, redefine f(c) = lim (x->c) f(x)];
    H --> G;
    C --> I[Cannot be made continuous by redefining one point];

Core explanation

Hello! I'm Saavi, and I'm here to guide you through one of the most satisfying ideas in early calculus: fixing broken functions.

Before we can fix something, we need to know what "broken" means. Remember our three conditions for continuity at a point x = c:

The function must be defined at c. (f(c) exists.)
The limit of the function as x approaches c must exist.
The function's value must equal the limit. (f(c) = lim (x→c) f(x))

When a function fails one of these conditions, it's discontinuous at c. But not all discontinuities are created equal.

Removable Discontinuities: The Pothole in the Road

A removable discontinuity is the most polite type of break. It's a single point missing from an otherwise well-behaved graph. This happens when the limit exists (Condition #2 is met), but either the function isn't defined at the point (Condition #1 fails) or the function's value is something different from the limit (Condition #3 fails).

Think of it like a bridge with a single wooden plank missing. You can't cross right there, but you can see exactly where the plank should go. The two sides of the bridge are perfectly aligned.

Mathematically, this "hole" often appears in rational functions—fractions with polynomials.

Consider the function: f(x) = (x² - 4) / (x - 2)

At x = 2, the denominator is zero, so the function is undefined. We have a discontinuity. But what kind? Let's check the limit as x approaches 2.

lim (x→2) [ (x² - 4) / (x - 2) ]

Let's factor the numerator: lim (x→2) [ (x - 2)(x + 2) / (x - 2) ]

Aha! We have a common factor of (x - 2) in the numerator and denominator. Since the limit is about the values near x = 2 (not at x = 2), we know x - 2 is not zero, so we can safely cancel.

lim (x→2) [ x + 2 ] = 2 + 2 = 4

So, the limit exists and is equal to 4! We have a hole at (2, 4). The function is heading toward a y-value of 4 as x approaches 2 from both sides.

"Removing" the discontinuity means defining a new function that is identical to the old one everywhere except at the hole, where we fill it in.

We can create a new, continuous function, let's call it g(x), like this:

g(x) = { (x² - 4) / (x - 2), if x ≠ 2 { 4, if x = 2

All we did was "plug the hole." We explicitly defined the value of the function at x = 2 to be the value of the limit. We've successfully paved over the pothole.

Making Piecewise Functions Continuous

The other common scenario involves piecewise functions. These are functions built from different equations on different parts of the domain.

Imagine two separate ramps in a skatepark, one for beginners and one for experts. For you to be able to ride smoothly from one to the other, the end of the first ramp must meet the beginning of the second ramp at the exact same point and height. If they don't, you're going to have a bad time—that's a jump discontinuity.

Our job is to adjust one of the ramps so they connect perfectly.

Let's look at a function h(x): h(x) = { x² + 1, if x < 3 { 2x + k, if x ≥ 3

This function is made of a parabola and a line. It's continuous everywhere except possibly at the boundary x = 3, where the rule changes. For h(x) to be continuous at x = 3, the two pieces must meet.

This means the limit from the left must equal the limit from the right.

1
Find the limit from the left (using the "before 3" rule)
lim (x→3⁻) h(x) = lim (x→3⁻) [x² + 1] = 3² + 1 = 10 The parabola piece is heading towards the point (3, 10).
2
Find the limit from the right (using the "3 and after" rule)
lim (x→3⁺) h(x) = lim (x→3⁺) [2x + k] = 2(3) + k = 6 + k The line piece is heading towards the point (3, 6 + k).
3
Set them equal
For the overall limit to exist, the left-hand limit must equal the right-hand limit. 10 = 6 + k
4
Solve for k
k = 4

If k = 4, the function becomes: h(x) = { x² + 1, if x < 3 { 2x + 4, if x ≥ 3

Now, the limit from the left is 10, and the limit from the right is 2(3) + 4 = 10. The value of the function at x = 3 is also 2(3) + 4 = 10. Since the limit and the function value are both 10, the function is now continuous everywhere! We've adjusted the second ramp to connect perfectly with the first.

Worked examples

Let's put this into practice with a couple of typical AP-style problems.

Example 1

Plugging a Hole

Problem: Let f(x) be the function defined below. Find the value of c that makes f(x) continuous at x = -1.

f(x) = { (x² + 3x + 2) / (x + 1), if x ≠ -1 { c, if x = -1

Solution Walkthrough:

1
Identify the Goal
We need the function to be continuous at x = -1. This means the value of the function, f(-1), must be equal to the limit of the function as x approaches -1.
2
Find the Function Value
The problem tells us that f(-1) = c. Our job is to find what c must be.
3
Find the Limit
We need to calculate lim (x→-1) f(x). We use the top piece of the function for this, since the limit describes the behavior near x = -1, not at x = -1. lim (x→-1) [ (x² + 3x + 2) / (x + 1) ] Plugging in -1 gives (1 - 3 + 2) / (-1 + 1) = 0/0. This is our signal to do more algebra!
4
Simplify by Factoring
Let's factor the numerator. We're looking for two numbers that multiply to 2 and add to 3. That's 1 and 2. lim (x→-1) [ (x + 1)(x + 2) / (x + 1) ]
5
Cancel the Common Factor
We can cancel the (x + 1) terms. lim (x→-1) [ x + 2 ]
6
Evaluate the Limit
Now we can substitute x = -1 into the simplified expression. -1 + 2 = 1 So, the limit of f(x) as x approaches -1 is 1.
7
Set the Value Equal to the Limit
For continuity, we need f(-1) = lim (x→-1) f(x). c = 1

Final Answer: The value c = 1 makes the function continuous. This is because if c=1, the point (-1, 1) fills the hole in the graph perfectly.

Example 2

Connecting Piecewise Parts

Problem: Let g(x) be the function defined below. For what value of the constant a is g(x) continuous for all real numbers?

g(x) = { ax - 5, if x ≤ 2 { x² + a, if x > 2

Solution Walkthrough:

1
Identify the Potential Break
The function g(x) is made of two polynomials, which are continuous everywhere on their own. The only place a discontinuity could occur is at the boundary point x = 2.
2
State the Condition for Continuity
For g(x) to be continuous at x = 2, the limit from the left must equal the limit from the right, and this must also be the value of the function at that point. In short, the two pieces must meet at x=2.
3
Calculate the Left-Side Value
We use the top rule (ax - 5) because it applies for x ≤ 2. Value from the left at x=2: a(2) - 5 = 2a - 5
4
Calculate the Right-Side Value
We use the bottom rule (x² + a) because it applies for x > 2. We're technically finding the limit from the right, but since it's a polynomial, we can just substitute. Value from the right at x=2: (2)² + a = 4 + a
5
Set the Pieces Equal
To make the function continuous, the value from the left must equal the value from the right. 2a - 5 = 4 + a
6

Solve for a: Now we just have a simple algebra problem. Subtract a from both sides: a - 5 = 4 Add 5 to both sides: a = 9

Final Answer: When a = 9, the function is continuous.

The left piece becomes 9x - 5. At x=2, its value is 9(2)-5 = 13.
The right piece becomes x² + 9. As x approaches 2 from the right, its value approaches 2²+9 = 13. Since both pieces meet at the point (2, 13), the function is continuous.

Try it yourself

Ready to try a couple on your own? Take your time and think through the steps.

Problem 1: The function f(x) is defined below. What value of k makes f(x) continuous at x = 4?

f(x) = { (x² - 16) / (x - 4), if x ≠ 4 { k, if x = 4

Hint: What is the limit of the top expression as x approaches 4? The value of k must be equal to that limit.

Problem 2: Find the value of the constant b that makes the function h(x) continuous on the entire real number line.

h(x) = { b*x², if x ≤ 2 { 2x + b, if x > 2

Hint: The two pieces must have the same exact value when x is 2. Set the first expression (with x=2 plugged in) equal to the second expression (with x=2 plugged in) and solve for b.

TL;DR

In simple terms, removing discontinuities is about finding and fixing "holes" or "breaks" in a graph to make it a single, unbroken line.

Syntax pattern

lim (x→2) [ (x² - 4) / (x - 2) ]

Key terms

AP Calculus Removing discontinuities Continuous function Piecewise function Limit Removable discontinuity Hole in graph Calculus help AP Calc AB Topic 1.13 ===END===

You can now…

LIM-2.C: Determine values of x or solve for parameters that make discontinuous functions continuous, if possible.

Essential knowledge (exam-tested)

LIM-2.C.1: If the limit of a function exists at a discontinuity in its graph, then it is possible to remove the discontinuity by defining or redefining the value of the function at that point, so it equals the value of the limit of the function as x approaches that point.
LIM-2.C.2: In order for a piecewise-defined function to be continuous at a boundary to the partition of its domain, the value of the expression defining the function on one side of the boundary must equal the value of the expression defining the other side of the boundary, as well as the value of the function at the boundary.

Concept map

flowchart TD
    A[Start: Analyze function f(x) at x=c] --> B{Does lim (x->c) f(x) exist?};
    B -- No --> C[Non-removable discontinuity (Jump or Infinite)];
    B -- Yes --> D{Does f(c) exist?};
    D -- No --> E[Removable Discontinuity (Hole)];
    D -- Yes --> F{Is f(c) == lim (x->c) f(x)?};
    F -- No --> E;
    F -- Yes --> G[Function is continuous at x=c];
    E --> H[To remove, redefine f(c) = lim (x->c) f(x)];
    H --> G;
    C --> I[Cannot be made continuous by redefining one point];

Read what Saavi narrates

Hi there, it's Saavi from Shrutam. Let's talk about fixing broken graphs.

Imagine you're driving on a brand-new highway. The road is perfectly smooth, except for one spot where a crew forgot to pave a single, tiny section, leaving a jarring pothole. You can see the road on the other side, you know exactly where you should be, but there's a gap. In calculus, we call that a "removable discontinuity." It's a hole in the graph. The good news is, just like a road crew can fill a pothole, we can fix these mathematical holes.

This topic is all about making a function continuous. We'll learn how to plug a single hole in a graph, and how to stitch two different function pieces together so they meet perfectly.

Let's work through an example. Imagine we have a function called g of x. It's a piecewise function. The rule is, g of x equals 'a' times 'x' minus 5, if x is less than or equal to 2. And, g of x equals 'x' squared plus 'a', if x is greater than 2. We need to find the value of the constant 'a' that makes the function continuous everywhere.

The only place this function might have a break is at the boundary, x equals 2. For the function to be continuous, the two pieces have to meet at the exact same point. So, let's see what the value of each piece is at x equals 2.

The left piece is 'a' times 'x' minus 5. At x equals 2, that's 2 times 'a' minus 5.
The right piece is 'x' squared plus 'a'. At x equals 2, that's 2 squared plus 'a', which is 4 plus 'a'.

Now, we just set them equal to each other. Two 'a' minus five... must equal... four plus 'a'.
Let's solve for 'a'. We can subtract 'a' from both sides, which gives us 'a' minus 5 equals 4. Then, add 5 to both sides. We get 'a' equals 9. That's it! When 'a' is 9, the two pieces meet perfectly at the point (2, 13), and our function is continuous.

A common mistake here is to confuse a removable hole with a vertical asymptote. A hole happens when a factor cancels out from the top and bottom of a fraction. An asymptote happens when a factor is left in the denominator. You can fill a hole, but you can't get rid of a vertical asymptote. Always check if the factor cancels completely.

You're building a really strong foundation here. Keep practicing, and you'll find that patching up these functions becomes second nature. You've got this.

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