# Defining continuity at a point

> AP Calculus AB · Unit 1: Limits and Continuity
> Source: https://www.owlsprep.com/study/ap-calculus-ab-u1-defining-continuity-at-a-point/

This guide covers the formal definition of continuity at a point, one-sided continuity for domain endpoints, checking continuity for piecewise functions, and exam-accepted justifications for AP Calculus AB questions.

**Prerequisites:** How to evaluate one-sided and two-sided limits; How to evaluate function values at a given point; How to interpret piecewise function definitions

## Learning objectives

- State the three conditions for continuity at an interior point of a domain
- Apply one-sided continuity definitions for domain endpoints
- Check continuity of piecewise functions at boundary points
- Justify continuity correctly to earn full credit on AP Calculus AB FRQs

## Definition of Continuity at an Interior Point

Informally, a function is continuous at a point $x=a$ if you can draw the graph of the function through $x=a$ without lifting your pencil off the paper. Continuity is a foundational property that underpins nearly all advanced results in calculus, from the Intermediate Value Theorem to derivatives and the Fundamental Theorem of Calculus.

**Continuity at an interior point $x=a$** — Three independent conditions must all hold: 1. $f(a)$ is defined ($a$ is in the domain of $f$), 2. $\lim_{x \to a} f(x)$ exists (the two one-sided limits are equal and finite), 3. $\lim_{x \to a} f(x) = f(a)$. All three conditions are captured by the single equality:

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

*Example:* Verified for any point inside the function's domain

$$\lim_{x \to a} f(x) = f(a)$$

Each condition serves a distinct purpose: the first confirms we have a function value to compare, the second confirms the graph approaches the same value from both sides, and the third confirms the point actually lies on the approaching curve.

**Worked example:** Let $f(x) = \begin{cases} \frac{x^2 - 4}{x-2} & x \neq 2 \\ 3 & x = 2 \end{cases}$. Does $f$ satisfy all three conditions for continuity at $x=2$?

1. Check if $f(2)$ is defined: per the piecewise definition, $f(2) = 3$, so condition 1 holds.
2. Evaluate the two-sided limit by factoring the numerator:
3. $$\frac{(x-2)(x+2)}{x-2} = x+2 \quad \text{for } x \neq 2$$
4. So $\lim_{x \to 2} f(x) = \lim_{x \to 2} (x+2) = 4$. Left and right limits are equal, so the limit exists, condition 2 holds.
5. Compare the limit to the function value: $\lim_{x \to 2} f(x) = 4 \neq 3 = f(2)$, so condition 3 fails. Conclusion: $f$ is not continuous at $x=2$.

> **Exam tip:** On AP FRQs, always explicitly check all three conditions when justifying continuity. Skipping any condition will cost you a point, even if your final conclusion is correct.

## One-Sided Continuity for Domain Endpoints

For functions defined on closed intervals $[a,b]$, a two-sided limit cannot exist at the endpoints, since the function is not defined outside the interval. For this reason, we use one-sided continuity to define continuity at endpoints.

**One-sided continuity** — A function is right-continuous at $x=a$ if $\lim_{x \to a^+} f(x) = f(a)$, used for left endpoints of domains. A function is left-continuous at $x=b$ if $\lim_{x \to b^-} f(x) = f(b)$, used for right endpoints of closed domains.

A function is continuous on the full closed interval $[a,b]$ if it is continuous at all interior points, right-continuous at the left endpoint, and left-continuous at the right endpoint.

**Worked example:** Let $f(x) = \sqrt{9 - x^2}$, which has domain $[-3, 3]$. Is $f$ continuous at the left endpoint $x=-3$? Justify your answer.

1. Confirm $f(-3)$ is defined: $f(-3) = \sqrt{9 - (-3)^2} = \sqrt{0} = 0$, so the function value exists.
2. Since $x=-3$ is the left endpoint of the domain, only the right-hand limit needs to be checked (no values of $x < -3$ are in the domain).
3. Evaluate the right-hand limit:
4. $$\lim_{x \to -3^+} \sqrt{9 - x^2} = \sqrt{ \lim_{x \to -3^+} (9 - x^2) } = \sqrt{9 - 9} = 0$$
5. Compare: $\lim_{x \to -3^+} f(x) = 0 = f(-3)$, so $f$ satisfies the definition of continuity at an endpoint. Conclusion: $f$ is continuous at $x=-3$.

> **Exam tip:** Never check for a two-sided limit at a domain endpoint. AP exam graders will penalize you for referencing a non-existent limit outside the function's domain.

## Checking Continuity at Piecewise Function Boundaries

The most common AP exam question on this topic asks you to check continuity at the boundary point where the definition of a piecewise function changes. Each individual piece of a piecewise function is almost always continuous (polynomials, rationals, trig functions are continuous on their domains), so the only possible discontinuity is at the boundary between pieces.

1. Find $f(a)$ from the piece that includes $x=a$.
2. Calculate the left-hand limit $\lim_{x \to a^-} f(x)$ using the piece defined for $x < a$.
3. Calculate the right-hand limit $\lim_{x \to a^+} f(x)$ using the piece defined for $x > a$.
4. Confirm both one-sided limits are equal (so the two-sided limit exists), then confirm the limit equals $f(a)$.

**Worked example:** Let $g(x) = \begin{cases} 2x + 1 & x < 1 \\ x^2 + 2 & x \geq 1 \end{cases}$. Determine if $g$ is continuous at $x=1$, the boundary between the two pieces. Justify your answer.

1. Check $g(1)$: from the second piece ($x \geq 1$), $g(1) = 1^2 + 2 = 3$, so $g(1)$ is defined.
2. Evaluate the left-hand limit (use the first piece for $x < 1$):
3. $$\lim_{x \to 1^-} (2x + 1) = 2(1) + 1 = 3$$
4. Evaluate the right-hand limit (use the second piece for $x > 1$):
5. $$\lim_{x \to 1^+} (x^2 + 2) = 1^2 + 2 = 3$$
6. Since left and right limits are equal, $\lim_{x \to 1} g(x) = 3$, which equals $g(1) = 3$. All three conditions hold. Conclusion: $g(x)$ is continuous at $x=1$.

> **Exam tip:** Always match the direction of the limit to the inequality of the piece. Swapping left and right pieces is the most common mistake on these problems.

## AP-Style Worked Practice Problems

**Worked example:** Let $h(x) = \begin{cases} \dfrac{\sin(2x)}{x} & x \neq 0 \\ k & x = 0 \end{cases}$. For what value of $k$ is $h(x)$ continuous at $x=0$?
A) $0$
B) $1$
C) $2$
D) $\frac{1}{2}$

1. For continuity at $x=0$, we require $\lim_{x \to 0} h(x) = h(0) = k$. Using the standard trigonometric limit $\lim_{x \to 0} \frac{\sin(ax)}{x} = a$, we get:
2. $$\lim_{x \to 0} \frac{\sin(2x)}{x} = 2$$
3. For all three conditions to hold, $k$ must equal 2: $h(0)=2$ is defined, the limit exists, and $\lim_{x \to 0} h(x) = h(0)$. The correct answer is C.

**Worked example:** Let $f(x) = \begin{cases} x^3 + ax + 4 & x < 2 \\ bx^2 - 3x & x \geq 2 \end{cases}$.
(a) Find all values of $a$ and $b$ such that $f(x)$ is continuous at $x=2$ if $a = b$.
(b) Suppose instead that $a = 1$, what value of $b$ makes $f$ continuous at $x=2$?
(c) Is $f$ continuous at $x=0$ for any values of $a$ and $b$? Justify your answer.

1. Part (a): First, calculate $f(2)$ from the second piece:
2. $$f(2) = b(2)^2 - 3(2) = 4b - 6$$
3. Next, calculate the left limit as $x \to 2^-$:
4. $$\lim_{x \to 2^-} (x^3 + ax + 4) = 8 + 2a + 4 = 12 + 2a$$
5. Set the limit equal to $f(2)$ for continuity: $12 + 2a = 4b - 6$. Substitute $a=b$: $12 + 2a = 4a - 6 \implies 18 = 2a \implies a=9$, so $b=9$.
6. Part (b): For $a=1$, the left limit becomes $12 + 2(1) = 14$. Set equal to $4b - 6$: $14 = 4b - 6 \implies 20 = 4b \implies b=5$.
7. Part (c): $x=0$ is an interior point of the interval $x < 2$, where $f(x)$ is a polynomial. All polynomials are continuous at all real numbers, so $\lim_{x \to 0} f(x) = 0 + 0 + 4 = 4 = f(0)$. Therefore, $f$ is continuous at $x=0$ for all values of $a$ and $b$.

**Worked example:** A coffee shop models the price of a large coffee over a 10-day period with the function $P(d) = \begin{cases} 4.50 & 0 \leq d < 4 \\ 4.50 + 0.25(d - 4) & 4 \leq d \leq 10 \end{cases}$, where $P(d)$ is the price in dollars on day $d$, for $0 \leq d \leq 10$. Is the price function continuous at $d=4$, the day the shop starts a gradual price increase? Justify your answer, and explain what continuity means in this context.

1. Check $P(4)$: using the second piece, $P(4) = 4.50 + 0.25(4-4) = 4.50$, so $P(4)$ is defined.
2. Evaluate the left-hand limit:
3. $$\lim_{d \to 4^-} P(d) = 4.50$$
4. Evaluate the right-hand limit:
5. $$\lim_{d \to 4^+} \left(4.50 + 0.25(d-4)\right) = 4.50 + 0 = 4.50$$
6. Compare: $\lim_{d \to 4} P(d) = 4.50 = P(4)$, so all three conditions are satisfied. Conclusion: The price function is continuous at $d=4$. In context, this means there is no abrupt jump in the price of coffee on day 4; the price increases gradually starting from the original \$4.50 price after day 4, rather than jumping immediately to a higher price.

## Common pitfalls

- **Wrong:** For $f(x) = \frac{x^2 - 9}{x-3}$ for $x \neq 3$, $f(3)=6$, concluding $f$ is discontinuous at $x=3$ because the original expression is undefined at $x=3$.
  - Why it fails: Students confuse the unsimplified algebraic expression with the given function's definition, forgetting the function explicitly defines $f(3)$.
  - Correct: Always use the given function definition to check if $f(a)$ is defined, not the unsimplified expression.
- **Wrong:** Checking only that $\lim_{x \to a} f(x)$ exists, then concluding continuity at $a$.
  - Why it fails: Students forget that $f(a)$ could be undefined, or could be a different value than the limit.
  - Correct: Always explicitly check all three conditions in order when justifying continuity on an FRQ.
- **Wrong:** Claiming a function is discontinuous at an endpoint $x=a$ because the two-sided limit doesn't exist.
  - Why it fails: Students confuse two-sided continuity for interior points with the definition of continuity at endpoints.
  - Correct: For any endpoint of the function's domain, only check the one-sided limit that exists (right for left endpoint, left for right endpoint) and compare it to the function value.
- **Wrong:** For a piecewise function with boundary $x=a$, calculating the left limit using the right piece and vice versa.
  - Why it fails: Students rush and misread the inequality signs defining the pieces.
  - Correct: Underline the inequality for each piece before evaluating one-sided limits, and match the direction of the limit to the inequality.
- **Wrong:** Concluding $f$ is continuous at interior point $x=a$ because $\lim_{x \to a^-} f(x) = f(a)$, without checking the right-hand limit.
  - Why it fails: Students mix up one-sided continuity for endpoints with the requirements for interior points.
  - Correct: For any interior point, always confirm that both one-sided limits exist, are equal, and equal $f(a)$.

## Cheatsheet

| Category | Formula/Rule | Notes |
| --- | --- | --- |
| Continuity at interior point $x=a$ | $\lim_{x \to a} f(x) = f(a)$ | Requires 3 conditions: $f(a)$ defined, $\lim_{x \to a} f(x)$ exists, equality holds |
| Right-continuous at $x=a$ | $\lim_{x \to a^+} f(x) = f(a)$ | Used for left endpoints of domains, only requires the right-hand limit |
| Left-continuous at $x=b$ | $\lim_{x \to b^-} f(x) = f(b)$ | Used for right endpoints of domains, only requires the left-hand limit |
| Continuity at endpoint | One-sided limit equals function value | No two-sided limit required, since the function is undefined outside the interval |
| Piecewise boundary check | Left limit from left piece, right limit from right piece | Match the limit direction to the inequality defining the piece |
| Polynomial continuity | All polynomials are continuous at all real points | Any interior point in a polynomial piece is automatically continuous |
| Rational function continuity | Continuous at all points where denominator $\neq 0$ | Discontinuous at points where the denominator equals zero |
| Radical function continuity | Continuous at all points in domain | Check one-sided continuity at endpoints of the domain |

## What's next

Defining continuity at a point is the foundational prerequisite for almost all of the rest of calculus. Next, you will extend this definition to continuity over an interval, which is required to apply the Intermediate Value Theorem (IVT), one of the most frequently tested theorems on the AP Calculus AB exam. You will also use this definition to classify different types of discontinuities and solve for constants that make piecewise functions continuous, a common problem on both MCQ and FRQ sections.

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

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