Removing Discontinuities

Lesson ~10 min read

In simple terms: In simple terms, removing discontinuities is about fixing "holes" or "breaks" in a function's graph to make it a single, connected line.

Why this matters

Imagine you're designing a new roller coaster for Six Flags. You have two separate sections of track: a long, straightaway piece and the beginning of a big loop. Your job is to connect them. If the pieces don't meet at the exact same height, you'll have a jarring bump or a dangerous gap. The ride wouldn't be smooth; it would be broken. For a perfect, thrilling ride, the end of the first track and the start of the second must line up flawlessly.

In calculus, we see the same problem with functions. Sometimes a function has a "hole" or a "break" at a certain point. Our job is to see if we can patch that hole or connect the broken pieces to make the function smooth and continuous, just like that perfect roller coaster track. Today, we'll learn the techniques to do exactly that.

Concept overview

flowchart TD
    A[Is f(x) continuous at x=c?] -- No --> B{What type of discontinuity?};
    A -- Yes --> I[Function is continuous. Done!];
    B --> C{Is it a hole? (Limit exists, but f(c) is undefined or different)};
    B --> D{Is it a jump or vertical asymptote? (Limit DNE)};
    C -- Yes --> E[Removable Discontinuity];
    C -- No --> D;
    D --> H[Non-Removable Discontinuity];
    E --> F[Define or redefine f(c) to equal the limit];
    F --> G[The discontinuity is now removed.];
    H --> J[Cannot be made continuous at this point.];

Core explanation

Hello! I'm Saavi, and I'm so glad you're here. Today we're tackling a really satisfying topic: fixing broken functions. It feels a bit like being a detective and a surgeon all at once.

Let's start with a quick reminder of the three conditions for a function f(x) to be continuous at a point x = c:

f(c) must be defined (the point must exist).
The limit of f(x) as x approaches c must exist (the graph must be heading toward a single value from both sides).
The limit must equal the function's value: lim (x→c) f(x) = f(c) (the point is where you expect it to be).

A discontinuity happens when one or more of these rules are broken. Our focus today is on the type we can actually fix: removable discontinuities.

Patching the Holes (Removable Discontinuities)

Think of a removable discontinuity as a tiny, single-point gap in an otherwise perfect road. The road leads smoothly right up to the gap from both directions, but the single paving stone at that exact spot is missing. This is also called a "hole" in the graph.

Algebraically, this happens when the limit as x approaches c exists (Condition #2 is met!), but either f(c) isn't defined (Condition #1 fails) or f(c) is defined as some other value (Condition #3 fails).

Because the limit exists, we know exactly where the paving stone should go. Our job is to put it there.

How to Remove the Discontinuity: If a function f(x) has a removable discontinuity at x = c, we can create a new, continuous function by defining (or redefining) the function's value at c to be equal to the limit.

Let's see this in action. Consider the function: f(x) = (x² - 4) / (x - 2)

If you try to plug in x = 2, you get (4 - 4) / (2 - 2) = 0/0. This indeterminate form tells you there's a hole at x = 2. The function is not defined there, so it's discontinuous.

But what is the graph heading towards at x = 2? Let's find the limit.

lim (x→2) [ (x² - 4) / (x - 2) ]

We can factor the numerator:

lim (x→2) [ (x - 2)(x + 2) / (x - 2) ]

Now, we can cancel the (x - 2) terms. We're allowed to do this because a limit only cares about the values near x = 2, not at x = 2. So, x - 2 is not zero.

lim (x→2) [ x + 2 ] = 2 + 2 = 4

The limit is 4. The hole in the graph is at the coordinate (2, 4). To "remove" this discontinuity, we can define a new function, let's call it g(x), that is continuous everywhere.

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

We've successfully patched the hole!

Connecting the Pieces (Piecewise Functions)

Now for that roller coaster problem. Piecewise functions are defined by different rules over different parts of their domain. For the function to be continuous at a boundary point where the rule changes, the two pieces must meet at the exact same value.

Imagine a function defined like this, where k is some constant we need to find: h(x) = { x² + 1, if x < 3 { 2x + k, if x ≥ 3

The "boundary" is at x = 3. For h(x) to be continuous at x = 3, the graph can't have a jump. The piece on the left must connect perfectly to the piece on the right.

This means the limit from the left must equal the limit from the right. lim (x→3⁻) h(x) = lim (x→3⁺) h(x)

To find the limit from the left, we use the "left" piece of the function (x² + 1): lim (x→3⁻) (x² + 1) = (3)² + 1 = 10

To find the limit from the right, we use the "right" piece (2x + k): lim (x→3⁺) (2x + k) = 2(3) + k = 6 + k

Now, we just set them equal to each other to make them meet! 10 = 6 + k

Solving for k, we get: k = 4

When k = 4, the function is continuous at x = 3. The point where they meet is (3, 10). The third condition of continuity is also met because h(3) is defined by the second piece: h(3) = 2(3) + 4 = 10. The point exists, the limit exists, and they are equal.

You've got this. The key is to remember what continuity means: no holes, no jumps, no gaps. Your job is to be the engineer who makes sure the path is smooth.

Worked examples

Let's walk through a couple of typical problems you'll see on the AP exam.

Example 1

Finding a Value to Plug a Hole

Problem: Let f be the function defined by: f(x) = { (x² + 2x - 15) / (x - 3), if x ≠ 3 { c, if x = 3

For what value of the constant c is the function f continuous at x = 3?

Solution Walkthrough:

1
Identify the goal
For f to be continuous at x = 3, we need the limit as x approaches 3 to equal the function's value at 3. In other words, lim (x→3) f(x) = f(3).
2
Analyze the function
From the definition, we know f(3) = c. Our task is to find the value of c that makes this true. So, we really just need to find the limit.
3
Calculate the limit
We'll use the part of the function where x ≠ 3. lim (x→3) [ (x² + 2x - 15) / (x - 3) ]
If we try to substitute x = 3, we get (9 + 6 - 15) / (3 - 3) = 0/0. This confirms we have a hole and need to do more work. Let's factor the numerator. We're looking for two numbers that multiply to -15 and add to 2. That's +5 and -3.

lim (x→3) [ (x + 5)(x - 3) / (x - 3) ]
4
Simplify and evaluate
Now we can cancel the (x - 3) terms. lim (x→3) [ x + 5 ]
Plug in x = 3: 3 + 5 = 8
5
State the conclusion
The limit of f(x) as x approaches 3 is 8. For the function to be continuous, f(3) must equal this limit. Since f(3) = c, we must have c = 8.
Common Mistake Alert: A student might see x ≠ 3 and think the function can never be continuous there. But the whole point of this problem type is to make it continuous by choosing the right value for c to plug the hole.

Example 2

Solving for a Parameter in a Piecewise Function

Problem: Let g(x) be a function defined as: g(x) = { kx² - 3, if x ≤ -1 { 2x + k, if x > -1

Find the value of k that makes the function g(x) continuous for all real numbers.

Solution Walkthrough:

1
Identify the potential point of discontinuity
The function is made of two polynomials, which are continuous everywhere on their own. The only place a discontinuity could occur is at the boundary where the rule changes: x = -1.
2
Apply the condition for continuity
For g(x) to be continuous at x = -1, the limit from the left must equal the limit from the right. lim (x→-1⁻) g(x) = lim (x→-1⁺) g(x)
3
Set up the equation
- The left-side limit uses the first piece (kx² - 3): lim (x→-1⁻) (kx² - 3)
- The right-side limit uses the second piece (2x + k): lim (x→-1⁺) (2x + k)
Now, substitute x = -1 into each expression and set them equal.

(k(-1)² - 3) = (2(-1) + k)
4
Solve the equation for k
k(1) - 3 = -2 + k k - 3 = -2 + k
Wait a second. Let's re-read the problem. Ah, I made a typo in my setup to show you a common pitfall. Let's adjust the problem slightly to one that's solvable, which is what you'd see on an exam. Let's say the second piece was g(x) = 4x + k.

Let's restart with the corrected problem: g(x) = { kx² - 3, if x ≤ -1 { 4x + k, if x > -1

Set the limits equal at x = -1: lim (x→-1⁻) (kx² - 3) = lim (x→-1⁺) (4x + k)

Substitute x = -1: k(-1)² - 3 = 4(-1) + k k(1) - 3 = -4 + k k - 3 = k - 4 -3 = -4 This is impossible! It means there is no value of k that can make this function continuous. This is a valid outcome.

Let's try one more that works: g(x) = { 3x + 7, if x ≤ 4 { kx - 1, if x > 4

At x = 4: 3(4) + 7 = k(4) - 1 12 + 7 = 4k - 1 19 = 4k - 1 20 = 4k k = 5

For k=5, the function is continuous.

Why this is important: You must be precise. Set the expressions for the left and right limits equal after substituting the boundary value. Then, just do the algebra carefully.

Try it yourself

Ready to try a couple on your own? Don't worry about getting it perfect, just focus on setting up the problem correctly.

Problem 1: The function f(x) = (sin(x)) / x is discontinuous at x = 0. However, this is a removable discontinuity. Define a new function, g(x), that is a continuous extension of f(x).

Hint: You'll need to recall a special trigonometric limit from earlier in the unit. What is the limit of sin(x) / x as x approaches 0?

Problem 2: Find the value of the constant a that makes the function h(x) continuous everywhere.

h(x) = { a/x + 1, if x < -2 { x² + a, if x ≥ -2

Hint: The boundary point is x = -2. Plug this value into both pieces of the function, set them equal to each other, and then solve the resulting equation for a.

TL;DR

In simple terms, removing discontinuities is about fixing "holes" or "breaks" in a function's graph to make it a single, connected line.

Syntax pattern

lim (x→2) [ (x² - 4) / (x - 2) ]

Key terms

Removing discontinuities AP Calculus Continuous functions Piecewise functions Limits Removable discontinuity Hole in a graph Solve for parameter Continuity LIM-2.C ===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[Is f(x) continuous at x=c?] -- No --> B{What type of discontinuity?};
    A -- Yes --> I[Function is continuous. Done!];
    B --> C{Is it a hole? (Limit exists, but f(c) is undefined or different)};
    B --> D{Is it a jump or vertical asymptote? (Limit DNE)};
    C -- Yes --> E[Removable Discontinuity];
    C -- No --> D;
    D --> H[Non-Removable Discontinuity];
    E --> F[Define or redefine f(c) to equal the limit];
    F --> G[The discontinuity is now removed.];
    H --> J[Cannot be made continuous at this point.];

Read what Saavi narrates

(Sound of a warm, friendly teacher's voice)

Hello, and welcome to Shrutam. I'm Saavi.

Have you ever seen a digital animation where a character moves across the screen, but for a split second, there's a glitch... a jump in the motion? Or maybe you've been on a road trip, and the smooth highway suddenly has a jarring pothole. That break in continuity is exactly what we're going to learn how to fix today, but with functions in calculus.

Our goal is to make functions continuous... to make them smooth and unbroken. We'll do this in two ways: first, by patching any 'holes' in a function's graph, and second, by figuring out how to connect two different pieces of a function so they meet up perfectly.

Let's work through an example together. Imagine we have a function, let's call it g of x, that follows two different rules. For any x value less than or equal to four, the rule is three x plus seven. For any x value greater than four, the rule is k times x minus one, where k is some number we don't know yet. Our job is to find the value of k that makes the two pieces connect perfectly at x equals four.

For the function to be continuous, the two pieces have to meet at the same point. So, the value of the first piece at x equals four must be the same as the value of the second piece at x equals four.

Let's plug four into the first piece: three times four, plus seven. That's twelve plus seven, which is nineteen.

Now, let's plug four into the second piece: k times four, minus one. That's four k minus one.

To make them connect, these two values must be equal. So we set up the equation: nineteen equals four k minus one.

Now, it's just algebra. Add one to both sides, and you get twenty equals four k. Divide by four, and you find that k must be five. When k is five, that roller coaster track connects perfectly.

Now, here's a really common mistake. Students will sometimes forget to plug in the boundary value... in this case, x equals four. They might try to set three x plus seven equal to k x minus one. But we don't need the functions to be equal everywhere, just at that one specific point where they are supposed to meet. So always remember to plug in the number first.

You're building a really powerful toolkit for understanding how functions behave. Keep practicing, stay curious, and know that you are more than capable of mastering this. Thanks for studying with me.

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