Unit 1 · Topic 1.14 · about 30 minutes · 7 parts · 12 scored checks, 3 of them multiple choice

AP Calculus AB 1.14: Connecting Infinite Limits and Vertical Asymptotes

Q: Is x a f(x) = a limit that exists?

No. A limit exists only when it is a real number, and is not one. Writing = is still useful: it says the limit fails to exist because the values grow without bound, rather than because they oscillate or the two sides disagree.

Q: What if the numerator and the denominator are both 0 at a?

Then 00 tells you nothing yet. Factor and cancel, and find the limit again. A real number means a hole; a number that is not 0 over a denominator approaching 0 means an asymptote. x - 2(x - 2)2 gives 00 at 2 and still has an asymptote there.

Q: Can I find vertical asymptotes from a calculator graph?

A graph can suggest one, but a window can hide behavior near an input, and a steep stretch can look like an asymptote that is not there. Confirm each one with a limit from the formula.

Q: Can a graph touch its vertical asymptote?

At most at one point, because a function has at most one output at a; a function defined in pieces can give f(a) a value. What the graph cannot do is stay bounded near a on the side where the limit is infinite.

Q: Do only rational functions have vertical asymptotes?

No. x 0+ x = -, so x = 0 is a vertical asymptote of y = x, and x has one at every odd multiple of 2. The same one-sided limit test applies.

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 involving infinity to describe and explain how a function behaves near an input where its values grow without bound, including at a vertical asymptote.

Key vocabulary

Term	What it means	The exam trap
Infinite limit	A limit statement such as $\lim_{x \to a} f (x) = \infty$ or $- \infty$ , which says the values of $f$ increase or decrease without bound as $x$ approaches $a$ .	Read both halves of the statement: the arrow names the input $x$ approaches, and the right side says what the values of $f$ do.
Unbounded	A function is unbounded near $a$ when, for inputs close enough to $a$ , its values pass any number you name in size.	Unbounded describes the outputs of the function; the inputs stay close to one number.
One-sided limit	What $f (x)$ approaches as $x$ approaches $a$ from one side only, written $\lim_{x \to a^{-}} f (x)$ from the left and $\lim_{x \to a^{+}} f (x)$ from the right.	The $+$ or $-$ in $a^{+}$ or $a^{-}$ names the side the input comes from, not the sign of the answer.
Vertical asymptote	A line $x = a$ for which at least one one-sided limit of $f$ at $a$ is $\infty$ or $- \infty$ .	Decide with a limit at that input, not with a glance at the formula or a calculator graph, which can hide behavior near $a$ .

Graph of f(x) = 1/(x - 2) for x from -1 to 5 with a dashed vertical line at x = 2. Right of 2 the marked points are (2.5, 2) and (2.1, 10), so the graph climbs as x nears 2. Left of 2 they are (1.5, -2) and (1.9, -10), so the graph falls.

Part 1 Think first Where does the image land?

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

Predict first

A camera lens has a focal length of 10 centimeters. When an object is $d$ centimeters in front of it, with $d > 10$ , the sharp image forms $v (d) = \frac{10 d}{d - 10}$ centimeters behind the lens. With the object 11 centimeters away, the image is 110 centimeters behind the lens. The object is moved in to 10.1 centimeters. About how far behind the lens is the image now?

In short. An infinite limit, such as $\lim_{x \to a^{+}} f (x) = \infty$ , says the values of $f$ grow without bound as $x$ approaches $a$ , and the line $x = a$ is a vertical asymptote when at least one one-sided limit there is $\infty$ or $- \infty$ .

$\infty$ is not a real number, so a limit written as $\infty$ or $- \infty$ does not exist, and the symbol says how it fails: the function is unbounded.
When a numerator approaches a number that is not 0 and the denominator approaches 0, the signs of the two parts on each side give the sign of the infinite limit.
A zero of the denominator is only a candidate: a limit that is a real number there means a hole in the graph, not an asymptote.

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

14a. When the values grow without bound

Some functions have no limit at an input because their values never settle: they grow without bound. Take $f (x) = \frac{1}{x^{2}}$ near 0. At $x = 0.1$ and at $x = - 0.1$ it is 100. At $x = 0.01$ and at $x = - 0.01$ it is 10,000. Closer inputs give larger outputs, with no ceiling.

Calculus extends limit notation to say this. We write

\lim_{x \to 0} \frac{1}{x^{2}} = \infty

to mean that $f (x)$ can be made larger than any number you name by taking $x$ close enough to 0, but not equal to 0. This is an infinite limit. Writing $- \infty$ instead means the values decrease without bound.

Two things to keep straight. First, $\infty$ is not a real number. The statement $\lim_{x \to 0} \frac{1}{x^{2}} = \infty$ says the limit does not exist, and it says how the limit fails: the function is unbounded near 0.

Second, an infinite limit can be one-sided. On the graph at the top, $\frac{1}{x - 2}$ climbs past 10 just right of 2 and falls past $- 10$ just left of it, so

\lim_{x \to 2^{+}} \frac{1}{x - 2} = \infty and \lim_{x \to 2^{-}} \frac{1}{x - 2} = - \infty .

The two sides disagree, so the two-sided limit at 2 does not exist, and it is neither $\infty$ nor $- \infty$ .

Quick check · Multiple choice

No calculator. A function $g$ has $\lim_{x \to 4^{-}} g (x) = - \infty$ . What does this statement say about $g$ ?

14b. Finding the sign from the formula

A few values suggest an infinite limit; the formula settles it. When the numerator of a quotient approaches a number that is not 0 and the denominator approaches 0, the quotient is unbounded. Its sign on each side comes from the signs of the two parts.

Take $\lim_{x \to 3^{+}} \frac{x + 1}{x - 3}$ . The numerator approaches 4. The denominator approaches 0, and because $x > 3$ it is positive. A number near 4 divided by a small positive number is a large positive number, so the limit is $\infty$ . (At $x = 3.01$ the quotient is 401.)

From the left, $x < 3$ makes $x - 3$ negative, so

\lim_{x \to 3^{-}} \frac{x + 1}{x - 3} = - \infty .

(At $x = 2.99$ the quotient is $- 399$ .)

A squared factor such as $(x - 3)^{2}$ is positive on both sides of 3, so it never flips the sign. With $(x - 3)^{2}$ in the denominator and a numerator near 4, the quotient is large and positive on both sides, and the two-sided limit is $\infty$ .

That is the rule for a two-sided infinite limit: it is $\infty$ only when both one-sided limits are $\infty$ , and $- \infty$ only when both are $- \infty$ .

Quick check · Sort it

No calculator. Tap a card, then tap the bin that says what the limit on it is.

The limit is $\infty$

The limit is $- \infty$

The two sides disagree

14c. Vertical asymptotes, explained with limits

The line $x = a$ is a vertical asymptote of the graph of $f$ when at least one one-sided limit at $a$ is $\infty$ or $- \infty$ . One side is enough. Near the line the graph runs alongside it, unbounded.

For a quotient, the zeros of the denominator are where to look, but a zero of the denominator is not yet an asymptote. Take

f (x) = \frac{x^{2} - 4}{x^{2} - x - 2} = \frac{(x - 2) (x + 2)}{(x - 2) (x + 1)} .

The denominator is 0 at $x = 2$ and at $x = - 1$ .

At $x = 2$ the numerator is 0 too. For $x \neq 2$ , $f (x) = \frac{x + 2}{x + 1}$ , so $\lim_{x \to 2} f (x) = \frac{4}{3}$ . That limit is a real number, so $f$ stays bounded near 2, and the graph has a hole there, not an asymptote.

At $x = - 1$ the numerator $x + 2$ approaches 1 and the denominator $x + 1$ approaches 0, through positive values from the right. So $\lim_{x \to - 1^{+}} f (x) = \infty$ , and the graph has a vertical asymptote at $x = - 1$ .

The limit is the explanation. An asymptote is a claim about unbounded behavior, and an infinite one-sided limit is the evidence for it.

Quick check · Decide and justify

No calculator. Let $q (x) = \frac{x^{2} - 5 x + 6}{x - 2}$ . A classmate says the graph of $q$ has a vertical asymptote at $x = 2$ , because the denominator is 0 there. Is the classmate right?

Which reason supports your answer?

Common mistakes

The mistake	Why it is wrong	Do this instead
Calling every zero of the denominator a vertical asymptote.	If the numerator is 0 there too, a common factor can cancel and leave a limit that is a real number. Then the graph has a hole, not an asymptote.	Find the limit at each zero of the denominator. Only an infinite one-sided limit makes $x = a$ an asymptote.
Saying a limit exists and equals $\infty$ .	$\infty$ is not a real number. $\lim_{x \to a} f (x) = \infty$ says the limit does not exist because the values of $f$ grow without bound.	Read $= \infty$ as a description: the values increase without bound as $x$ approaches $a$ .
Taking the sign of a one-sided infinite limit from the numerator alone.	The sign of the denominator on that side counts just as much: a positive numerator over a small negative denominator gives $- \infty$ .	On each side, write the sign of the numerator and the sign of the denominator, then combine them.

Part 3 Watch it done Every vertical asymptote of a quotient

Watch each step, then do the last one yourself.

Let $f (x) = \frac{x^{2} + 2 x - 3}{x^{2} - 4 x + 3}$ . Find every vertical asymptote of the graph of $f$ , and justify each answer with a limit.

Step 1 · Factor and list the candidates
$f (x) = \frac{(x + 3) (x - 1)}{(x - 1) (x - 3)}$ . The denominator is 0 at $x = 1$ and $x = 3$ , so those are the only inputs where an asymptote can be.
Step 2 · Test $x = 1$
For $x \neq 1$ , $f (x) = \frac{x + 3}{x - 3}$ , so $\lim_{x \to 1} f (x) = \frac{1 + 3}{1 - 3} = - 2$ . The limit is a real number, so $x = 1$ is not a vertical asymptote. The graph has a hole at $(1, - 2)$ .
Why this step?
Why does a limit of $- 2$ at $x = 1$ rule out an asymptote there?
A vertical asymptote at $x = 1$ needs the values of $f$ to grow without bound as $x$ approaches 1 from at least one side. Values that approach $- 2$ from both sides stay bounded near 1, so neither one-sided limit is infinite.
Step 3 · Test $x = 3$ from the right
As $x \to 3^{+}$ , the numerator $x + 3$ approaches 6 and the denominator $x - 3$ approaches 0 through positive values. So $\lim_{x \to 3^{+}} f (x) = \infty$ , and that one side is enough: the graph of $f$ has a vertical asymptote at $x = 3$ .
Your step · Fill the statement
No calculator. Finish it. Tap a blank, then tap an answer.

For $f (x) = \frac{x + 3}{x - 3}$ near 3, as $x$ approaches 3 from the left the numerator approaches and the denominator approaches 0 through values. So $\lim_{x \to 3^{-}} f (x) =$ .

Part 4 Try it A new quotient and a new context

Try it · Calculate

No calculator. How many vertical asymptotes does the graph of $g (x) = \frac{x^{2} - 9}{x^{3} - x^{2} - 6 x}$ have?

Your answer

Try it · Multiple choice

No calculator. A model for a small lamp gives the brightness $d$ meters from it as $I (d) = \frac{50}{d^{2}}$ lux, for $d > 0$ . Which statement interprets $\lim_{d \to 0^{+}} I (d) = \infty$ in this context?

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

Free-response practice · Question 6 style

No calculator. A tank holds 600 liters of water when full. While it is being filled, water leaks out at a constant rate of 20 liters per hour. A pump that delivers $r$ liters per hour, with $r > 20$ , fills the empty tank in $T (r) = \frac{600}{r - 20}$ hours.

(a) Find $T (50)$ . Using correct units, interpret the meaning of $T (50)$ in the context of the problem.

(b) Find $\lim_{r \to 20^{+}} T (r)$ . Using correct units, explain what this limit says about filling the tank.

(c) The leak rate can change. With a pump that delivers 50 liters per hour and a leak of $L$ liters per hour, the empty tank fills in $F (L) = \frac{600}{50 - L}$ hours, for $0 \leq L < 50$ . Find $\lim_{L \to 50^{-}} F (L)$ . Using correct units, explain what this limit says about filling the tank.

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

Model answer

(a) $T (50) = \frac{600}{50 - 20} = 20$ . A pump that delivers 50 liters per hour fills the empty tank in 20 hours.

(b) As $r \to 20^{+}$ , the numerator stays 600 and the denominator $r - 20$ approaches 0 through positive values, so $\lim_{r \to 20^{+}} T (r) = \infty$ . As the pump's rate decreases toward the leak rate of 20 liters per hour, the time needed to fill the tank increases without bound.

(c) As $L \to 50^{-}$ , the numerator stays 600 and the denominator $50 - L$ approaches 0 through positive values, because $L < 50$ . So $\lim_{L \to 50^{-}} F (L) = \infty$ . As the leak rate increases toward the pump's 50 liters per hour, the time needed to fill the tank increases without bound.

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

Gets $T (50) = 20$ and says a pump delivering 50 liters per hour fills the tank in 20 hours.
Gets $\lim_{r \to 20^{+}} T (r) = \infty$ , with the reason: 600 over a denominator approaching 0 through positive values.
Says that as the pump rate decreases toward 20 liters per hour, the fill time increases without bound, with units.
Gets $\lim_{L \to 50^{-}} F (L) = \infty$ , with the reason: 600 over a denominator $50 - L$ approaching 0 through positive values.
Says that as the leak rate increases toward 50 liters per hour, the fill time increases without bound, with units.

Part 6 Check yourself One quotient, two suspects

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

Let $r (x) = \frac{2 x^{2} - 2 x - 4}{x^{2} - 3 x + 2}$ . Its denominator is 0 at $x = 1$ and at $x = 2$ , so $r$ is not defined at either input. The table gives some values of $r$ near them.

Values of r(x) near x = 1 and x = 2
$x$	$r (x)$
0.9	$- 38$
0.99	$- 398$
1.01	402
1.1	42
1.9	6.444
1.99	6.040
2.01	5.960
2.1	5.636

Case question 1 · Match

No calculator. Use the formula for $r$ , with the table as a guide. Tap a limit, then what it is.

Case question 2 · Multiple choice

No calculator. Where does the graph of $r$ have a vertical asymptote?

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

Check yourself · Spot the error

No calculator. A student looks for the vertical asymptotes of $h (x) = \frac{x^{2} - 1}{x^{2} + x - 2}$ . Tap the first line with a mistake.

Check yourself · Calculate

No calculator. Let $f (x) = \frac{x + k}{x^{2} - 9}$ , where $k$ is a constant. For what value of $k$ does the graph of $f$ have a vertical asymptote at $x = - 3$ but not at $x = 3$ ?

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 · Calculate

No calculator. Let $f (x) = k x + 1$ for $x < 2$ and $f (x) = x^{2} - k$ for $x \geq 2$ , where $k$ is a constant. For what value of $k$ is $f$ continuous at $x = 2$ ?

Your answer

Bring it back · Match

No calculator. Tap a limit, then what is true of it.

In three lines

An infinite limit, written with $\infty$ or $- \infty$ , says the values of a function grow without bound, and that the limit does not exist.
Near a zero of a denominator, the signs of the numerator and the denominator on each side give the sign of each one-sided infinite limit.
The line $x = a$ is a vertical asymptote when at least one one-sided limit at $a$ is infinite; a real-number limit there means a hole.

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.

To justify a vertical asymptote at $x = a$ , write one one-sided limit at $a$ that is $\infty$ or $- \infty$ , with its reason: the numerator approaches a number that is not 0 while the denominator approaches 0 from a known side. To rule an asymptote out, show that the limit at $a$ is a real number.

Questions students ask

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

a limit that exists?

No. A limit exists only when it is a real number, and $\infty$ is not one. Writing $= \infty$ is still useful: it says the limit fails to exist because the values grow without bound, rather than because they oscillate or the two sides disagree.

What if the numerator and the denominator are both 0 at

a

Then $\frac{0}{0}$ tells you nothing yet. Factor and cancel, and find the limit again. A real number means a hole; a number that is not 0 over a denominator approaching 0 means an asymptote. $\frac{x - 2}{(x - 2)^{2}}$ gives $\frac{0}{0}$ at 2 and still has an asymptote there.

Can I find vertical asymptotes from a calculator graph?

A graph can suggest one, but a window can hide behavior near an input, and a steep stretch can look like an asymptote that is not there. Confirm each one with a limit from the formula.

Can a graph touch its vertical asymptote?

At most at one point, because a function has at most one output at $a$ ; a function defined in pieces can give $f (a)$ a value. What the graph cannot do is stay bounded near $a$ on the side where the limit is infinite.

Do only rational functions have vertical asymptotes?

No. $\lim_{x \to 0^{+}} \ln x = - \infty$ , so $x = 0$ is a vertical asymptote of $y = \ln x$ , and $\tan x$ has one at every odd multiple of $\frac{π}{2}$ . The same one-sided limit test applies.

Course alignment, for teachers

AP Calculus AB topic 1.14, Unit 1: Limits and Continuity.

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.