Unit 1 · Topic 1.7 · about 30 minutes · 7 parts · 12 scored checks, 3 of them multiple choice

AP Calculus AB 1.7: Selecting Procedures for Determining Limits

Q: Is 00 ever the answer to a limit?

No. It is not a number. It says the quotient theorem cannot be used yet, and the limit could turn out to be any number or fail to exist. Rewrite and substitute again.

Q: If substitution gives a number, do I still need to rewrite?

No. When no denominator approaches 0 and the input is inside the domain of every part, the number you get is the limit. Rewriting is for 00.

Q: How do I know which rewrite to use?

Read what the expression is made of. Polynomials factor, a square root in a difference takes its conjugate, a quotient built from 2 x and x takes a Pythagorean identity, and a fraction inside a fraction takes a common denominator.

Q: Can a limit exist where the function has no value?

Yes. The limit uses only inputs near c. x2 - 1x - 1 has no value at 1, but for x 1 it equals x + 1, so its limit there is 2.

Q: Why use one-sided limits for an absolute value?

Because |x - c| follows one rule for x > c and another for x < c. Each one-sided limit uses one rule, and the two-sided limit exists only if they agree.

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

Key vocabulary

Term	What it means	The exam trap
Limit	A real number $R$ is $\lim_{x \to c} f (x)$ if $f (x)$ can be made arbitrarily close to $R$ by taking $x$ sufficiently close to $c$ , but not equal to $c$ .	Keep writing $\lim_{x \to c}$ on every line until you substitute; without it, a line states a value, not a limit.
One-sided limit	The value $f (x)$ approaches as $x$ approaches $c$ from one side only: $\lim_{x \to c^{-}} f (x)$ from the left and $\lim_{x \to c^{+}} f (x)$ from the right.	The minus sign in $c^{-}$ means from the left of $c$ , not a negative number.
Limit theorems	Rules that give the limit of a sum, difference, product, quotient or composite function from the limits of its parts.	Each theorem needs the limits of the parts to exist, so say what they are before you combine them.
Equivalent form	An expression equal to the original at every input near $c$ other than $c$ itself, such as a quotient after a common factor is canceled.	A canceled form can differ from the original at $c$ itself, so use it for the limit, not for the value at $c$ .
Conjugate	For a sum or difference with a square root, such as $\sqrt{x + 1} - 1$ , the same two terms joined by the opposite sign, $\sqrt{x + 1} + 1$ .	Multiply the top and the bottom by the conjugate, never only one of them, or the value changes.

Graph of a function with two rules. For x less than 2 it follows y = x squared minus 1, rising from (-1, 0) toward an open circle at (2, 3). For x at least 2 it follows y = 7 minus 3x, starting at a filled dot at (2, 1) and falling to (4, -5). The left side approaches 3 and the right side approaches 1.

Part 1 Think first Is an error message the answer?

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

Predict first

Jordan is checking his homework. To find $\lim_{x \to 3} \frac{x^{2} - x - 6}{x - 3}$ , he types $x = 3$ into his calculator and gets an error, because the top and the bottom are both 0 there. What does that tell him about the limit?

In short. Selecting a procedure for a limit means reading what you are given, a graph, a table or a formula, and what substitution gives, and letting that choose the method.

Given only a graph or a table, estimate the limit from outputs at inputs near $c$ on both sides, not from the output at $c$ .
Given a formula, substitute $x = c$ first; the kind of result you get decides the next step.
Where the rule changes at $c$ , as in a piecewise function or an absolute value, find each one-sided limit and compare them.

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

7a. Start from what you are given

A limit question hands you a graph, a table or a formula, and that decides your first move.

With a graph or a table there is nothing to compute. Read the outputs at inputs closer and closer to $c$ from both sides, and say what they approach. The output at $c$ itself plays no part: $\lim_{x \to c} f (x)$ depends only on inputs near $c$ , never on $f (c)$ .

With a formula, substitute $x = c$ first. For polynomials, and for sums, products, quotients and composites built from them, the limit theorems come down to substitution, as long as no denominator approaches 0. Roots, sine and cosine also work by substitution at any input inside their domain, a fact you will confirm when you study continuity. What you get tells you what to do next.

A number. That number is the limit. A top of 0 is fine: $\frac{0}{8} = 0$ .
A nonzero number over 0. The top stays near that number while the bottom shrinks toward 0, so the size of the quotient grows without bound, positive or negative. The function is unbounded near $c$ , and the limit does not exist.
0 over 0. The quotient theorem needs a bottom whose limit is not 0, so it cannot be used, and $\frac{0}{0}$ is not a value. Rewrite the expression in an equivalent form, then substitute again.

Quick check · Sort it

No calculator. Substitute the input into each expression. Tap a card, then tap what the substitution tells you about the limit.

Substitution gives the limit

0 over 0: rewrite first

Nonzero over 0: no limit

7b. Zero over zero: let the form choose the rewrite

When substitution gives $\frac{0}{0}$ , the top and the bottom both approach 0, and the quotient could approach any number, or none. To find out, replace the expression with an equivalent form: one equal to it at every input near $c$ other than $c$ itself. The two have the same limit, because a limit never looks at $x = c$ .

Which rewrite to use depends on what the expression is made of.

Polynomials on top and bottom. Both are 0 at $c$ , so both have the factor $x - c$ . Factor, cancel it, and substitute again.
A square root in a sum or difference, such as $\sqrt{x + 1} - 1$ . Multiply the top and the bottom by the conjugate, $\sqrt{x + 1} + 1$ . The product $(\sqrt{x + 1} - 1) (\sqrt{x + 1} + 1) = x + 1 - 1 = x$ has no root left.
Some trigonometric quotients. When the top and the bottom are built from $\sin^{2} x$ , $\cos x$ and $1 - \cos x$ , a Pythagorean identity such as $\sin^{2} x = 1 - \cos^{2} x$ reaches a form that factors.
A fraction inside a fraction. Combine the small fractions over a common denominator, then cancel.

After the rewrite, read the new substitution the same way as before. If it is still $\frac{0}{0}$ , rewrite again.

Quick check · Match

No calculator. Each limit gives $\frac{0}{0}$ when you substitute. Tap a limit, then the rewrite that suits it.

7c. Two rules at one input

Some functions change their rule at $c$ . A piecewise function can switch formulas there, and an absolute value such as $| x - 3 |$ is $x - 3$ for $x > 3$ but $- (x - 3)$ for $x < 3$ . No single formula then describes both sides, so find the two one-sided limits, each from the rule on its own side:

\lim_{x \to c^{-}} f (x) and \lim_{x \to c^{+}} f (x) .

If they are the same number, the limit is that number. If they differ, the limit does not exist. A change of rule alone settles nothing. The graph at the top changes rule at $x = 2$ and its sides approach 3 and 1, but two other rules could meet at one value.

Here are three ways a limit can fail to exist, and each shows on a graph or in a table:

the function is unbounded near $c$ ;
it oscillates near $c$ , as $\cos (\frac{π}{x})$ swings between $- 1$ and 1 again and again as $x \to 0$ ;
the limit from the left and the limit from the right are different numbers.

Quick check · Decide and justify

No calculator. A function is defined by $f (x) = x^{2} + 1$ for $x < 2$ and $f (x) = 3 x - 1$ for $x \geq 2$ . A classmate says $\lim_{x \to 2} f (x)$ does not exist, because $f$ uses one formula to the left of 2 and another to the right. Is the classmate right?

Which reason supports your answer?

Common mistakes

The mistake	Why it is wrong	Do this instead
Writing $\frac{0}{0}$ as the answer, calling it 0, or taking it to mean the limit does not exist.	$\frac{0}{0}$ only means the quotient theorem cannot be used yet. The limit could be any number, or fail to exist.	Rewrite the expression in an equivalent form and substitute again.
Canceling $\| x - c \|$ as if it were $x - c$ .	$\| x - c \|$ equals $x - c$ only for $x > c$ ; for $x < c$ it is $- (x - c)$ .	Split the work into the two one-sided limits and simplify each side with its own sign.
Using $f (c)$ as the limit when it is given separately, in a table or by its own rule.	The limit depends only on inputs near $c$ ; $f (c)$ can be any value, or have no value at all.	Read the outputs on both sides of $c$ , and leave the output at $c$ out of it.

Part 3 Watch it done Choosing the rewrite, step by step

Watch each step, then do the last one yourself.

Find $\lim_{x \to - 2} \frac{x^{3} + 8}{x^{2} + x - 2}$ .

Step 1 · Substitute first
At $x = - 2$ the top is $(- 2)^{3} + 8 = 0$ and the bottom is $(- 2)^{2} + (- 2) - 2 = 0$ . That is $\frac{0}{0}$ , so the quotient theorem cannot be used yet.
Step 2 · Let the form choose
Both parts are polynomials, and both are 0 at $x = - 2$ , so both have the factor $x + 2$ . The top is a sum of cubes, $x^{3} + 8 = (x + 2) (x^{2} - 2 x + 4)$ , and the bottom is $x^{2} + x - 2 = (x + 2) (x - 1)$ .
Why this step?
Why must $x + 2$ be a factor of both polynomials?
A polynomial that is 0 at $x = - 2$ has $x - (- 2)$ , which is $x + 2$ , as a factor. So $\frac{0}{0}$ from two polynomials tells you exactly which factor to look for.
Step 3 · Cancel, then substitute again
For $x \neq - 2$ , $\frac{(x + 2) (x^{2} - 2 x + 4)}{(x + 2) (x - 1)} = \frac{x^{2} - 2 x + 4}{x - 1}$ . At $x = - 2$ the new bottom is $- 3$ , not 0, so substitution now gives the limit.
Your step · Calculate

No calculator. Finish it. For $x \neq - 2$ , $\frac{x^{3} + 8}{x^{2} + x - 2} = \frac{x^{2} - 2 x + 4}{x - 1}$ . What is $\lim_{x \to - 2} \frac{x^{3} + 8}{x^{2} + x - 2}$ ?

Your answer

Part 4 Try it Two new limits

Try it · Calculate

No calculator. Find $\lim_{x \to π} \frac{\sin^{2} x}{1 + \cos x}$ .

Your answer

Try it · Multiple choice

No calculator. What is $\lim_{x \to 3} \frac{\frac{1}{x} - \frac{1}{3}}{x - 3}$ ?

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 4, from free-response Part B (calculator not permitted).

Free-response practice · Question 4 style

No calculator. A cyclist rides away from a trailhead along a straight path. Her distance from the trailhead, in meters, $t$ seconds after she starts is

s (t) = {\begin{matrix} 20 \sqrt{t} & for 0 \leq t \leq 4 \\ t^{2} - 3 t + 36 & for 4 < t \leq 6 \end{matrix}

For $t \neq 4$ , the function $A (t) = \frac{s (t) - s (4)}{t - 4}$ gives her average velocity, in meters per second, between time 4 and time $t$ .

(a) Find $\lim_{t \to 4^{-}} A (t)$ . Show the work that leads to your answer.

(b) Find $\lim_{t \to 4^{+}} A (t)$ . Show the work that leads to your answer.

(c) Does $\lim_{t \to 4} A (t)$ exist? Justify your answer. If it exists, interpret its value in the context of the problem, using correct units.

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

Model answer

(a) $s (4) = 20 \sqrt{4} = 40$ . For $0 \leq t < 4$ , $A (t) = \frac{20 \sqrt{t} - 40}{t - 4}$ , which gives $\frac{0}{0}$ at $t = 4$ . Since $t - 4 = (\sqrt{t} - 2) (\sqrt{t} + 2)$ , $A (t) = \frac{20 (\sqrt{t} - 2)}{(\sqrt{t} - 2) (\sqrt{t} + 2)} = \frac{20}{\sqrt{t} + 2}$ for $t \neq 4$ . So $\lim_{t \to 4^{-}} A (t) = \frac{20}{\sqrt{4} + 2} = 5$ .

(b) For $4 < t \leq 6$ , $A (t) = \frac{t^{2} - 3 t + 36 - 40}{t - 4} = \frac{(t - 4) (t + 1)}{t - 4} = t + 1$ for $t \neq 4$ . So $\lim_{t \to 4^{+}} A (t) = 4 + 1 = 5$ .

(c) Yes. The left-hand and right-hand limits of $A$ at $t = 4$ are both 5, so $\lim_{t \to 4} A (t) = 5$ . As the time interval on either side of $t = 4$ shrinks, her average velocity approaches 5 meters per second, so at the instant $t = 4$ seconds the cyclist is moving away from the trailhead at 5 meters per second.

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

Writes $A (t) = \frac{20 \sqrt{t} - 40}{t - 4}$ for $t < 4$ and rewrites it, by the conjugate or by factoring $t - 4$ , before substituting.
Gets $\lim_{t \to 4^{-}} A (t) = 5$ .
Simplifies $\frac{t^{2} - 3 t - 4}{t - 4}$ to $t + 1$ and gets $\lim_{t \to 4^{+}} A (t) = 5$ .
Says the limit exists and equals 5 because both one-sided limits are 5.
Interprets 5 as her velocity at the instant $t = 4$ seconds, 5 meters per second, with units.

Part 6 Check yourself A function known only from a table

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

A function $g$ is defined for all real numbers. Its formula is not given, but $g (2) = 5$ , and the table shows $g (x)$ at inputs close to 2.

Values of g near x = 2
$x$	$g (x)$
1.9	11.41
1.99	11.9401
1.999	11.994001
2.001	12.006001
2.01	12.0601
2.1	12.61

Case question 1 · Calculate

No calculator. Use the table to estimate $\lim_{x \to 2} g (x)$ .

Your answer

Case question 2 · Multiple choice

No calculator. Which statement about $g$ near $x = 2$ do the table and the value $g (2) = 5$ support?

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

Check yourself · Spot the error

No calculator. A student finds $\lim_{x \to 3} \frac{x^{2} - 9}{| x - 3 |}$ . Tap the first line with a mistake.

Check yourself · Fill the statement

No calculator. Tap a blank, then tap an answer.

As $x \to 0$ , $\sin (\frac{1}{x})$ has no limit because it . As $x \to 1$ , $\frac{1}{(x - 1)^{2}}$ has no limit because it . As $x \to 0$ , $\frac{| x |}{x}$ has no limit because it .

Part 7 Keep it Bring back earlier ideas

Two earlier ideas that every procedure here relies on. Pulling them back up is what makes them last.

Bring it back · Calculate

No calculator. Suppose $\lim_{x \to 2} f (x) = 4$ and $\lim_{x \to 2} g (x) = - 3$ . Find $\lim_{x \to 2} \frac{[f (x)]^{2} - g (x)}{f (x) + g (x)}$ .

Your answer

Bring it back · Multiple choice

No calculator. A runner's distance from the start line is known at every time $t$ . How does calculus use limits to define the runner's speed at the instant $t = 5$ ?

In three lines

Read what you are given first: a graph or a table means estimate, and a formula means substitute.
A number from substitution is the limit, a nonzero number over 0 means no limit, and 0 over 0 means rewrite in an equivalent form.
Where the rule changes at the input, find both one-sided limits and compare them.

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.

On a free-response limit, show the substitution that gives $\frac{0}{0}$ , then the rewrite, and keep $\lim_{x \to c}$ on every line until the final substitution. Where a function changes rule at the input, write both one-sided limits and say whether they agree before you state the limit.

Questions students ask

\frac{0}{0}

ever the answer to a limit?

No. It is not a number. It says the quotient theorem cannot be used yet, and the limit could turn out to be any number or fail to exist. Rewrite and substitute again.

If substitution gives a number, do I still need to rewrite?

No. When no denominator approaches 0 and the input is inside the domain of every part, the number you get is the limit. Rewriting is for $\frac{0}{0}$ .

How do I know which rewrite to use?

Read what the expression is made of. Polynomials factor, a square root in a difference takes its conjugate, a quotient built from $\sin^{2} x$ and $\cos x$ takes a Pythagorean identity, and a fraction inside a fraction takes a common denominator.

Can a limit exist where the function has no value?

Yes. The limit uses only inputs near $c$ . $\frac{x^{2} - 1}{x - 1}$ has no value at 1, but for $x \neq 1$ it equals $x + 1$ , so its limit there is 2.

Why use one-sided limits for an absolute value?

Because $| x - c |$ follows one rule for $x > c$ and another for $x < c$ . Each one-sided limit uses one rule, and the two-sided limit exists only if they agree.

Course alignment, for teachers

AP Calculus AB topic 1.7, Unit 1: Limits and Continuity.

Skill 1.C: Identify an appropriate mathematical rule or procedure based on the classification of a given expression (e.g., Use the chain rule to find the derivative of a composite function).

This lesson maps the topic's 0 learning objectives, since the course framework gives it no essential knowledge statements, to the part that teaches each one: 0 are checked by a scored question. The map is in the lesson file.