Unit 1 · Topic 1.6 · about 30 minutes · 7 parts · 12 scored checks, 3 of them multiple choice

AP Calculus AB 1.6: Determining Limits Using Algebraic Manipulation

Q: Why may I divide out a factor that is 0 at the input?

Because the limit never uses the input itself. At every x near c other than c, the factor is not 0, so dividing it out is allowed there, and those are the only inputs the limit looks at.

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

Determine a limit by rewriting the function as an equivalent expression, by factoring, by multiplying by a conjugate or by using a trigonometric identity, when substitution gives $\frac{0}{0}$ .

Key vocabulary

Term	What it means	The exam trap
Equivalent expressions	Two expressions that give the same output at every input in a set, such as $\frac{x^{2} - 1}{x - 1}$ and $x + 1$ at every $x \neq 1$ .	Each time you replace one form with the other, write the inputs where they agree, such as $x \neq 1$ , so no step claims more than the algebra gives.
Common factor	A factor of the whole numerator and of the whole denominator, such as $x - 1$ in $\frac{(x - 1) (x + 1)}{x - 1}$ , which can be divided out.	Factor completely before you divide anything out; a piece that is added to other terms is not a factor.
Conjugate	For a sum or difference with a square root, such as $\sqrt{x} - 5$ , the same two terms with the opposite sign between them, $\sqrt{x} + 5$ .	The conjugate changes the sign between the terms, not the sign in front of the square root.
Alternate form of a trigonometric function	The same trigonometric expression written another way by an identity, such as $1 - \cos^{2} x$ for $\sin^{2} x$ , or $\frac{\sin x}{\cos x}$ for $\tan x$ .	Choose the form that makes a factor appear in both the numerator and the denominator. A form that exposes no common factor leaves the $\frac{0}{0}$ where it was.

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Part 1 Think first A cell the spreadsheet cannot fill

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

Predict first

Priya fills a spreadsheet column with $y = \frac{x^{2} - 3 x}{x - 3}$ . Every cell works except the one for $x = 3$ , which shows a divide-by-zero error, because there the formula is $\frac{9 - 9}{3 - 3} = \frac{0}{0}$ . As $x$ gets closer and closer to 3, what value does $y$ approach?

In short. When substituting into a limit gives $\frac{0}{0}$ , rewrite the function as an equivalent expression that agrees with it near the input, and find the limit of that expression instead.

$\frac{0}{0}$ from substitution is not a value. It says the expression has to be rewritten before the limit can be found.
Factoring and dividing out a common factor, multiplying by a conjugate, and switching to another form of a trigonometric function each give an equivalent expression.
An expression that equals the function for every input near $c$ , other than $c$ itself, has the same limit as $x \to c$ .

Part 2 Learn it Three ways to rewrite, each checked as you go

2a. When substitution gives 0/0, factor

In 1.5 the limit theorems let you find many limits by substituting. Try that on $\lim_{x \to 2} \frac{x^{2} + 2 x - 8}{x - 2}$ , the function graphed at the top, and you get $\frac{0}{0}$ . That is not an answer, and it does not mean the limit fails to exist. It means the top and the bottom share a factor that is 0 at $x = 2$ , and your job is to find it.

Factor, then divide out the common factor:

\frac{x^{2} + 2 x - 8}{x - 2} = \frac{(x + 4) (x - 2)}{x - 2} = x + 4 for x \neq 2.

The two expressions are equivalent for every $x$ except 2. A limit as $x \to 2$ looks only at inputs near 2, never at 2 itself, so the two have the same limit:

\lim_{x \to 2} \frac{x^{2} + 2 x - 8}{x - 2} = \lim_{x \to 2} (x + 4) = 6.

The graph shows it: the quotient is the line $y = x + 4$ with an open circle at $(2, 6)$ .

Two cautions. Divide out only a factor of the whole top and the whole bottom, never one term of a sum. And if substitution gives a number, such as $\frac{0}{5} = 0$ , you are already done. Only $\frac{0}{0}$ calls for a rewrite.

Quick check · Calculate

No calculator. Find $\lim_{x \to 4} \frac{x^{2} - 16}{x^{2} - 4 x}$ .

Your answer

2b. Square roots: multiply by the conjugate

A square root in a sum or difference hides its common factor. Take $\lim_{x \to 0} \frac{\sqrt{x + 9} - 3}{x}$ . Substitution gives $\frac{3 - 3}{0} = \frac{0}{0}$ , and nothing factors in sight.

The conjugate of $\sqrt{x + 9} - 3$ is $\sqrt{x + 9} + 3$ : the same two terms with the sign between them changed. Their product has no square root in it, because $(a - b) (a + b) = a^{2} - b^{2}$ :

(\sqrt{x + 9} - 3) (\sqrt{x + 9} + 3) = (x + 9) - 9 = x .

Multiply the top and the bottom by the conjugate. That is multiplying by $\frac{\sqrt{x + 9} + 3}{\sqrt{x + 9} + 3}$ , which is 1, so the expression stays equivalent:

\frac{\sqrt{x + 9} - 3}{x} \cdot \frac{\sqrt{x + 9} + 3}{\sqrt{x + 9} + 3} = \frac{x}{x (\sqrt{x + 9} + 3)} = \frac{1}{\sqrt{x + 9} + 3}

for $x \neq 0$ . Now substitution works, and the limit is $\frac{1}{3 + 3} = \frac{1}{6}$ .

On the side without the square root, leave the conjugate as a factor in parentheses. Expanding it there gains nothing, and the factor you divide out comes from the other side.

Quick check · Multiple choice

No calculator. What is $\lim_{x \to 16} \frac{\sqrt{x} - 4}{x - 16}$ ?

2c. Trigonometric limits: switch to another form

With sines and cosines, the rewrite is an identity you already know:

$\sin^{2} x + \cos^{2} x = 1$ , so $\sin^{2} x = 1 - \cos^{2} x$ and $\cos^{2} x = 1 - \sin^{2} x$ ;
$\sin (2 x) = 2 \sin x \cos x$ ;
$\tan x = \frac{\sin x}{\cos x}$ .

Pick the form that makes one factor appear in both the top and the bottom. For $\lim_{x \to 0} \frac{\sin^{2} x}{1 - \cos x}$ , substitution gives $\frac{0}{1 - 1} = \frac{0}{0}$ . Rewrite the top with the Pythagorean identity, and it factors as a difference of squares:

\frac{\sin^{2} x}{1 - \cos x} = \frac{1 - \cos^{2} x}{1 - \cos x} = \frac{(1 - \cos x) (1 + \cos x)}{1 - \cos x} = 1 + \cos x

for every $x$ near 0 other than 0. So the limit is $1 + \cos 0 = 2$ .

The identity did not find the limit by itself. It exposed a common factor, and from there the work is the same as in a rational function. As everywhere in calculus, the angles are in radians.

Quick check · Calculate

No calculator. Find $\lim_{x \to π} \frac{\sin (2 x)}{\sin x}$ .

Your answer

Common mistakes

The mistake	Why it is wrong	Do this instead
Writing $\frac{0}{0}$ as the limit, or calling it 0 or 1.	$\frac{0}{0}$ is not a number. It only says the top and the bottom both approach 0, and the limit can still be any number.	Treat $\frac{0}{0}$ as a signal: rewrite the expression, then substitute into the rewritten form.
Canceling a term that is added to others, such as the $x^{2}$ in $\frac{x^{2} - 16}{x^{2} - 4 x}$ .	Dividing the top and the bottom by the same thing is allowed only when it is a factor of each whole expression.	Factor the top and the bottom completely, and divide out only a factor that appears in both.
Multiplying only the numerator by the conjugate.	That multiplies the expression by something other than 1, so the new expression is not equivalent and its limit can differ.	Multiply by the conjugate over itself, so the top and the bottom both change and the value does not.

Part 3 Watch it done A conjugate in the denominator

Watch each step, then do the last one yourself.

Find $\lim_{x \to 5} \frac{x^{2} - 25}{\sqrt{x + 4} - 3}$ .

Step 1 · Try substitution
The top is $25 - 25 = 0$ and the bottom is $\sqrt{9} - 3 = 0$ , so substitution gives $\frac{0}{0}$ . The square root is in the bottom, so its conjugate, $\sqrt{x + 4} + 3$ , is the tool.
Step 2 · Multiply by the conjugate over itself
$\frac{x^{2} - 25}{\sqrt{x + 4} - 3} \cdot \frac{\sqrt{x + 4} + 3}{\sqrt{x + 4} + 3} = \frac{(x^{2} - 25) (\sqrt{x + 4} + 3)}{(x + 4) - 9} = \frac{(x^{2} - 25) (\sqrt{x + 4} + 3)}{x - 5}$
Why this step?
Why does multiplying by $\frac{\sqrt{x + 4} + 3}{\sqrt{x + 4} + 3}$ leave the limit the same?
For every $x$ near 5 that fraction is a positive number over itself, which is 1. Multiplying by 1 gives an equivalent expression, so its limit is the same. The conjugate is chosen because $(\sqrt{x + 4} - 3) (\sqrt{x + 4} + 3) = (x + 4) - 9$ has no square root left.
Step 3 · Factor and divide out
$x^{2} - 25 = (x - 5) (x + 5)$ , so for $x \neq 5$ the expression is $\frac{(x - 5) (x + 5) (\sqrt{x + 4} + 3)}{x - 5} = (x + 5) (\sqrt{x + 4} + 3)$ .
Your step · Calculate

No calculator. Finish it. For $x \neq 5$ , $\frac{x^{2} - 25}{\sqrt{x + 4} - 3} = (x + 5) (\sqrt{x + 4} + 3)$ . What is $\lim_{x \to 5} \frac{x^{2} - 25}{\sqrt{x + 4} - 3}$ ?

Your answer

Part 4 Try it A claim to judge, and a fraction inside a fraction

Try it · Decide and justify

No calculator. Let $f (x) = \frac{x^{2} - 2 x}{x - 2}$ . A classmate says $\lim_{x \to 2} f (x)$ does not exist, because $f (2)$ is $\frac{0}{0}$ and has no value. Is the classmate right?

Which reason supports your answer?

Try it · Calculate

No calculator. Find $\lim_{x \to 0} \frac{\frac{1}{x + 2} - \frac{1}{2}}{x}$ . Give an exact decimal.

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. Water drains from a tank. The depth of the water, in centimeters, $t$ minutes after a valve is opened is $D (t) = 100 - 20 \sqrt{t}$ for $0 \leq t \leq 25$ . For $t \neq 4$ , let $R (t) = \frac{D (t) - D (4)}{t - 4}$ .

(a) Explain why $\lim_{t \to 4} R (t)$ cannot be found by substituting $t = 4$ into $R (t)$ .

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

(c) In part (b), the limit of $R (t)$ was found from a different expression, one that has a value at $t = 4$ . $R (4)$ is not defined. Explain why the two expressions must have the same limit as $t \to 4$ .

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

Model answer

(a) $D (4) = 100 - 20 \cdot 2 = 60$ . At $t = 4$ the numerator of $R$ is $D (4) - D (4) = 0$ and the denominator is $4 - 4 = 0$ , so substitution gives $\frac{0}{0}$ , which is not a value. $R (t)$ has to be rewritten first.

(b) For $t \neq 4$ , $R (t) = \frac{(100 - 20 \sqrt{t}) - 60}{t - 4} = \frac{- 20 (\sqrt{t} - 2)}{t - 4}$ . Multiply the numerator and the denominator by $\sqrt{t} + 2$ :

R (t) = \frac{- 20 (\sqrt{t} - 2) (\sqrt{t} + 2)}{(t - 4) (\sqrt{t} + 2)} = \frac{- 20 (t - 4)}{(t - 4) (\sqrt{t} + 2)} = \frac{- 20}{\sqrt{t} + 2} .

So $\lim_{t \to 4} R (t) = \frac{- 20}{2 + 2} = - 5$ centimeters per minute.

(c) For every $t \neq 4$ in $0 \leq t \leq 25$ , the work in (b) shows $R (t) = \frac{- 20}{\sqrt{t} + 2}$ . A limit as $t \to 4$ uses only the values at inputs near 4, never the value at 4 itself, so two expressions that agree at every $t$ near 4 other than 4 have the same limit there. That $R (4)$ is not defined does not change the limit.

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

Shows that the numerator and the denominator of $R$ are both 0 at $t = 4$ , so substitution gives $\frac{0}{0}$ .
Multiplies by $\sqrt{t} + 2$ over itself and divides out $t - 4$ , reaching $\frac{- 20}{\sqrt{t} + 2}$ .
Gets $\lim_{t \to 4} R (t) = - 5$ centimeters per minute.
Says the two expressions are equal for every $t$ near 4 other than 4, and that the limit does not use $t = 4$ .

Part 6 Check yourself One function, two inputs

The two case questions use the same function. Answer them from its definition.

A function $g$ is defined by one formula for every $x$ except 3 and a separate value at 3, as the table shows.

The definition of g
Input	$g (x)$
$x \neq 3$	$\frac{x^{2} - x - 6}{x - 3}$
$x = 3$	$1$

Case question 1 · Calculate

No calculator. Find $\lim_{x \to 3} g (x)$ .

Your answer

Case question 2 · Multiple choice

No calculator. What is $\lim_{x \to - 2} g (x)$ ?

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

Check yourself · Sort it

No calculator. Substitute first, and rewrite only when substitution gives $\frac{0}{0}$ . Tap a card, then tap what that rule says to do for its limit.

Factor polynomials and divide out a common factor

Multiply by the conjugate

Rewrite with a trigonometric identity

Substitute; no rewrite is needed

Check yourself · Spot the error

No calculator. A student finds $\lim_{x \to π / 2} \frac{\cos^{2} x}{1 - \sin x}$ . Tap the first line with a mistake.

Part 7 Keep it Bring back earlier ideas

This lesson stands on two earlier ideas: the limit theorems, and what a limit statement means. Pulling them back up is what makes them last.

Bring it back · Calculate

No calculator. Suppose $\lim_{x \to 1} f (x) = 3$ and $\lim_{x \to 1} g (x) = - 2$ . Find $\lim_{x \to 1} \frac{[f (x)]^{2} - 3 g (x)}{f (x) + g (x)}$ .

Your answer

Bring it back · Multiple choice

No calculator. Which statement says what $\lim_{x \to 2} f (x) = 7$ means?

In three lines

Substitution that gives $\frac{0}{0}$ is a signal to rewrite, not an answer.
Factor and divide out, multiply by a conjugate, or use a trigonometric identity to reach an equivalent expression.
The rewritten expression agrees with the function near the input, so its limit there is the function's limit.

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 no-calculator question, write the substitution first and say what it gives. If it is $\frac{0}{0}$ , show the rewrite itself: the factoring, the conjugate or the identity. Keep $\lim_{x \to c}$ in front of every line until you substitute, because a line without it says the expression equals the number, which is not true.

Questions students ask

Does

\frac{0}{0}

mean the limit does not exist?

No. It means substitution has told you nothing yet. The limit may be any number, or it may fail to exist; rewriting the expression is how you find out.

Why may I divide out a factor that is 0 at the input?

Because the limit never uses the input itself. At every $x$ near $c$ other than $c$ , the factor is not 0, so dividing it out is allowed there, and those are the only inputs the limit looks at.

What if substitution gives a nonzero number over 0?

Then rewriting will not help. Near that input the quotient is unbounded, so the limit does not exist as a real number. Topic 1.14 shows how to describe that behavior with $\infty$ and $- \infty$ . The rewrites in this lesson are for $\frac{0}{0}$ .

How do I know which rewrite to use?

Read the expression. Polynomials on the top and the bottom: factor. A square root in a sum or difference: use the conjugate. Sines, cosines and tangents: look for an identity that exposes a common factor.

Can't I just use a table of values instead?

A table gives an estimate, as in 1.4, and it can mislead. The algebra gives the limit exactly, and when a question asks you to find a limit without a calculator, the algebra is the work to show.

Course alignment, for teachers

AP Calculus AB topic 1.6, 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 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.