Is there a rule for limits of quotients like x3ex?

Topic 4.7 adds one, L'Hospital's Rule, once you have derivatives. Here, use the order of growth for large x: x, then powers of x, then exponentials with base greater than 1.

Unit 1 · Topic 1.15 · about 30 minutes · 7 parts · 12 scored checks, 3 of them multiple choice

AP Calculus AB 1.15: Connecting Limits at Infinity and Horizontal Asymptotes

Q: If x f(x) = , does the limit exist?

No. Writing describes how the limit fails to exist: the outputs grow without bound and approach no real number. That end of the graph has no horizontal asymptote.

Q: Can a function have two different horizontal asymptotes?

Yes, one at each end. 4xx2 + 9 has y = 4 on the right and y = -4 on the left. It cannot have more than two, because there are only two ends.

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

Use limits as $x \to \infty$ and as $x \to - \infty$ to describe a function's end behavior, find the horizontal asymptotes of its graph and compare how fast two functions grow.

Key vocabulary

Term	What it means	The exam trap
Limit at infinity	A limit such as $\lim_{x \to \infty} f (x)$ or $\lim_{x \to - \infty} f (x)$ : the value the outputs approach as the input grows or decreases without bound.	$\infty$ says which way $x$ is heading; it is never a number to substitute into the formula.
Horizontal asymptote	A line $y = L$ such that $\lim_{x \to \infty} f (x) = L$ or $\lim_{x \to - \infty} f (x) = L$ .	It describes the ends of the graph only; in the middle the graph may cross it.
End behavior	What the outputs of a function do as $x \to \infty$ and as $x \to - \infty$ , stated with limits at infinity.	State each end with its own limit, in limit notation, rather than describing the graph in words alone.
Relative magnitude	How large one function is compared with another for large inputs, read from the limit of their quotient.	Say which function is on top before you read the limit: $\frac{f}{g}$ and $\frac{g}{f}$ lead to opposite conclusions.

Graph of f(x) = (x squared + 3x) over (x squared + 1) for x from -12 to 12, with the dashed line y = 1. The graph crosses y = 1 at x = 1/3, swings up to about 2.08 and down to about -1.08 near the origin, and then flattens toward y = 1 at both ends, through the points (-10, 0.69) and (10, 1.29).

Part 1 Think first What does a shirt cost in the long run?

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

Predict first

A print shop charges a 600 dollar setup fee for a design, plus 8 dollars for each shirt it prints. For an order of $n$ shirts, the average cost per shirt is $A (n) = \frac{600 + 8 n}{n}$ dollars, which is 14 dollars a shirt for 100 shirts. What happens to the average cost per shirt as the orders get larger and larger?

In short. A limit at infinity, such as $\lim_{x \to \infty} f (x)$ , describes what the outputs of $f$ approach as $x$ grows without bound, and when that limit is a number it gives a horizontal asymptote.

If $\lim_{x \to \infty} f (x) = L$ or $\lim_{x \to - \infty} f (x) = L$ for a number $L$ , the line $y = L$ is a horizontal asymptote of the graph of $f$ .
For a rational function, dividing the top and the bottom by the highest power of $x$ in the denominator shows its end behavior.
The limit of a quotient $\frac{f (x)}{g (x)}$ as $x \to \infty$ compares how fast $f$ and $g$ grow.

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

15a. Limits at infinity and horizontal asymptotes

A limit at infinity asks what the outputs of $f$ approach as the input grows without bound. We write $\lim_{x \to \infty} f (x) = L$ when $f (x)$ can be made as close to $L$ as we like by taking $x$ large enough, and $\lim_{x \to - \infty} f (x) = L$ when the same happens as $x$ decreases without bound. The symbol $\infty$ is not a number to substitute. It says which way $x$ is heading.

Take $f (x) = 5 + \frac{2}{x}$ . At $x = 10$ it is 5.2, at 100 it is 5.02, at 1000 it is 5.002. The piece $\frac{2}{x}$ shrinks toward 0, so $\lim_{x \to \infty} f (x) = 5$ . On the left, $f (- 10) = 4.8$ and $f (- 100) = 4.98$ , so $\lim_{x \to - \infty} f (x) = 5$ as well.

A few end behaviors to know cold. For any positive integer $n$ , $\frac{1}{x^{n}} \to 0$ at both ends. As $x \to - \infty$ , $e^{x} \to 0$ . As $x \to \infty$ , both $e^{x}$ and $\ln x$ grow without bound, which is written $\lim_{x \to \infty} e^{x} = \infty$ .

When $\lim_{x \to \infty} f (x) = L$ or $\lim_{x \to - \infty} f (x) = L$ for a number $L$ , the line $y = L$ is a horizontal asymptote of the graph of $f$ . It describes the ends of the graph, not the middle: the graph at the top crosses its asymptote $y = 1$ near the origin.

Quick check · Multiple choice

No calculator. Suppose $\lim_{x \to \infty} f (x) = 4$ . Which statement must be true?

15b. End behavior of rational functions

End behavior is what a function does as $x \to \infty$ and as $x \to - \infty$ , and limits at infinity are how calculus states it. For a rational function, divide the numerator and the denominator by the highest power of $x$ in the denominator. Every term that is left with $x$ in its denominator then goes to 0.

For $f (x) = \frac{6 x^{2} - x}{3 x^{2} + 4}$ , divide by $x^{2}$ :

\lim_{x \to \infty} \frac{6 - \frac{1}{x}}{3 + \frac{4}{x^{2}}} = \frac{6 - 0}{3 + 0} = 2,

so $y = 2$ is a horizontal asymptote. A value agrees: $f (100) \approx 1.9964$ .

The same division gives three cases, by comparing the degree of the top with the degree of the bottom:

top degree lower: the limit is 0;
equal degrees: the limit is the quotient of the leading coefficients, signs included;
top degree higher: the outputs grow or decrease without bound, so that end has no horizontal asymptote.

The leading coefficient belongs to the highest power, wherever that term is written. In $\frac{5 - 3 x}{x + 2}$ it is $- 3$ on top, so the limit is $\frac{- 3}{1} = - 3$ .

The same move works beyond rational functions: for $\frac{3 e^{x} - 1}{2 e^{x} + 4}$ , divide the top and the bottom by $e^{x}$ , and the limit as $x \to \infty$ is $\frac{3}{2}$ .

When a rational function has a horizontal asymptote, it has the same one at both ends. Other functions need not, so work out $x \to \infty$ and $x \to - \infty$ separately.

Quick check · Sort it

No calculator. Each card is a function $f (x)$ . Tap a card, then tap what $f (x)$ does as $x \to \infty$ .

Approaches 0

Approaches a nonzero number

Grows or decreases without bound

15c. Which function grows faster?

Limits at infinity also compare two functions. Suppose $f (x)$ and $g (x)$ are positive and grow without bound as $x \to \infty$ . Look at $\lim_{x \to \infty} \frac{f (x)}{g (x)}$ :

if it is 0, $g$ grows faster, and $f$ becomes negligible next to $g$ ;
if it grows without bound, $f$ grows faster;
if it is a positive number, the two grow at the same rate.

For example, $\frac{x^{2} + 5 x}{x^{2}}$ is 1.5 at $x = 10$ , 1.05 at 100 and 1.005 at 1000. Its limit is 1, so next to $x^{2}$ the $5 x$ stops mattering. That is why only the leading terms decided the limits in the last idea.

For large $x$ the common families line up in order. $\ln x$ grows more slowly than any positive power $x^{p}$ , and every power of $x$ grows more slowly than an exponential $b^{x}$ with $b > 1$ . So $\lim_{x \to \infty} \frac{x^{3}}{e^{x}} = 0$ and $\lim_{x \to \infty} \frac{\ln x}{x} = 0$ .

A few values can mislead. At $t = 5$ , $3 t^{2} = 75$ is larger than $2^{t} = 32$ , but the quotient $\frac{3 t^{2}}{2^{t}}$ is about 0.29 at $t = 10$ and 0.0011 at $t = 20$ , and its limit is 0. The same limit compares rates of change: if two tanks fill at $3 t^{2}$ and $2^{t}$ liters per minute, the second tank's rate eventually leaves the first one's far behind.

Quick check · Decide and justify

No calculator. A classmate compares $x^{5}$ and $2^{x}$ . At $x = 10$ , $x^{5} = 100,000$ and $2^{x} = 1024$ , so the classmate concludes that $x^{5}$ grows faster and that $\frac{x^{5}}{2^{x}}$ grows without bound as $x \to \infty$ . Is the classmate right?

Which reason supports your answer?

Common mistakes

The mistake	Why it is wrong	Do this instead
Taking the leading coefficient from the term written first, as in reading 1 as the top's leading coefficient in $\frac{1 + 2 x^{2}}{3 - x^{2}}$ .	The leading coefficient belongs to the highest power, wherever it sits. Here it is 2 on top and $- 1$ on the bottom, so the limit is $- 2$ .	Rewrite each polynomial with its powers in descending order before you compare the degrees.
Assuming the limit as $x \to - \infty$ is the same as the limit as $x \to \infty$ .	Rational functions with a horizontal asymptote behave that way, but functions built from $e^{x}$ or square roots need not: $\sqrt{x^{2}}$ is $- x$ , not $x$ , when $x$ is negative.	Work out each end on its own, and write a separate limit statement for each.
Deciding which function grows faster from a few values in a table.	A power can lead an exponential for a while: $3 t^{2}$ is larger than $2^{t}$ at $t = 5$ , yet $\frac{3 t^{2}}{2^{t}} \to 0$ .	Find the limit of the quotient, or use the order for large $x$ : $\ln x$ , then powers of $x$ , then exponentials with base greater than 1.

Part 3 Watch it done Two different ends

Watch each step, then do the last one yourself.

Find every horizontal asymptote of the graph of $f (x) = \frac{4 x}{\sqrt{x^{2} + 9}}$ .

Step 1 · Divide by the highest power, for $x > 0$
Under the root the highest power is $x^{2}$ , so divide by $\sqrt{x^{2}}$ . For $x > 0$ , $\sqrt{x^{2}} = x$ , so divide the top and the bottom by $x$ : the bottom becomes $\frac{\sqrt{x^{2} + 9}}{x} = \sqrt{\frac{x^{2} + 9}{x^{2}}} = \sqrt{1 + \frac{9}{x^{2}}}$ , and $f (x) = \frac{4}{\sqrt{1 + \frac{9}{x^{2}}}}$ for $x > 0$ .
Step 2 · Let $x$ grow
As $x \to \infty$ , $\frac{9}{x^{2}} \to 0$ , so $\lim_{x \to \infty} f (x) = \frac{4}{\sqrt{1 + 0}} = 4$ , and $y = 4$ is a horizontal asymptote.
Step 3 · Rewrite for $x < 0$
For $x < 0$ , $\sqrt{x^{2}} = - x$ , so divide the top and the bottom by $- x$ . The top becomes $\frac{4 x}{- x} = - 4$ , and the bottom becomes $\frac{\sqrt{x^{2} + 9}}{- x} = \sqrt{1 + \frac{9}{x^{2}}}$ . So $f (x) = \frac{- 4}{\sqrt{1 + \frac{9}{x^{2}}}}$ for $x < 0$ .
Why this step?
Why divide by $- x$ , and not by $x$ , when $x$ is negative?
To move the divisor inside the square root, it has to equal $\sqrt{x^{2}}$ , and $\sqrt{x^{2}} = | x |$ is never negative. For $x < 0$ , $| x | = - x$ . Dividing by $x$ instead would quietly treat $\sqrt{x^{2}}$ as $x$ and flip the sign of the bottom, which is how the left end gets the wrong asymptote.
Your step · Calculate

No calculator. Finish it. For $x < 0$ , $f (x) = \frac{4 x}{\sqrt{x^{2} + 9}}$ can be written $\frac{- 4}{\sqrt{1 + \frac{9}{x^{2}}}}$ . What is $\lim_{x \to - \infty} f (x)$ ?

Your answer

Part 4 Try it Two new functions

Try it · Calculate

No calculator. Find $\lim_{x \to \infty} \frac{5 x^{3} - 2 x}{4 - 2 x^{3}}$ . Give your answer as an exact decimal.

Your answer

Try it · Multiple choice

No calculator. Which of these limits is equal to 0?

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 large tank holds 200 liters of pure water. Starting at time $t = 0$ , salt water flows into the tank at 10 liters per minute, and each liter of it carries 0.3 kilogram of salt. Nothing flows out, and the tank stays well mixed. For $t \geq 0$ minutes, the concentration of salt in the tank, in kilograms per liter, is

C (t) = \frac{3 t}{200 + 10 t} .

(a) Find $\lim_{t \to \infty} C (t)$ . Show the work that leads to your answer.

(b) The graph of $C$ has a horizontal asymptote. Write its equation. Using correct units, interpret the meaning of this asymptote in the context of the problem.

(d) Is there a time $t \geq 0$ at which $C (t) = 0.3$ ? Explain your reasoning.

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

Model answer

(a) For $t > 0$ , divide the numerator and the denominator by $t$ : $C (t) = \frac{3}{\frac{200}{t} + 10}$ . As $t \to \infty$ , $\frac{200}{t} \to 0$ , so $\lim_{t \to \infty} C (t) = \frac{3}{0 + 10} = 0.3$ .

(b) $y = 0.3$ . As time goes on, the concentration of salt in the tank gets as close as we like to 0.3 kilogram per liter, the concentration of the salt water flowing in.

(d) No. $C (t) = 0.3$ would need $3 t = 0.3 (200 + 10 t) = 60 + 3 t$ , that is, $0 = 60$ , which is false. The concentration approaches 0.3 kilogram per liter but never reaches it, because $0.3 - C (t) = \frac{60}{200 + 10 t} > 0$ for every $t \geq 0$ .

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

Divides the numerator and the denominator by $t$ (or compares the leading terms $3 t$ and $10 t$ ) and gets $\lim_{t \to \infty} C (t) = 0.3$ .
Writes $y = 0.3$ and says the concentration approaches 0.3 kilogram per liter as time increases, with units.
Solves $\frac{3 t}{200 + 10 t} = 0.29$ and gets $t = 580$ minutes.
Shows that $C (t) = 0.3$ leads to $0 = 60$ (or that $0.3 - C (t) > 0$ ), so the concentration never equals 0.3.

Part 6 Check yourself A function with two different ends

The two case questions use the same function. The table shows only its right end, so use the formula for the left.

Let $f (x) = \frac{5 e^{x} + 2}{e^{x} + 1}$ for all real numbers $x$ . Some of its values, rounded to four decimal places, are in the table.

Values of f(x)
$x$	$f (x)$
0	3.5000
2	4.6424
5	4.9799
10	4.9999

Case question 1 · Multiple choice

No calculator. Which line is the horizontal asymptote of the graph of $f$ at its left end, as $x \to - \infty$ ?

Case question 2 · Calculate

No calculator. Find $\lim_{x \to \infty} f (x)$ .

Your answer

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

Check yourself · Spot the error

No calculator. A student finds the horizontal asymptote of the graph of $g (x) = \frac{6 x^{2} - 5}{3 x^{3} + x}$ . Tap the first line with a mistake.

Check yourself · Fill the statement

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

Suppose $f (x)$ and $g (x)$ are positive and grow without bound as $x \to \infty$ . If $\lim_{x \to \infty} \frac{f (x)}{g (x)} = 0$ , then the function grows faster. If that limit is , then $f$ and $g$ grow at the same rate.

Part 7 Keep it Bring back earlier ideas

Two earlier ideas this lesson builds on. Answer them from memory before you look back.

Bring it back · Calculate

No calculator. Let $f (x) = \frac{x + 3}{x^{2} - 9}$ . The graph of $f$ has exactly one vertical asymptote. At what value of $x$ is it?

Your answer

Bring it back · Calculate

No calculator. Suppose $\lim_{x \to 1} f (x) = 4$ and $\lim_{x \to 1} g (x) = - 3$ . Find $\lim_{x \to 1} [2 f (x) - {(g (x))}^{2}]$ .

Your answer

In three lines

A limit at infinity tells what the outputs approach as $x$ grows or decreases without bound.
A finite limit at either end gives a horizontal asymptote $y = L$ , and each end is worked out on its own.
The limit of $\frac{f (x)}{g (x)}$ as $x \to \infty$ shows which function grows faster, or that they keep pace.

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 for a horizontal asymptote, write the limit that produces it, as in " $\lim_{x \to \infty} f (x) = 2$ , so $y = 2$ is a horizontal asymptote," and give the asymptote as the equation of a line. For a function that is not rational, find the limit as $x \to \infty$ and as $x \to - \infty$ separately.

Questions students ask

Can a graph cross its horizontal asymptote?

Yes. A horizontal asymptote is about the ends of the graph. The graph at the top of this lesson crosses $y = 1$ at $x = \frac{1}{3}$ and still closes in on it at both ends.

\lim_{x \to \infty} f (x) = \infty

, does the limit exist?

No. Writing $\infty$ describes how the limit fails to exist: the outputs grow without bound and approach no real number. That end of the graph has no horizontal asymptote.

Can a function have two different horizontal asymptotes?

Yes, one at each end. $\frac{4 x}{\sqrt{x^{2} + 9}}$ has $y = 4$ on the right and $y = - 4$ on the left. It cannot have more than two, because there are only two ends.

Why do only the leading terms matter for a rational function?

Because every other term is negligible next to the leading one for large $x$ . The quotient $\frac{x^{2} + 5 x}{x^{2}}$ approaches 1, so far out $x^{2} + 5 x$ and $x^{2}$ are nearly the same size compared with each other.

Is there a rule for limits of quotients like

\frac{x^{3}}{e^{x}}

Topic 4.7 adds one, L'Hospital's Rule, once you have derivatives. Here, use the order of growth for large $x$ : $\ln x$ , then powers of $x$ , then exponentials with base greater than 1.

Course alignment, for teachers

AP Calculus AB topic 1.15, Unit 1: Limits and Continuity.

Skill 2.D: Identify how mathematical characteristics or properties of functions are related in different representations.

This lesson maps all 3 essential knowledge statements of the topic to the part that teaches each one: 3 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 What does a shirt cost in the long run?

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

15a. Limits at infinity and horizontal asymptotes

15b. End behavior of rational functions

15c. Which function grows faster?

Part 3 Watch it done Two different ends

Part 4 Try it Two new functions

Part 5 Explain it Write it the way the exam asks

Part 6 Check yourself A function with two different ends

Part 7 Keep it Bring back earlier ideas

Exam strategy for this topic

Questions students ask

Past exam questions on this topic