Unit 1 · Topic 1.11 · about 30 minutes · 7 parts · 12 scored checks, 3 of them multiple choice

AP Calculus AB 1.11: Defining Continuity at a Point

Q: If f(c) does not exist, do I still need to find the limit?

No. The first condition already fails, so f is not continuous at c. Say that f(c) does not exist and why, for example because the formula divides by 0 there.

Q: Is a function with a sharp corner, like |x| at 0, continuous there?

Yes. |0| = 0 and both sides approach 0, so all three conditions hold. A corner changes the direction of the graph, not its height.

Q: Why does the limit not use the value f(c)?

A limit as x c is about inputs close to c but not equal to c. That is exactly why continuity needs a separate condition comparing the limit with f(c).

Built by Tanner Crow from College Board's course framework · Every answer key independently checked

Unit 1: 10 to 15% of the multiple-choice section

What you'll be able to do

Justify whether a function is continuous at a point by checking the three conditions of the definition of continuity, with values, before stating the conclusion.

Key vocabulary

Term	What it means	The exam trap
Continuous at $x = c$	A function $f$ is continuous at $x = c$ provided that $f (c)$ exists, $\lim_{x \to c} f (x)$ exists, and $\lim_{x \to c} f (x) = f (c)$ .	Continuity is a property at one input. A function can be continuous at one input and not at another, so always say which $x = c$ you mean.
One-sided limit	The value $f (x)$ approaches as $x$ approaches $c$ from one side only, written $\lim_{x \to c^{-}} f (x)$ from the left and $\lim_{x \to c^{+}} f (x)$ from the right.	The minus and plus signs name the side $x$ comes from, not the sign of the output.
Discontinuity	An input where a function fails to be continuous, such as a removable discontinuity, a jump discontinuity, or a discontinuity due to a vertical asymptote.	Naming the type is not a justification. Say which condition of the definition fails, with values.

Graph of a function f in two pieces: y = x + 1 for x up to 2 and y = (x - 4) squared minus 1 for x greater than 2. A dashed line marks x = 2, where both pieces reach height 3, and a filled point shows f(2) = 3.

Part 1 Think first A glitch in the temperature log

Guess before you learn. A wrong guess still makes the answer stick.

Predict first

A data logger records the temperature of a water tank, $T (t)$ degrees Celsius at time $t$ minutes. Near $t = 5$ the readings are 39.8 at $t = 4.9$ , 39.98 at $t = 4.99$ , 40.02 at $t = 5.01$ and 40.2 at $t = 5.1$ . At exactly $t = 5$ a reading failed, and the logger saved $T (5) = 0$ . Is the function $T$ the logger saved continuous at $t = 5$ ?

In short. A function $f$ is continuous at $x = c$ provided that $f (c)$ exists, $\lim_{x \to c} f (x)$ exists, and $\lim_{x \to c} f (x) = f (c)$ .

The three conditions are checked in order, and failing any one of them means $f$ is not continuous at $c$ .
Where a function changes formula at $c$ , the limit exists only if the limits from the left and from the right are the same number.
A justification states each condition with its value and then names the function in the conclusion.

Part 2 Learn it Three short ideas, each checked as you go

11a. The definition: three conditions

A function $f$ is continuous at $x = c$ provided that all three of these are true:

$f (c)$ exists.
$\lim_{x \to c} f (x)$ exists.
$\lim_{x \to c} f (x) = f (c)$ .

The second condition is about values of $f$ near $c$ . The first is about the value at $c$ . The third asks whether they agree. Each one can fail on its own:

No value. $\frac{x^{2} - 1}{x - 1}$ has no value at $x = 1$ , because the formula divides by 0 there, even though its outputs near 1 close in on 2.
No limit. The outputs may approach one number from the left and a different number from the right, or grow without bound as $x$ nears $c$ .
A value and a limit that differ. The tank logger in Think first had $T (5) = 0$ while the readings pointed to a limit of 40.

Use the conditions as a checklist, in order. If the first one fails, the function is not continuous at $c$ , and you can stop there; the third condition cannot hold when the second fails, because there is no limit to compare. When one of them fails, $f$ has a discontinuity at $c$ , one of the types of discontinuity from the last topic.

Quick check · Sort it

No calculator. Each card describes a function $f$ at and near an input $c$ . Tap a card, then the first condition of the definition, in order, that fails, or Continuous at $c$ if none fails.

Continuous at $c$

$f (c)$ does not exist

The limit does not exist

The limit is not $f (c)$

11b. Checking a function given in pieces

A function defined in pieces is where continuity most often needs checking, at the input where the formula changes. Take the graph at the top:

f (x) = {\begin{matrix} x + 1, & x \leq 2 \\ (x - 4)^{2} - 1, & x > 2 \end{matrix}

The value. Read the inequality signs. $x = 2$ belongs to the piece with $x \leq 2$ , so $f (2) = 2 + 1 = 3$ .

The limit. Near 2, $f$ uses one formula on the left and another on the right, so find each one-sided limit from its own piece. With the limit properties from earlier in the unit, a polynomial piece's limit is its formula evaluated at 2:

\lim_{x \to 2^{-}} f (x) = 2 + 1 = 3, \lim_{x \to 2^{+}} f (x) = (2 - 4)^{2} - 1 = 3.

The two sides agree, so $\lim_{x \to 2} f (x) = 3$ . If they had disagreed, the limit would not exist and you would stop.

The comparison. $\lim_{x \to 2} f (x) = 3 = f (2)$ , so $f$ is continuous at $x = 2$ .

When the piece that holds $c$ also covers inputs on one side of $c$ , as $x \leq 2$ does here, it gives $f (c)$ and the limit from that side. The other side needs its own formula, and a value defined at $c$ alone gives neither limit.

Quick check · Multiple choice

No calculator. Let $f (x) = x^{2} + 2$ for $x < 2$ and $f (x) = 3 x - 1$ for $x \geq 2$ . Which statement about $f$ at $x = 2$ is true?

11c. Writing the justification

When a question says justify, the conclusion earns credit only with the conditions behind it. Write each condition with its value, then the conclusion, and name the function.

For the graph at the top, a complete answer is:

" $f (2) = 3$ . $\lim_{x \to 2^{-}} f (x) = 3$ and $\lim_{x \to 2^{+}} f (x) = 3$ , so $\lim_{x \to 2} f (x) = 3$ . Because $\lim_{x \to 2} f (x) = f (2)$ , $f$ is continuous at $x = 2$ ."

For a function that is not continuous, name the condition that fails and show the numbers that make it fail. For the function in the last check: " $\lim_{x \to 2^{-}} f (x) = 6$ and $\lim_{x \to 2^{+}} f (x) = 5$ , so $\lim_{x \to 2} f (x)$ does not exist, and $f$ is not continuous at $x = 2$ ."

Three shortcuts lose the point. "The graph has no break" gives no values. " $f (2)$ exists" is one condition of three. And one side of a limit is not the limit: a left-hand limit that matches $f (c)$ says nothing about the right.

Quick check · Decide and justify

No calculator. Let $h (x) = x^{2}$ for $x \leq 3$ and $h (x) = 2 x + 3$ for $x > 3$ . Asked to justify that $h$ is continuous at $x = 3$ , a student writes: " $h (3) = 9$ and $\lim_{x \to 3^{-}} h (x) = 9$ , so $h$ is continuous at $x = 3$ ." Is that justification complete?

Which reason supports your answer?

Common mistakes

The mistake	Why it is wrong	Do this instead
Stopping once $f (c)$ exists.	A value at $c$ is only the first condition. A function can have a value at $c$ and still jump or have its limit somewhere else.	Check the conditions in order, writing the value each one gives, until one fails or all three hold.
Checking only one side at the input where a piecewise function changes formula.	The limit at $c$ exists only when the limit from the left and the limit from the right are the same number.	Find $\lim_{x \to c^{-}} f (x)$ from the left piece and $\lim_{x \to c^{+}} f (x)$ from the right piece, and compare them.
Reading $f (c)$ from the wrong piece.	The inequality signs decide which piece holds $c$ . With $x \leq 2$ on the left piece, $f (2)$ comes from the left formula, not the right.	Before any arithmetic, say which piece's condition includes $c$ , and evaluate that piece.

Part 3 Watch it done A function with a value of its own at $x = 3$

Watch each step, then do the last one yourself.

Let $f (x) = x^{2} - 1$ for $x < 3$ , $f (3) = 5$ , and $f (x) = 2 x + 2$ for $x > 3$ . Is $f$ continuous at $x = 3$ ? Justify your answer using the definition of continuity.

Step 1 · The value at 3
Only the rule $f (3) = 5$ covers $x = 3$ itself, so $f (3) = 5$ . The first condition holds.
Step 2 · The limit from each side
$\lim_{x \to 3^{-}} f (x) = 3^{2} - 1 = 8$ and $\lim_{x \to 3^{+}} f (x) = 2 (3) + 2 = 8$ . Both sides give 8, so $\lim_{x \to 3} f (x) = 8$ . The second condition holds.
Why this step?
Why does neither one-sided limit use the value $f (3) = 5$ ?
A limit as $x \to 3$ looks at inputs close to 3 but never equal to 3. Near 3 on the left the function is $x^{2} - 1$ and near 3 on the right it is $2 x + 2$ ; the single value at 3 itself plays no part in either limit.
Your step · Fill the statement
No calculator. Finish it. Tap a blank, then tap an answer.

The third condition compares $\lim_{x \to 3} f (x) =$ with $f (3) =$ . So $f$ is at $x = 3$ .

Part 4 Try it Two new functions

Try it · Spot the error

No calculator. Let $f (x) = x^{2} + 1$ for $x \leq 1$ and $f (x) = 4 - x$ for $x > 1$ . A student tests whether $f$ is continuous at $x = 1$ . Tap the first line with a mistake.

Try it · Calculate

No calculator. Let $g (x) = \frac{x^{2} - 2 x - 8}{x - 4}$ for $x \neq 4$ , and $g (4) = 3$ . How many of the three conditions in the definition of continuity does $g$ meet at $x = 4$ ?

Your answer

Part 5 Explain it Write it the way the exam asks

On the exam free response is handwritten, so work on paper first. Nothing you type here leaves this page. Then compare with the model answer and score each point.

Free response is 50% of the exam score. This one is shaped like Question 3, from free-response Part B (calculator not permitted).

Free-response practice · Question 3 style

No calculator. A lab sample is heated, then cooled, then moved to a water bath. Its temperature, in degrees Celsius, $t$ minutes after the start is modeled by

T (t) = {\begin{matrix} 20 + 3 t, & 0 \leq t \leq 5 \\ 60 - 5 t, & 5 < t \leq 8 \\ 30 - t, & 8 < t \leq 10 \end{matrix}

(a) Find $\lim_{t \to 5^{-}} T (t)$ and $\lim_{t \to 5^{+}} T (t)$ . Using correct units, interpret the meaning of $\lim_{t \to 5^{+}} T (t)$ in the context of the problem.

(b) Is $T$ continuous at $t = 5$ ? Justify your answer using the definition of continuity.

Your answer stays on this page. It is never sent or saved.

Model answer

(a) $\lim_{t \to 5^{-}} T (t) = 20 + 3 (5) = 35$ and $\lim_{t \to 5^{+}} T (t) = 60 - 5 (5) = 35$ . As $t$ approaches 5 minutes from the right, through times just after 5 minutes, the modeled temperature of the sample approaches 35 degrees Celsius.

(b) $T (5) = 20 + 3 (5) = 35$ , so $T (5)$ exists. From part (a) the two one-sided limits are both 35, so $\lim_{t \to 5} T (t) = 35$ exists. Because $\lim_{t \to 5} T (t) = T (5)$ , $T$ is continuous at $t = 5$ .

(c) $T (8) = 60 - 5 (8) = 20$ . $\lim_{t \to 8^{-}} T (t) = 60 - 5 (8) = 20$ and $\lim_{t \to 8^{+}} T (t) = 30 - 8 = 22$ . The one-sided limits differ, so $\lim_{t \to 8} T (t)$ does not exist, and $T$ is not continuous at $t = 8$ .

Score your own answer. This is a self-check, not a grade. Tick each point your answer makes.

Finds both one-sided limits at $t = 5$ from the correct pieces, 35 and 35.
Says that as $t$ approaches 5 minutes from the right, the modeled temperature approaches 35 degrees Celsius, with units.
States $T (5) = 35$ , and that $\lim_{t \to 5} T (t) = 35$ exists because the one-sided limits agree.
Concludes $T$ is continuous at $t = 5$ because $\lim_{t \to 5} T (t) = T (5)$ , naming $T$ .
Finds the one-sided limits 20 and 22 at $t = 8$ , says the limit does not exist, and concludes $T$ is not continuous at $t = 8$ .

Part 6 Check yourself A table of values and limits

The two case questions use the same table. Answer them from it.

For a function $f$ , the table gives the value $f (c)$ and the two one-sided limits at five inputs $c$ . "Not defined" means $f$ has no value at that input.

Values and one-sided limits of f
$c$	$f (c)$	$\lim_{x \to c^{-}} f (x)$	$\lim_{x \to c^{+}} f (x)$
1	2	2	2
2	Not defined	4	4
3	1	5	5
4	3	3	6
5	0	0	0

Case question 1 · Calculate

No calculator. At how many of the five inputs in the table is $f$ continuous?

Your answer

Case question 2 · Multiple choice

No calculator. At which inputs in the table does $\lim_{x \to c} f (x)$ fail to exist?

The rest stand on their own and do not use the case.

Check yourself · Match

No calculator. Each function is tested for continuity at $x = 2$ . Tap a function, then the first condition of the definition, in order, that fails, or Continuous if none fails.

Check yourself · Decide and justify

No calculator. Let $k (x) = | x - 1 | + 2$ . A classmate says $k$ is not continuous at $x = 1$ , because its graph has a sharp corner there. Is the classmate right?

Which reason supports your answer?

Part 7 Keep it Bring back earlier ideas

Two earlier ideas this lesson leans on. Pulling them back up is what makes them last.

Bring it back · Multiple choice

No calculator. Let $f (x) = \frac{x^{2} - 1}{x^{2} - 3 x + 2}$ . Which describes the discontinuities of $f$ ?

Bring it back · Calculate

No calculator. Find $\lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4}$ . Give an exact decimal.

Your answer

In three lines

A function $f$ is continuous at $x = c$ when $f (c)$ exists, $\lim_{x \to c} f (x)$ exists, and the two are equal.
At an input where a piecewise function changes formula, find the limit from each side with its own piece.
A justification gives the value for each condition, then names the function in its conclusion.

Exam strategy for this topic

Section I Part A, Multiple Choice (calculator not permitted): 29 questions in 62 minutes.
Section I Part B, Multiple Choice (graphing calculator required): 13 questions in 38 minutes.
Section II Part A, Free Response (graphing calculator required): 2 questions in 30 minutes.
Section II Part B, Free Response (calculator not permitted): 4 questions in 60 minutes.

When a question asks whether a function is continuous at a point, write the three conditions as three short lines: the value $f (c)$ , the limit with both sides shown where the formula changes, and the comparison. End with a sentence that names the function: "Because $\lim_{x \to c} f (x) = f (c)$ , $f$ is continuous at $x = c$ ."

Questions students ask

Do I always have to check the limit from both sides?

You need the two-sided limit. Where a single formula holds on both sides of $c$ , one calculation gives it. Where the formula changes at $c$ , as in a piecewise function, find each one-sided limit from its own piece and check that they agree.

f (c)

does not exist, do I still need to find the limit?

No. The first condition already fails, so $f$ is not continuous at $c$ . Say that $f (c)$ does not exist and why, for example because the formula divides by 0 there.

Is a function with a sharp corner, like

| x |

at 0, continuous there?

Yes. $| 0 | = 0$ and both sides approach 0, so all three conditions hold. A corner changes the direction of the graph, not its height.

Can I just say the graph has no holes or breaks?

That is a good way to picture continuity, but it is not a justification. A justification uses the definition: the value, the limit and the comparison, each with its number.

Why does the limit not use the value

f (c)

A limit as $x \to c$ is about inputs close to $c$ but not equal to $c$ . That is exactly why continuity needs a separate condition comparing the limit with $f (c)$ .

Course alignment, for teachers

AP Calculus AB topic 1.11, Unit 1: Limits and Continuity.

Skill 3.C: Confirm whether hypotheses or conditions of a selected definition, theorem, or test have been satisfied.

This lesson maps the one essential knowledge statement of the topic to the part that teaches each one: 1 is checked by a scored question. The map is in the lesson file.

What you'll be able to do

Key vocabulary

Part 1 Think first A glitch in the temperature log

Part 2 Learn it Three short ideas, each checked as you go

11a. The definition: three conditions

11b. Checking a function given in pieces

11c. Writing the justification

Part 3 Watch it done A function with a value of its own at x=3

Part 4 Try it Two new functions

Part 5 Explain it Write it the way the exam asks

Part 6 Check yourself A table of values and limits

Part 7 Keep it Bring back earlier ideas

Exam strategy for this topic

Questions students ask

Part 3 Watch it done A function with a value of its own at $x = 3$