Defining continuity at a point

AP Calculus BC· 25 min read

1. The Three-Part Definition of Continuity★☆☆☆☆⏱ 10 min

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.

📘 Definition

Continuous at a point

$f (x)$ is continuous at $x = a$

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).

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.

2. Testing Continuity Algebraically★★☆☆☆⏱ 15 min

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.

📐 Worked Example

Is the piecewise function $f (x) = ⎩ ⎨ ⎧ 3 x + 1 4 x^{2} + 3 x < 1 x = 1 x > 1$ 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 $x = 1$ .

3. Classifying Types of Discontinuities★★☆☆☆⏱ 12 min

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.

4. Continuity and AP Exam Applications★★☆☆☆⏱ 10 min

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.

✓ Quick check

Test your understanding of core continuity rules:

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
Reveal answer
A —
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 $a$ .
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
Reveal answer
B —
One-sided limits exist but are not equal, so this is a non-removable jump discontinuity.

5. Common Pitfalls

Wrong move:

Stopping after checking $f (a)$ is defined and concluding continuity

Why:

You still need to confirm the limit exists and matches the function value

Correct move:

Check all three conditions in order before concluding continuity

Wrong move:

Classifying a vertical asymptote discontinuity as removable

Why:

The two-sided limit does not exist for infinite discontinuities, so they cannot be removable

Correct move:

Only classify a discontinuity as removable if the two-sided limit exists at the point

Wrong move:

Assuming all piecewise functions are discontinuous at boundary points

Why:

Many piecewise functions are designed to be continuous at the boundary

Correct move:

Always test all three conditions at the boundary, do not assume discontinuity

Wrong move:

Claiming a function cannot be continuous at an endpoint of its domain

Why:

One-sided continuity at endpoints counts as continuous for interval continuity

Correct move:

At domain endpoints, confirm the one-sided limit equals the function value to verify continuity

6. Quick Reference Cheatsheet

Condition	Description	Result of failure
$f (a)$ is defined	Point exists on domain	Discontinuity (removable if limit exists)
$lim_{x \to a} f (x)$ exists	Left/right limits are equal	Non-removable discontinuity
$lim_{x \to a} f (x) = f (a)$	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

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · MC
Determine continuity at $x = 2$
2021 · FRQ
Verify continuity at a point
2023 · MC
Identify discontinuity type

Going deeper

syllabusCollege Board AP Calculus BC CEDPage 51, Limits and Continuity section

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.