Defining Continuity at a Point

Lesson ~10 min read

In simple terms: In simple terms, continuity at a point means you can draw a function's graph through that specific spot without lifting your pencil.

Why this matters

Imagine you're on a cross-country road trip from Boston to Seattle. You've planned your route perfectly. But when you get to a bridge over a river in South Dakota, you find it's closed for repairs. The road leading to the bridge is fine, and the road picking up after the bridge is fine, but you can't cross at that exact point. The path is broken.

In calculus, we call this a "discontinuity." It's a break in the function's path. While "drawing without lifting your pencil" is a great starting point, the AP exam will ask you to prove whether a function is continuous at a specific point. You can't just say, "the pencil didn't lift!" You need a formal, mathematical way to justify your conclusion. In this lesson, we'll build that exact three-step justification you'll use on the exam.

Concept overview

flowchart TD
    A[Start: Is f(x) continuous at x=c?] --> B{1. Does f(c) exist?};
    B -- No --> F[Not Continuous];
    B -- Yes --> C{2. Does lim f(x) as x->c exist?};
    C -- No --> F;
    C -- Yes --> D{3. Does lim f(x) = f(c)?};
    D -- No --> F;
    D -- Yes --> G[Continuous!];

Core explanation

Hello! I'm Saavi, and I'm excited to walk you through one of the foundational ideas in calculus: continuity.

Intuitively, you already know what a continuous function looks like. It's a smooth, unbroken curve. You can trace it from left to right without your pencil ever leaving the paper. A function with gaps, jumps, or holes is discontinuous.

But in AP Calculus, "it looks continuous" isn't a valid justification. We need a rigorous, mathematical definition. This brings us to the Three-Part Test for Continuity.

For a function f(x) to be continuous at a single point x = c, it must satisfy all three of the following conditions. If even one condition fails, the function is discontinuous at that point.

Let's break them down one by one.

The Three-Part Continuity Test

Imagine our function is a path, and x = c is a specific location on our map.

Condition 1: The Point Must Exist (`f(c)` is defined)

Does the destination actually exist on the map? Is there a solid point on the graph at x = c?

In mathematical terms, we ask: Does f(c) exist?

This means when you plug c into the function, you get a real number back. You don't get an error like division by zero or the square root of a negative number.

Continuous
f(x) = x^2 at x = 3. f(3) = 9. The point exists.
Discontinuous
g(x) = 1/(x-2) at x = 2. g(2) = 1/0, which is undefined. There's a vertical asymptote here. The destination doesn't exist, so the function is not continuous at x=2.

Condition 2: The Limit Must Exist (`lim f(x)` as `x` approaches `c` exists)

Are the roads from both sides heading toward the same place?

As you approach the point x = c from the left side and the right side, do the y-values of the function close in on the same number?

In mathematical terms: Does lim f(x) as x → c exist?

Remember from our lessons on limits, this is only true if the left-hand limit equals the right-hand limit: lim f(x) as x → c⁻ = lim f(x) as x → c⁺

This is where piecewise functions often try to trick you. You have to check the approach from both sides. If they don't meet, you have a "jump discontinuity."

Limit Exists
For f(x) = x^2 at x = 3, the limit from the left is 9, and the limit from the right is 9. They match, so the limit is 9.
Limit Fails
Consider a piecewise function where f(x) = x for x < 2 and f(x) = x + 1 for x ≥ 2. As you approach x=2 from the left, the function heads toward 2. From the right, it heads toward 3. Since 2 ≠ 3, the limit does not exist. The roads don't meet.

Condition 3: The Point and the Limit Must Be the Same (`lim f(x) = f(c)`)

Is the bridge intact? Does the road actually connect to the destination, or is there a sinkhole right in the middle?

The value the function is approaching (the limit) must be the same as the function's actual value at that point.

In mathematical terms: Does lim f(x) as x → c equal f(c)?

This condition catches "holes" in the graph, also called removable discontinuities. You might have a point defined (Condition 1 passes), and the limit might exist (Condition 2 passes), but the point is in the wrong place!

Continuous
For f(x) = x^2 at x = 3, we found lim f(x) = 9 and f(3) = 9. Since 9 = 9, Condition 3 passes. The function is continuous at x=3.
Discontinuous
Consider a function g(x) where g(x) = (x^2 - 4)/(x - 2) for x ≠ 2, but we define g(2) = 5.
- Condition 1: g(2) = 5. It exists.
- Condition 2: The limit as x approaches 2 is 4. (You can find this by factoring the numerator). The limit exists.
- Condition 3: Does the limit equal the function value? Does 4 = 5? No. The bridge is out. The function is heading toward a y-value of 4, but someone plopped the actual point up at y=5.

Justifying Your Answer

On the AP Exam, a question might ask, "Is f(x) continuous at x=c? Justify your answer."

A complete justification requires you to address all three parts of the test. You can't just say "yes" or "no." You have to show your work by checking each condition.

// A perfect justification looks like this:

The function f(x) is continuous at x = c because:
1. f(c) is defined and equals [some value].
2. The limit of f(x) as x approaches c exists because the limit from the left equals the limit from the right, and thus the limit is [some value].
3. The limit of f(x) as x approaches c equals f(c).

// Or, if it's discontinuous:

The function f(x) is NOT continuous at x = c because [state the FIRST condition that fails].
For example: "...because the limit as x approaches c does not exist."

You only need to find one failed condition to prove discontinuity. Once a condition fails, the test is over.

Master this three-step process. It's your key to unlocking all sorts of problems in calculus, from the Intermediate Value Theorem to differentiability. You've got this.

Worked examples

Let's put the three-part test into action. The key is to be methodical and check every condition in order.

Example 1

A Continuous Piecewise Function

Problem: Let f(x) be the function defined below. Is f(x) continuous at x = 1? Justify your answer. f(x) = { 5 - x^2, if x < 1 { 2x + 2, if x ≥ 1 }

Solution: We need to use our three-part test for continuity at x = 1.

Step 1: Check if f(1) exists. To find the value of f(1), we look at the function definition. The second piece, 2x + 2, is used for x ≥ 1. f(1) = 2(1) + 2 = 4. Since we got a real number, f(1) exists. Condition 1 passes.

Step 2: Check if lim f(x) as x → 1 exists. For a piecewise function, we must check the limit from the left and the right.

Left-hand limit
We use the piece for x < 1. lim (5 - x^2) as x → 1⁻ = 5 - (1)^2 = 4.
Right-hand limit
We use the piece for x ≥ 1. lim (2x + 2) as x → 1⁺ = 2(1) + 2 = 4. Since the left-hand limit (4) equals the right-hand limit (4), the overall limit exists and is 4. lim f(x) as x → 1 = 4. Condition 2 passes.

Step 3: Check if lim f(x) as x → 1 equals f(1). From our previous steps, we found:

lim f(x) as x → 1 = 4
f(1) = 4 Since 4 = 4, the limit equals the function value. Condition 3 passes.

Example 2

A Function with a Hole

Problem: Let g(x) = (x^2 + x - 6) / (x - 2). Is g(x) continuous at x = 2? Justify.

Solution: Let's apply the three-part test at x = 2.

Step 1: Check if g(2) exists. We plug x = 2 into the function: g(2) = (2^2 + 2 - 6) / (2 - 2) = (4 + 2 - 6) / 0 = 0 / 0. Division by zero is undefined. Therefore, g(2) does not exist.

This is a critical point: As soon as a condition fails, the test is over. We don't need to check the other two conditions.

Try it yourself

Ready to try on your own? Use the three-part test to justify your answers. Remember to show your work for each condition.

Problem 1: Let h(x) be the function defined below. Is h(x) continuous at x = -3? Justify your answer. h(x) = { 2, if x = -3 { (x^2 - 9)/(x+3), if x ≠ -3 }

Problem 2: The fee for parking in a downtown Dallas garage is given by the function C(t), where t is the time in hours. C(t) = { $5, if 0 < t ≤ 1 { $10, if 1 < t ≤ 2 { $15, if 2 < t ≤ 3 } Is the function C(t) continuous at t = 1? Justify your answer.

TL;DR

In simple terms, continuity at a point means you can draw a function's graph through that specific spot without lifting your pencil.

Syntax pattern

// A perfect justification looks like this:

The function f(x) is continuous at x = c because:
1. f(c) is defined and equals [some value].
2. The limit of f(x) as x approaches c exists because the limit from the left equals the limit from the right, and thus the limit is [some value].
3. The limit of f(x) as x approaches c equals f(c).

// Or, if it's discontinuous:

The function f(x) is NOT continuous at x = c because [state the FIRST condition that fails].
For example: "...because the limit as x approaches c does not exist."

Key terms

Continuity at a point Ap calculus Three-part continuity test Limits Continuity definition Lim-2.a Calculus unit 1 Continuous function Discontinuity Justifying continuity

You can now…

LIM-2.A: Justify conclusions about continuity at a point using the definition.

Essential knowledge (exam-tested)

LIM-2.A.2: A function f is continuous at x = c provided that f(c) exists, lim f(x) exists, and lim f(x) = f(c).

Concept map

flowchart TD
    A[Start: Is f(x) continuous at x=c?] --> B{1. Does f(c) exist?};
    B -- No --> F[Not Continuous];
    B -- Yes --> C{2. Does lim f(x) as x->c exist?};
    C -- No --> F;
    C -- Yes --> D{3. Does lim f(x) = f(c)?};
    D -- No --> F;
    D -- Yes --> G[Continuous!];

Read what Saavi narrates

Hi there, I'm Saavi. Let's talk about an idea in calculus that seems simple at first, but has some really important details: continuity.

Imagine you're on a road trip, following a route on your GPS. Suddenly, the road is gone. A bridge is out. Even though the road leading up to the river was perfect, and the road continues on the other side, there's a break... a discontinuity. You can't get across at that exact point.

In calculus, we have a formal, three-step test to check for these "breaks" in a function's graph at any given point. It's more precise than just saying "you have to lift your pencil."

Let's work through an example. Imagine a piecewise function, f of x. For x values less than 1, the function is 5 minus x-squared. For x values greater than or equal to 1, the function is 2x plus 2. Is this function continuous at x equals 1?

First, we check condition one: does the point exist? We need to find f of 1. Since 1 is included in the "greater than or equal to 1" piece, we use 2x plus 2. So, f of 1 is 2 times 1, plus 2... which is 4. Yes, the point exists.

Second, condition two: does the limit exist? We have to check from both sides. From the left, as x approaches 1, we use the 5 minus x-squared piece. The limit is 5 minus 1-squared, which is 4. From the right, we use the 2x plus 2 piece. The limit is 2 times 1, plus 2... which is also 4. Since the left and right limits match, the overall limit is 4. So, yes, the limit exists.

Finally, condition three: does the limit equal the point's value? Our limit was 4, and our function's value at 1 was also 4. Since 4 equals 4, this condition passes too.

Because all three conditions passed, we can say with certainty that the function is continuous at x equals 1.

A really common mistake is to stop after one or two steps. For example, some students think if the limit exists, the function must be continuous. But that's not always true. You could have a hole in the graph where the limit exists, but the point itself is undefined. You must check all three conditions, every time.

Mastering this three-part test is a huge step. It's a skill you'll use all year. Keep practicing, and you'll be able to spot those "broken bridges" with confidence. You can do this.

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