Unit 4 · Topic 4.7 · about 30 minutes · 7 parts · 13 scored checks, 3 of them multiple choice

AP Calculus AB 4.7: Using L’Hospital’s Rule for Determining Limits of Indeterminate Forms

Q: Is L'Hospital's Rule the same as the quotient rule?

No. The quotient rule finds the derivative of f(x)g(x). L'Hospital's Rule finds a limit, using f'(x) over g'(x), each found on its own.

Q: Why can't I write that a limit equals 00?

Because 00 is not a number. It names what the top and the bottom are doing, so write "the limit has the form 00" instead.

Q: Does the rule work for limits as x goes to infinity?

Yes. If the quotient tends to 00 or as x or x -, the rule applies in the same way.

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

Unit 4: 10 to 15% of the multiple-choice section

What you'll be able to do

Determine limits of quotients that result in the indeterminate forms $\frac{0}{0}$ and $\frac{\infty}{\infty}$ , using L'Hospital's Rule after checking that it applies.

Key vocabulary

Term	What it means	The exam trap
Indeterminate form	The form of a limit of a quotient whose numerator and denominator both approach 0, or both become infinite, written $\frac{0}{0}$ or $\frac{\infty}{\infty}$ .	The form is a label for a situation, not a value, so never write it after an equal sign.
The form $\frac{0}{0}$	A quotient whose numerator and denominator both approach 0.	Say the limit has the form $\frac{0}{0}$ ; never write $\frac{0}{0}$ as its value.
The form $\frac{\infty}{\infty}$	A quotient whose numerator and denominator both grow without bound, toward $\infty$ or $- \infty$ .	Do not read the form as a number; compare how fast the two parts grow.
L'Hospital's Rule	If $\frac{f (x)}{g (x)}$ has the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ as $x \to a$ , with $f$ and $g$ differentiable and $g^{'} (x) \neq 0$ near $a$ , then $\lim_{x \to a} \frac{f (x)}{g (x)} = \lim_{x \to a} \frac{f^{'} (x)}{g^{'} (x)}$ whenever the second limit exists.	Show that the rule applies before you use it, and differentiate the numerator and the denominator separately.

Graphs of y = 2 ln x and y = x cubed minus 1 for x from 0 to 2. Both curves pass through the point (1, 0), and a dashed tangent line at that point has slope 2 on the first curve and slope 3 on the second.

Part 1 Think first Two puddles drying out

Make a guess before the lesson starts. Even a wrong guess makes the answer easier to remember.

Predict first

After a rain shower, two puddles dry in the sun. At noon puddle A covers 36 square feet and puddle B covers 12. Their areas $t$ hours after noon are $A (t) = 36 - t^{2}$ and $B (t) = 12 - 2 t$ square feet, so both are gone at $t = 6$ . In the last moments before they vanish, about how many times the area of puddle B is puddle A?

In short. A limit of a quotient whose top and bottom both approach 0, or both become infinite, has an indeterminate form, and L'Hospital's Rule evaluates it as the limit of the derivative of the top over the derivative of the bottom.

The forms $\frac{0}{0}$ and $\frac{\infty}{\infty}$ are labels, not numbers: the form alone does not determine the limit.
Check the form and the rule's conditions before using it; on any other form it can give a wrong value.
Differentiate the top and the bottom separately, then check the new form; if it is still indeterminate, check the conditions and use the rule again.

Part 2 Learn it The form, the rule and the second look

2a. Two indeterminate forms: $\frac{0}{0}$ and $\frac{\infty}{\infty}$

Some limits of a quotient $\frac{f (x)}{g (x)}$ cannot be found by substituting. If the top and the bottom both approach 0, the quotient tends to $\frac{0}{0}$ . If both grow without bound, toward $\infty$ or $- \infty$ , it tends to $\frac{\infty}{\infty}$ . These two are called indeterminate forms, because the form by itself does not determine the limit.

Three quotients near $x = 0$ show why. Each one tends to $\frac{0}{0}$ , yet $\frac{x}{x}$ approaches 1, $\frac{x^{2}}{x}$ approaches 0 and $\frac{5 x}{x}$ approaches 5. Same form, different limits.

A quotient that tends to anything else is not indeterminate. As $x \to 3$ , $\frac{x - 3}{x + 2}$ tends to $\frac{0}{5}$ , and its limit is 0. A top that approaches a number other than 0 over a bottom that approaches 0 is not indeterminate either: the quotient becomes unbounded, so the limit does not exist as a number.

The symbols $\frac{0}{0}$ and $\frac{\infty}{\infty}$ describe a situation; they are not values. Say "the limit has the form $\frac{0}{0}$ " in words, and never write $\frac{0}{0}$ after an equal sign.

Quick check · Sort it

No calculator. Substitute, or think about where the top and the bottom are headed. Tap a card, then tap whether its limit has an indeterminate form.

Indeterminate form

Not an indeterminate form

2b. The rule: differentiate the top and the bottom separately

L'Hospital's Rule evaluates a limit with one of these two forms. Suppose that near $x = a$ , though not necessarily at $a$ itself:

$f$ and $g$ are differentiable, and $g^{'} (x) \neq 0$ ;
$\lim_{x \to a} f (x) = 0$ and $\lim_{x \to a} g (x) = 0$ , or both limits are infinite;
$\lim_{x \to a} \frac{f^{'} (x)}{g^{'} (x)}$ exists.

Then

\lim_{x \to a} \frac{f (x)}{g (x)} = \lim_{x \to a} \frac{f^{'} (x)}{g^{'} (x)} .

The rule holds the same way for limits as $x \to \infty$ or $x \to - \infty$ .

The new quotient is the derivative of the top over the derivative of the bottom. It is not the derivative of the quotient, so the quotient rule plays no part.

The picture at the top shows why the rule works for $\frac{0}{0}$ . Near $x = 1$ each curve stays close to its tangent line, so $2 \ln x$ is about $2 (x - 1)$ and $x^{3} - 1$ is about $3 (x - 1)$ . Their quotient is then about $\frac{2}{3}$ , the quotient of the two slopes.

For $\lim_{x \to 0} \frac{\sin (3 x)}{x}$ : both parts approach 0 and both are differentiable, and the derivative of $x$ is 1, never 0. The derivatives give $\frac{3 \cos (3 x)}{1}$ , which approaches 3, so the limit is 3.

Quick check · Multiple choice

No calculator. The limit $\lim_{x \to 1} \frac{\ln x}{x^{2} - 1}$ has the form $\frac{0}{0}$ . Which limit does L'Hospital's Rule say is equal to it?

2c. Check the form before the rule, and again after it

The rule is only for $\frac{0}{0}$ and $\frac{\infty}{\infty}$ . Used on any other form, it can give a wrong answer. By substitution, $\lim_{x \to 0} \frac{x + 1}{x + 2} = \frac{1}{2}$ , but the quotient of the derivatives is $\frac{1}{1} = 1$ . So name the form before you differentiate anything.

Then look again. The new quotient can still be indeterminate, and $\frac{0}{0}$ is not 0 or 1 or any other answer. If the new top and bottom meet the conditions, use the rule again.

For $\lim_{x \to \infty} \frac{x^{2}}{e^{x}}$ , the form is $\frac{\infty}{\infty}$ . One use gives $\lim_{x \to \infty} \frac{2 x}{e^{x}}$ , still $\frac{\infty}{\infty}$ . A second use gives $\lim_{x \to \infty} \frac{2}{e^{x}}$ . Now the top stays at 2 while the bottom grows without bound, so this limit is 0 and the form is no longer indeterminate. Stop there: $\lim_{x \to \infty} \frac{x^{2}}{e^{x}} = 0$ .

Each use needs its own indeterminate form, so a written answer names the form at every step. And each equal sign holds only because the last limit exists.

Quick check · Decide and justify

No calculator. Maya finds $\lim_{x \to 0} \frac{\cos x}{x + 1}$ with L'Hospital's Rule and writes $\lim_{x \to 0} \frac{- \sin x}{1} = 0$ . Jordan says the rule does not apply to this limit. Is Jordan right?

Which reason supports your answer?

Common mistakes

The mistake	Why it is wrong	Do this instead
Using L'Hospital's Rule on a limit whose form is not $\frac{0}{0}$ or $\frac{\infty}{\infty}$ .	The rule's conclusion depends on that condition. Without it the quotient of derivatives can be a different number, as with $\frac{x + 1}{x + 2}$ at 0, which approaches $\frac{1}{2}$ while the derivatives give 1.	Substitute first. Write the limits of the top and the bottom, name the form, and use the rule only if it is indeterminate.
Differentiating the whole quotient $\frac{f (x)}{g (x)}$ with the quotient rule.	The rule replaces the quotient with $\frac{f^{'} (x)}{g^{'} (x)}$ , the derivative of the top over the derivative of the bottom. The derivative of the quotient is a different function.	Find $f^{'} (x)$ and $g^{'} (x)$ on their own, then write one over the other.
Stopping after one use of the rule when the new quotient still tends to $\frac{0}{0}$ or $\frac{\infty}{\infty}$ , and reporting 0 or 1.	An indeterminate form is not a value, so the first use has not produced an answer yet.	Name the form of every new quotient. If it is still indeterminate and the conditions hold, use the rule again.

Part 3 Watch it done A limit that needs the rule twice

Watch each step, then do the last one yourself.

Find $\lim_{x \to 0} \frac{e^{2 x} - 1 - 2 x}{x^{2}}$ .

Step 1 · Name the form
At $x = 0$ the top is $e^{0} - 1 - 0 = 0$ and the bottom is $0^{2} = 0$ , so the limit has the form $\frac{0}{0}$ . Both parts are differentiable everywhere, and the bottom's derivative, $2 x$ , is not 0 for $x \neq 0$ .
Step 2 · Use the rule once
$\lim_{x \to 0} \frac{e^{2 x} - 1 - 2 x}{x^{2}} = \lim_{x \to 0} \frac{2 e^{2 x} - 2}{2 x}$ , provided the new limit exists. The chain rule gives the 2 in front of $e^{2 x}$ .
Step 3 · Name the form again
At $x = 0$ the new top is $2 e^{0} - 2 = 0$ and the new bottom is $2 \cdot 0 = 0$ . The form is still $\frac{0}{0}$ , so there is no answer yet. The new top and bottom are differentiable, and the derivative of $2 x$ is 2, never 0.
Why this step?
Why is it all right to use the rule a second time?
The new quotient meets the same conditions the first one did: its top and bottom are differentiable, both approach 0, and the derivative of the bottom is not 0. Each use of the rule needs only its own check, so it can be used as many times as those checks pass.
Your step · Calculate

No calculator. Finish it. Use L'Hospital's Rule on $\lim_{x \to 0} \frac{2 e^{2 x} - 2}{2 x}$ . What is $\lim_{x \to 0} \frac{e^{2 x} - 1 - 2 x}{x^{2}}$ ?

Your answer

Part 4 Try it A limit at infinity, and a worked solution to check

Try it · Calculate

No calculator. Find $\lim_{x \to \infty} \frac{\ln (x^{2} + 1)}{\ln x}$ .

Your answer

Try it · Spot the error

No calculator. A student finds $\lim_{x \to 0} \frac{\sin x}{e^{x} - 1}$ . Tap the first line with a mistake.

Part 5 Explain it Answer it as a free-response question

Free response on the exam is written by hand, so work this on paper first. Nothing you type here leaves this page. Then open the model answer and score yourself against 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. Three drones fly along a straight north-south track that passes a beacon. Their positions, in meters north of the beacon, are $A (t)$ , $B (t)$ and $C (t)$ at time $t$ seconds; a negative position is south of the beacon. Each of the three functions is twice differentiable, and its second derivative is continuous. All three drones are at the beacon at $t = 4$ seconds, and

$A (4) = 0, A^{'} (4) = 8, A^{''} (4) = - 2,$ $B (4) = 0, B^{'} (4) = 5, B^{''} (4) = 3,$ $C (4) = 0, C^{'} (4) = 0, C^{''} (4) = 5.$

(a) Explain why $\frac{A (t)}{B (t)}$ tends to an indeterminate form as $t \to 4$ .

(b) Use L'Hospital's Rule to find $\lim_{t \to 4} \frac{A (t)}{B (t)}$ . Interpret the meaning of this value in the context of the problem.

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

Model answer

(a) $A$ and $B$ are differentiable, so they are continuous, and $\lim_{t \to 4} A (t) = A (4) = 0$ and $\lim_{t \to 4} B (t) = B (4) = 0$ . The quotient tends to $\frac{0}{0}$ , an indeterminate form.

(b) The form is $\frac{0}{0}$ and $A$ and $B$ are differentiable. $A^{'}$ and $B^{'}$ are continuous and $B^{'} (4) = 5 \neq 0$ , so $B^{'} (t) \neq 0$ near $t = 4$ and $\lim_{t \to 4} \frac{A^{'} (t)}{B^{'} (t)} = \frac{A^{'} (4)}{B^{'} (4)} = \frac{8}{5}$ exists. By L'Hospital's Rule, $\lim_{t \to 4} \frac{A (t)}{B (t)} = \lim_{t \to 4} \frac{A^{'} (t)}{B^{'} (t)} = \frac{8}{5} = 1.6$ . At times close to $t = 4$ seconds, drone A is about 1.6 times as far from the beacon as drone B, on the same side of it.

(c) $\lim_{t \to 4} C (t) = C (4) = 0$ and $\lim_{t \to 4} (t - 4)^{2} = 0$ , so the form is $\frac{0}{0}$ . $C$ and $(t - 4)^{2}$ are differentiable, and the derivative of $(t - 4)^{2}$ , $2 (t - 4)$ , is not 0 for $t \neq 4$ . One use of the rule gives $\lim_{t \to 4} \frac{C^{'} (t)}{2 (t - 4)}$ . Since $C^{'}$ is continuous, $C^{'} (t) \to C^{'} (4) = 0$ while $2 (t - 4) \to 0$ , so the form is still $\frac{0}{0}$ . $C^{'}$ is differentiable and the derivative of $2 (t - 4)$ is $2 \neq 0$ , so the rule applies again: $\lim_{t \to 4} \frac{C^{''} (t)}{2} = \frac{C^{''} (4)}{2} = \frac{5}{2}$ , because $C^{''}$ is continuous. So $\lim_{t \to 4} \frac{C (t)}{(t - 4)^{2}} = 2.5$ meters per second squared.

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

Uses continuity of $A$ and $B$ to say both approach 0, so the quotient tends to $\frac{0}{0}$ .
States that the rule applies and writes $\lim_{t \to 4} \frac{A^{'} (t)}{B^{'} (t)} = \frac{8}{5}$ , getting 1.6.
Interprets 1.6: near $t = 4$ , drone A is about 1.6 times as far from the beacon as drone B.
Shows that one use of the rule on $\frac{C (t)}{(t - 4)^{2}}$ still gives $\frac{0}{0}$ , and uses it a second time.
Gets $\frac{C^{''} (4)}{2} = 2.5$ , in meters per second squared.

Part 6 Check yourself Two functions in a table

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

The functions $f$ and $g$ are twice differentiable for all $x$ , and $f^{''}$ and $g^{''}$ are continuous. The table gives their values and the values of their first two derivatives at three inputs.

Values of f, g and their derivatives
$x$	$f^{'} (x)$	$f^{''} (x)$	$g (x)$	$g^{'} (x)$	$g^{''} (x)$
1	2	$- 3$	4	$- 1$	2
2	$- 9$	1	0	$- 2$	3
3	0	$- 6$	0	0	$- 4$

Case question 1 · Multiple choice

No calculator. What is $\lim_{x \to 1} \frac{f (x)}{g (x)}$ ?

Case question 2 · Calculate

No calculator. What is $\lim_{x \to 2} \frac{f (x)}{g (x)}$ ? Give an exact decimal.

Your answer

Case question 3 · Calculate

No calculator. What is $\lim_{x \to 3} \frac{f (x)}{g (x)}$ ? Give an exact decimal.

Your answer

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

Check yourself · Match

No calculator. Each limit has an indeterminate form. Tap a limit, then its value.

Check yourself · Fill the statement

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

L'Hospital's Rule can be used on $\lim_{x \to a} \frac{f (x)}{g (x)}$ when $f (x)$ and $g (x)$ both approach , or both become infinite, and $\lim_{x \to a} \frac{f^{'} (x)}{g^{'} (x)}$ . The original limit then equals .

Part 7 Keep it Bring back earlier ideas

This lesson leans on the chain rule and on limits at infinity. Recalling them now keeps them ready.

Bring it back · Calculate

No calculator. Let $h (x) = (x^{2} + 1)^{3}$ . Find $h^{'} (1)$ .

Your answer

Bring it back · Multiple choice

No calculator. Which line is a horizontal asymptote of the graph of $y = \frac{6 x^{2} - x}{3 x^{2} + 5}$ ?

In three lines

A limit of a quotient whose top and bottom both approach 0, or both become infinite, has an indeterminate form.
L'Hospital's Rule replaces such a quotient with the derivative of the top over the derivative of the bottom, once its conditions are checked.
After each use, name the new form: stop when it is not indeterminate, and use the rule again when it is.

Unit 4 check: 10 questions from every lesson in Unit 4

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.

Show that L'Hospital's Rule applies before you use it: write the limit of the numerator and the limit of the denominator, say in words that the form is $\frac{0}{0}$ or $\frac{\infty}{\infty}$ , and then write the quotient of the derivatives. An equation such as $\lim_{x \to a} \frac{f (x)}{g (x)} = \frac{0}{0}$ is wrong, because the form is not a value.

Questions students ask

Is L'Hospital's Rule the same as the quotient rule?

No. The quotient rule finds the derivative of $\frac{f (x)}{g (x)}$ . L'Hospital's Rule finds a limit, using $f^{'} (x)$ over $g^{'} (x)$ , each found on its own.

Can I use the rule more than once on the same limit?

Yes, as long as each new quotient is still indeterminate and meets the conditions. Check the form every time, and stop as soon as it is not indeterminate.

Do I have to use L'Hospital's Rule on a

\frac{0}{0}

limit?

No. Any correct method works. Factoring and canceling, from Unit 1, is often quicker for a polynomial quotient such as $\frac{x^{2} - 9}{x - 3}$ .

Why can't I write that a limit equals

\frac{0}{0}

Because $\frac{0}{0}$ is not a number. It names what the top and the bottom are doing, so write "the limit has the form $\frac{0}{0}$ " instead.

Does the rule work for limits as x goes to infinity?

Yes. If the quotient tends to $\frac{0}{0}$ or $\frac{\infty}{\infty}$ as $x \to \infty$ or $x \to - \infty$ , the rule applies in the same way.

Beyond the AP Exam

College Board sets this limit on what the AP Exam asks about this topic. Anything past it is not on the exam.

There are many other indeterminate forms, such as ∞ − ∞, for example, but these will not be assessed on either the AP Calculus AB or BC Exam. However, teachers may include these topics, if time permits.

Course alignment, for teachers

AP Calculus AB topic 4.7, Unit 4: Contextual Applications of Differentiation.

Skill 3.D: Apply an appropriate mathematical definition, theorem, or test.

This lesson maps all 2 essential knowledge statements of the topic to the part that teaches each one: 2 are 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 Two puddles drying out

Part 2 Learn it The form, the rule and the second look

2a. Two indeterminate forms: 00 and ∞∞

2b. The rule: differentiate the top and the bottom separately

2c. Check the form before the rule, and again after it

Part 3 Watch it done A limit that needs the rule twice

Part 4 Try it A limit at infinity, and a worked solution to check

Part 5 Explain it Answer it as a free-response question

Part 6 Check yourself Two functions in a table

Part 7 Keep it Bring back earlier ideas

Exam strategy for this topic

Questions students ask

2a. Two indeterminate forms: $\frac{0}{0}$ and $\frac{\infty}{\infty}$