Defining Continuity at a Point

Lesson ~10 min read

In simple terms: In simple terms, continuity at a point is about checking if a function's graph is unbroken at a specific spot, without any gaps, holes, or jumps.

Why this matters

Imagine you're planning a solo cross-country road trip, and your map is a bit old. You're looking at the route through a specific town, let's say Lincoln, Nebraska. To know if you can drive through without any trouble, you need to confirm three things.

First, does Lincoln actually exist on the map at that coordinate? Second, does the highway leading into Lincoln from the west seem to line up perfectly with the highway leaving it to the east? If they don't meet, you've got a problem. Finally, and most importantly, does the highway actually go through the town, or is there a massive, un-bridged canyon right in the middle?

This is exactly what we do when we test for continuity in calculus. We have a formal, three-step check to prove that a function's path is connected and drivable at a single point, with no surprises. Let's learn the official checklist.

Concept overview

flowchart TD
    A[Start: Test continuity of f(x) at x=c] --> B{1. Does f(c) exist?};
    B -- No --> F[Not Continuous at x=c];
    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 --> E[Continuous at x=c];

Core explanation

In everyday language, "continuous" means "unbroken." In calculus, we need a more precise definition than that. We need a way to prove, without a doubt, that a function's graph doesn't have a hole, a jump, or a gap at one specific x-value.

This brings us to the formal definition of continuity at a point.

A function f is continuous at a point x = c if and only if all three of the following conditions are met:

f(c) exists.
lim f(x) as x approaches c exists.
lim f(x) as x approaches c = f(c).

If even one of these conditions fails, the function is discontinuous at x = c. Let's break down what each condition really means.

The Continuity Checklist

Imagine we're engineers inspecting a bridge at a specific point c. The function f(x) represents the bridge's road.

Condition 1: `f(c)` exists (The Point Exists)

What it means
The function must have a defined value at x = c. You can plug c into the function and get a real number back.
Bridge Analogy
There must be a physical point on the bridge structure directly above or below x = c. There's a concrete value, a place to stand.
How it can fail
If plugging in c gives you a 0 in the denominator (like 1/0) or the square root of a negative number, f(c) is undefined. Graphically, this could be a hole or a vertical asymptote.

// Example of failure:
// For f(x) = 1/(x-2), f(2) is undefined.
// Condition 1 fails. The function is not continuous at x=2.

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

What it means
The function must be approaching the same y-value from both the left and the right side of c. Remember from our lessons on limits, for a limit to exist, the left-hand limit must equal the right-hand limit. lim f(x) as x→c⁻ = lim f(x) as x→c⁺
Bridge Analogy
The road approaching the point c from the left and the road approaching from the right must be heading toward the exact same spot. If one ramp leads to a height of 50 feet and the other leads to 30 feet, you have a disaster.
How it can fail
This is where we see a jump discontinuity. This often happens in piecewise functions where the two pieces don't meet up at the connection point. If the limit from the left is not equal to the limit from the right, the overall limit does not exist, and Condition 2 fails.

Condition 3: `lim f(x)` as `x` approaches `c` = `f(c)` (The Limit Equals the Point)

What it means
The value the function is approaching (the limit) must be the same as the value the function actually is at that point.
Bridge Analogy
The roads approaching from both sides (the limit) don't just line up with each other; they connect perfectly to the actual bridge structure at that point (f(c)). The place you were heading is the place you actually land.
How it can fail
This is the classic removable discontinuity, or a hole. You might have a limit that exists (the roads line up), but the actual point f(c) is either undefined (a literal hole in the bridge) or defined at a different y-value (the bridge pillar is there, but it's 10 feet below the road).

// Example of failure:
// A piecewise function like this:
// f(x) = { x^2  if x != 2
//        { 5    if x = 2
//
// lim f(x) as x->2 is 4. But f(2) is 5.
// Since 4 != 5, Condition 3 fails. Not continuous at x=2.

When you are asked to justify continuity on the AP Exam, you can't just say "the graph looks connected." You must explicitly state and check all three conditions. Writing them down is the best way to make sure you don't miss a step and to earn full credit.

Worked examples

Let's put the three-part test into practice. The key is to be systematic and show your work for each of the three conditions.

Example 1

A Piecewise Function

Is the function h(x) continuous at x = 1?

       { 5x - 2,  if x < 1
h(x) = {
       { x² + 2,  if x ≥ 1

Solution: We must check the three conditions for continuity at c = 1.

Does h(1) exist? Yes. To find h(1), we use the piece of the function where x ≥ 1. h(1) = (1)² + 2 = 3. The point exists, and its value is 3.
Does lim h(x) as x approaches 1 exist? To find out, we must check the limit from the left and the right.
- Left-hand limit
  We use the 5x - 2 piece for x < 1. lim (5x - 2) as x→1⁻ = 5(1) - 2 = 3.
- Right-hand limit
  We use the x² + 2 piece for x ≥ 1. lim (x² + 2) as x→1⁺ = (1)² + 2 = 3. Since the left-hand limit (3) equals the right-hand limit (3), the overall limit exists and is equal to 3.
Does lim h(x) as x→1 = h(1)? From our work above, we found:
- lim h(x) as x→1 = 3
- h(1) = 3 Yes, they are equal.

Example 2

A Rational Function with a Discontinuity

Is the function g(x) = (x² - 4x + 3) / (x - 1) continuous at x = 1?

Solution: Let's go through the checklist for c = 1.

Does g(1) exist? Let's plug in x = 1: g(1) = (1² - 4(1) + 3) / (1 - 1) = (1 - 4 + 3) / 0 = 0 / 0. This is an indeterminate form, which means the function is undefined at x = 1.

This is a critical point that saves you time: Once a condition fails, the test is over. You don't need to proceed to the other steps to prove discontinuity.

For our own understanding, though, let's see what the limit is. This helps us classify the discontinuity. lim (x² - 4x + 3) / (x - 1) as x→1 lim (x - 1)(x - 3) / (x - 1) as x→1 lim (x - 3) as x→1 = 1 - 3 = -2. The limit exists! Because the limit exists but the point g(1) does not, this is a removable discontinuity (a hole) at (1, -2).

Try it yourself

Time to try it yourself. Use the three-part definition to justify your answers.

Consider the function f(x) below. Is it continuous at x = 0?
```
       { (sin(3x))/x,  if x ≠ 0
f(x) = {
       { 3,           if x = 0
```
Hint
You'll need to use your knowledge of special trigonometric limits. Does the limit you find match the function's actual value at x=0?
Let k(x) = (x² + x - 6) / (x² - 4). Determine if k(x) is continuous at x = 2 and x = -2. Classify any discontinuities you find. Hint: Factor the numerator and denominator first. See what cancels and what remains. One point will be a hole, and the other will be a vertical asymptote. The three-part test will tell you which is which.

TL;DR

In simple terms, continuity at a point is about checking if a function's graph is unbroken at a specific spot, without any gaps, holes, or jumps.

Syntax pattern

// Example of failure:
// For f(x) = 1/(x-2), f(2) is undefined.
// Condition 1 fails. The function is not continuous at x=2.

Key terms

AP Calculus Continuity at a point Definition of continuity Limits Continuous function Removable discontinuity Jump discontinuity Calculus help AP Calc BC LIM-2.A ===END===

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) as x approaches c exists, and lim f(x) as x approaches c = f(c).

Concept map

flowchart TD
    A[Start: Test continuity of f(x) at x=c] --> B{1. Does f(c) exist?};
    B -- No --> F[Not Continuous at x=c];
    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 --> E[Continuous at x=c];

Read what Saavi narrates

(Audio begins with warm, gentle music that fades into the background)

Hi everyone, I’m Saavi, and welcome to Shrutam.

Have you ever planned a road trip, maybe looking at a map on your phone? You see the route, a nice blue line going from, say, Chicago to Minneapolis. But what if there's a problem right at Madison, Wisconsin? Is the bridge over the lake closed? Is the highway under construction? To be sure you can get through, you need to know that the road exists at that point, that the roads leading in and out connect, and that they connect *at* the right spot.

That's exactly what we're doing today with continuity. We're moving past just 'eyeballing' a graph and learning the formal, three-step checklist that proves a function is connected at a specific point.

This checklist has three conditions. One: the point has to exist. Two: the limit has to exist. And three: the point and the limit have to be equal. Let's see it in action.

Imagine we have a piecewise function, h of x. For x-values less than 1, the function is defined by the rule 5x minus 2. For x-values greater than or equal to 1, the function is defined by x-squared plus 2. Is this function continuous right at x equals 1?

Let's run the checklist.

First, does the point exist? What is h of 1? We use the second rule, since it includes x equals 1. So, 1-squared plus 2... is 3. Yes, condition one passes. The point exists at (1, 3).

Second, does the limit exist? We have to check the approach from both sides. From the left, as x approaches 1, we use 5x minus 2. That gives us... 5 times 1, minus 2, which is 3. From the right, we use x-squared plus 2. That gives us... 1-squared plus 2, which is also 3. Since the left and right sides both approach 3, the limit exists and is 3. Condition two passes.

Third, and finally, does the limit equal the point's value? Our limit is 3, and our point's value is 3. Yes, they match. Condition three passes.

Since it passed all three checks, we can say with certainty that the function is continuous at x equals 1.

Now, a common mistake here is to find that the limit exists and just stop, assuming the function is continuous. But that's not enough! You could have a situation where the limit is 3, but the point itself is defined somewhere else, like at 5. That would be a hole with a dot somewhere else, which is discontinuous. You must always check all three steps.

It might feel a little formal at first, but this three-part definition is your most powerful tool for understanding and proving continuity. You can do this.

(Music fades back in)

Back to AP Calculus BC Take a mock exam

Defining Continuity at a Point

Why this matters

Concept overview

Core explanation

The Continuity Checklist

Condition 1: f(c) exists (The Point Exists)

Condition 2: lim f(x) as x approaches c exists (The Limit Exists)

Condition 3: lim f(x) as x approaches c = f(c) (The Limit Equals the Point)

Worked examples

Try it yourself

Condition 1: `f(c)` exists (The Point Exists)

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

Condition 3: `lim f(x)` as `x` approaches `c` = `f(c)` (The Limit Equals the Point)