# Defining continuity at a point

> AP Calculus BC · Limits and Continuity
> Source: https://www.owlsprep.com/study/ap-calculus-bc-u1-defining-continuity-at-a-point/

This sub-topic covers the formal three-part definition of continuity at a point, how to test for continuity using graphs and algebraic limit calculations, and an introduction to the main types of discontinuities assessed on the AP exam.

**Prerequisites:** [One-sided and two-sided limits](https://www.owlsprep.com/study/ap-calculus-bc-u1-calculating-two-sided-limits/); [Limit properties](https://www.owlsprep.com/study/ap-calculus-bc-u1-limit-properties/)

## Learning objectives

- State the three-part formal definition of continuity at a point
- Test for continuity at a point algebraically and graphically
- Classify common types of discontinuities
- Identify when the continuity condition is satisfied for AP exam theorem applications

## The Three-Part Definition of Continuity

Intuitively, a function is continuous at a point if you can draw its graph through that point without lifting your pencil. Formally, this intuition translates to three required conditions, which you will use for all continuity checks on the AP exam.

**Continuous at a point** — A function satisfies all three of the following conditions: 1. $f(a)$ is defined (the point exists on the function), 2. $\lim_{x \to a} f(x)$ exists (left and right limits are equal), 3. $\lim_{x \to a} f(x) = f(a)$ (the limit equals the function value).

*Notation:* $f(x)$ is continuous at $x=a$

*Example:* $f(x) = x^2$ is continuous at $x=2$, since $f(2)=4$ and $\lim_{x \to 2} x^2 = 4$, so all conditions hold.

**Worked example:** Is $f(x) = \frac{x^2 - 4}{x - 2}$ continuous at $x=2$?

1. First, check the first condition: is $f(2)$ defined?
2. Evaluate $f(2)$:
3. $$f(2) = \frac{2^2 - 4}{2 - 2} = \frac{0}{0} = \text{undefined}$$
4. Since the first condition fails, we do not need to check the other conditions. The function is not continuous at $x=2$, even though the two-sided limit exists at this point.

> **Exam tip:** AP free-response questions require you to explicitly reference all three conditions when justifying continuity.

## Testing Continuity Algebraically

Piecewise functions are the most common scenario for algebraic continuity checks, since you need to compare one-sided limits from each side of the boundary point to confirm the two-sided limit exists.

> **tip**
>
> Check conditions in order. If any condition fails, you can immediately conclude discontinuity and stop checking.

**Worked example:** Is the piecewise function $f(x) = \begin{cases} 3x + 1 & x < 1 \\ 4 & x = 1 \\ x^2 + 3 & x > 1 \end{cases}$ continuous at $x=1$?

1. Step 1: Check if $f(1)$ is defined. From the function definition, $f(1) = 4$, so the first condition holds.
2. Step 2: Calculate one-sided limits to check if the two-sided limit exists:
3. $$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (3x + 1) = 3(1) + 1 = 4 \\ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x^2 + 3) = 1^2 + 3 = 4$$
4. The one-sided limits are equal, so $\lim_{x \to 1} f(x) = 4$ exists, so the second condition holds.
5. Step 3: Check if the limit equals the function value: $\lim_{x \to 1} f(x) = 4 = f(1)$. All three conditions are satisfied.
6. Conclusion: The function is continuous at $x=1$.

## Classifying Types of Discontinuities

Any function that fails the continuity test at a point is discontinuous there. Discontinuities are grouped into removable and non-removable categories based on whether the two-sided limit exists.

- **Removable**: Limit exists, discontinuity is caused by an undefined point or mismatched function value (appears as a hole in the graph)
- **Jump**: Non-removable; one-sided limits exist but are not equal, graph jumps between two values
- **Infinite**: Non-removable; one or both limits approach $\pm \infty$, usually from a vertical asymptote
- **Oscillating**: Non-removable; function oscillates near the point with no defined limit

**Worked example:** Classify the discontinuity of $f(x) = \frac{x^2 - 9}{x - 3}$ at $x=3$.

1. Step 1: Check if $f(3)$ is defined: $f(3) = \frac{9 - 9}{3 - 3} = \frac{0}{0}$, so $f(3)$ is undefined.
2. Step 2: Calculate the limit at $x=3$ by factoring and canceling:
3. $$\lim_{x \to 3} \frac{x^2 - 9}{x - 3} = \lim_{x \to 3} \frac{(x-3)(x+3)}{x-3} = \lim_{x \to 3} (x+3) = 6$$
4. The two-sided limit exists, but $f(3)$ is undefined. This matches the definition of a removable discontinuity.

## Continuity and AP Exam Applications

The definition of continuity at a point is the foundation for nearly all major theorems in AP Calculus. For example, the Intermediate Value Theorem (IVT) only applies to functions that are continuous at every point on a closed interval.

> **info**
>
> AP FRQs often require you to confirm continuity at a point as the first step to justifying that a theorem applies. Always reference the definition to earn full points.

**Check your understanding**

Test your understanding of core continuity rules:

1. If $\lim_{x \to a} f(x) = f(a)$, which of the following must be true?

   - A. $f(x)$ is continuous at $x=a$
   - B. Only the third continuity condition holds
   - C. The limit does not exist at $x=a$
   - D. $f(a)$ is undefined

   *Why:* If $\lim_{x \to a} f(x) = f(a)$, this implies the limit exists and $f(a)$ is defined, so all three conditions are satisfied. $f$ is continuous at $a$.

2. A function has $f(2) = 5$, $\lim_{x \to 2^-} f(x) = 3$, $\lim_{x \to 2^+} f(x) = 6$. What is the discontinuity type at $x=2$?

   - A. Removable
   - B. Jump
   - C. Infinite
   - D. No discontinuity

   *Why:* One-sided limits exist but are not equal, so this is a non-removable jump discontinuity.

## Common pitfalls

- **Wrong:** Stopping after checking $f(a)$ is defined and concluding continuity
  - Why it fails: You still need to confirm the limit exists and matches the function value
  - Correct: Check all three conditions in order before concluding continuity
- **Wrong:** Classifying a vertical asymptote discontinuity as removable
  - Why it fails: The two-sided limit does not exist for infinite discontinuities, so they cannot be removable
  - Correct: Only classify a discontinuity as removable if the two-sided limit exists at the point
- **Wrong:** Assuming all piecewise functions are discontinuous at boundary points
  - Why it fails: Many piecewise functions are designed to be continuous at the boundary
  - Correct: Always test all three conditions at the boundary, do not assume discontinuity
- **Wrong:** Claiming a function cannot be continuous at an endpoint of its domain
  - Why it fails: One-sided continuity at endpoints counts as continuous for interval continuity
  - Correct: At domain endpoints, confirm the one-sided limit equals the function value to verify continuity

## Cheatsheet

| Condition | Description | Result of failure |
| --- | --- | --- |
| 1. $f(a)$ is defined | Point exists on domain | Discontinuity (removable if limit exists) |
| 2. $\lim_{x \to a} f(x)$ exists | Left/right limits are equal | Non-removable discontinuity |
| 3. $\lim_{x \to a} f(x) = f(a)$ | Limit matches function value | Removable discontinuity |
| Removable | Limit exists, hole |  |
| Jump | One-sided limits not equal |  |
| Infinite | Limit approaches $\pm\infty$ |  |
| Oscillating | No limit from oscillation |  |

## What's next

Understanding the definition of continuity at a point is the foundational building block for all subsequent topics in AP Calculus, including continuity over an interval, the Intermediate Value Theorem, differentiability, and the Fundamental Theorem of Calculus. You will regularly use this definition to justify conclusions about function behavior on free-response questions, and to confirm that key calculus theorems apply to a given function on an interval. In particular, the relationship between differentiability and continuity relies entirely on this definition, since a function can only be differentiable at a point if it is first continuous there.

- [Confirming continuity over an interval](https://www.owlsprep.com/study/ap-calculus-bc-u1-confirming-continuity-over-an-interval/)
- [Removing Discontinuities](https://www.owlsprep.com/study/ap-calculus-bc-u1-removing-discontinuities/)
- [Connecting infinite limits and vertical asymptotes](https://www.owlsprep.com/study/ap-calculus-bc-u1-connecting-infinite-limits-and-vertical/)

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