Is an infinite limit the same as saying the limit does not exist?

Yes, formally. Since ∞ is not a real number, a limit equal to +∞ or -∞ technically does not exist. We still write =±∞ because it communicates the specific way the function diverges, which is more informative than just saying 'DNE'.

How do I know whether a denominator zero gives an asymptote or a hole?

Factor both numerator and denominator and cancel common factors. If a factor (x-a) remains in the denominator after canceling, you get a vertical asymptote at x=a. If the factor cancels completely, you get a removable hole and the limit is finite.

How do I decide between +∞ and −∞ for a one-sided limit?

Determine the sign of the numerator near the point, then determine the sign of the denominator as you approach from the chosen side. Positive over tiny positive is +∞; positive over tiny negative is -∞, and so on. Watch even versus odd powers of the vanishing factor.

Can non-rational functions have vertical asymptotes?

Absolutely. For example x has a vertical asymptote at x=0 because _x→ 0^+ x=-∞, and x has vertical asymptotes wherever x=0. Any function where values grow unbounded at a finite input qualifies.

AP-CALCBC-1.14

U1.14 Infinite Limits and Vertical Asymptotes

Master AP Calculus BC topic 1.14: identify infinite limits at finite x-values, connect them to vertical asymptotes, and nail the sign of one-sided limits.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on U1.14 Infinite Limits and Vertical Asymptotes, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

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What this lesson covers

When a function's values shoot toward

+\infty

-\infty

x

approaches a fixed number, an ordinary limit fails to exist in the usual sense — but calculus gives us a precise way to describe that explosion. This is the language of infinite limits, and it is the algebraic fingerprint of a vertical asymptote.

In this lesson you will learn to detect where a function blows up, decide whether it heads to

+\infty

-\infty

from each side, and translate those results into vertical asymptotes. This skill shows up constantly on the AP exam, both in multiple-choice sign-chasing questions and in justifying asymptotic behavior on free-response problems.

What an Infinite Limit Actually Means

An infinite limit describes unbounded behavior at a finite input. We write

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

to mean that as

x

gets arbitrarily close to

a

, the values

f(x)

grow without any upper bound. Similarly

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

means the values decrease without bound.

A crucial point that the exam loves to test: an infinite limit is a limit that does not exist. The symbol

+\infty

is not a number; it is a description of how the limit fails. If a question asks whether

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

exists and the honest answer is

+\infty

, then formally the limit does not exist, though we specify the direction of divergence to convey more information.

This unbounded growth typically happens when a denominator approaches

0

while the numerator approaches a nonzero number. As the denominator shrinks toward zero, the quotient's magnitude explodes. Contrast this with a limit like

\lim_{x\to 2}\frac{x-2}{x-2}

, where both parts vanish and you get a removable situation instead — no asymptote there. Recognizing the difference between "nonzero over zero" (infinite limit) and "zero over zero" (indeterminate, must simplify) is the single most important diagnostic skill in this topic.

Connecting Infinite Limits to Vertical Asymptotes

The graph

x = a

is a vertical asymptote of

f

if at least one of the one-sided limits at

a

is infinite. Formally,

x = a

is a vertical asymptote when any of the following holds:

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

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

.

Notice you only need one side to blow up. A function can have a vertical asymptote where it rises to

+\infty

on one side and falls to

-\infty

on the other (like

\frac{1}{x}

x=0

), or where both sides go the same direction (like

\frac{1}{x^2}

x=0

Behavior at $x=a$	Left limit	Right limit	Asymptote?
$f(x)=\frac{1}{x}$ at $0$	$-\infty$	$+\infty$	Yes
$f(x)=\frac{1}{x^2}$ at $0$	$+\infty$	$+\infty$	Yes
$f(x)=\frac{x^2-4}{x-2}$ at $2$	$4$	$4$	No (removable hole)

The last row is the classic trap: the factor cancels, so there is no asymptote — just a hole. Always simplify a rational function before declaring an asymptote.

Determining the Sign of a One-Sided Infinite Limit

Deciding between

+\infty

and

-\infty

comes down to a sign analysis near

x=a

. Use this reliable procedure. First, factor and reduce the function so only genuine asymptote factors remain in the denominator. Second, evaluate the sign of the numerator near

a

(plug in a value close to

a

). Third, determine the sign of the denominator as

x

approaches

a

from the chosen side — this is where the side matters.

For example, consider

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

. Near

x=3

the numerator

x+1 \approx 4 > 0

. From the right,

x-3

is a tiny positive number. Positive over tiny positive gives

+\infty

. From the left,

x-3

is a tiny negative number, so positive over tiny negative gives

-\infty

.

A helpful shortcut: track whether the denominator's factor

(x-a)

appears to an even or odd power. An even power (like

(x-3)^2

) keeps the denominator positive on both sides, so both one-sided limits share the numerator's sign. An odd power flips sign across

a

, producing opposite infinities. On the AP exam, the sign is frequently the entire point of the question, so never skip the side-by-side test.

Beyond Rational Functions and Common Mistakes

Vertical asymptotes are not exclusive to rational functions. The natural logarithm has one:

\lim_{x\to 0^+}\ln x = -\infty

, so

x=0

is a vertical asymptote of

y=\ln x

. The tangent function has infinitely many, at

x = \frac{\pi}{2} + k\pi

, where the cosine denominator hits zero. Any function of the form (bounded nonzero) over (something

\to 0

) is a candidate.

The most common errors are worth memorizing. Do not assume every zero of the denominator gives an asymptote — cancel common factors first, since a shared factor produces a removable discontinuity (hole) instead. Do not write that the limit "equals" a number when it diverges; report

\pm\infty

or state it does not exist. Do not confuse a vertical asymptote (infinite limit at a finite

x

) with a horizontal asymptote (limit as

x\to\pm\infty

), which is the next topic. Finally, always check both one-sided limits; a function may approach

+\infty

from one side and

-\infty

from the other, and stating only one gives an incomplete answer on free-response justifications.

Key terms

Infinite limit.: A statement that $f(x)$ grows or decreases without bound as $x$ approaches a finite value $a$ , written $\lim_{x\to a} f(x)=\pm\infty$ .
Vertical asymptote.: A vertical line $x=a$ that the graph approaches without touching, occurring when at least one one-sided limit at $a$ is $+\infty$ or $-\infty$ .
One-sided limit.: The value $f(x)$ approaches as $x$ nears $a$ strictly from the left ( $a^-$ ) or strictly from the right ( $a^+$ ).
Removable discontinuity.: A hole in the graph caused by a factor that cancels from numerator and denominator; it does not create a vertical asymptote.
Nonzero over zero.: The signature of an infinite limit: a numerator approaching a nonzero value divided by a denominator approaching zero.
Indeterminate form.: An expression like $\frac{0}{0}$ that requires algebraic simplification before the limit can be determined; not automatically an asymptote.

Worked example

Let

f(x)=\dfrac{x-1}{x^2-x-6}

. Find all vertical asymptotes of

f

and, for each, evaluate the one-sided limits to determine whether

f

approaches

+\infty

-\infty

Start by factoring the denominator:

x^2-x-6=(x-3)(x+2)

. So

f(x)=\dfrac{x-1}{(x-3)(x+2)}

.

Check for cancellation. The numerator

x-1

shares no factor with the denominator, so nothing cancels — both zeros of the denominator give genuine vertical asymptotes at

x=3

and

x=-2

.

Analyze

x=3

. Near

x=3

the numerator is

3-1=2>0

, and the factor

x+2\approx 5>0

. The deciding factor is

x-3

. From the right (

x\to 3^+

x-3

is a tiny positive, so

f\to \frac{2}{(+)(\text{small}+)}=+\infty

. From the left (

x\to 3^-

x-3

is a tiny negative, so

f\to -\infty

. Thus

\lim_{x\to 3^+}f(x)=+\infty

and

\lim_{x\to 3^-}f(x)=-\infty

.

Analyze

x=-2

. Near

x=-2

the numerator is

-2-1=-3<0

, and the factor

x-3\approx -5<0

. The deciding factor is

x+2

. From the right (

x\to -2^+

x+2

is a tiny positive:

f\to \frac{(-)}{(-)(\text{small}+)}=\frac{(-)}{(\text{small}-)}=+\infty

. From the left (

x\to -2^-

x+2

is a tiny negative:

f\to \frac{(-)}{(-)(\text{small}-)}=\frac{(-)}{(\text{small}+)}=-\infty

.

Conclusion: vertical asymptotes at

x=3

and

x=-2

, with the sign behavior determined above.

Practice questions

For

g(x)=\dfrac{x+4}{(x-1)^2}

, what is

\lim_{x\to 1^-} g(x)

$+\infty$
$-\infty$
$0$
$5$

Answer: $+\infty$

Near

x=1

the numerator

x+4\approx 5>0

. The denominator

(x-1)^2

is a square, so it is positive on both sides of

1

and approaches

0

. A positive number over a tiny positive number gives

+\infty

. Because the power is even, both one-sided limits equal

+\infty

, so the left-hand limit is

+\infty

Consider

h(x)=\dfrac{x^2-9}{x-3}

. Does

h

have a vertical asymptote at

x=3

? Justify your answer using limits.

Answer: No; there is a removable discontinuity (hole) at $x=3$ , not a vertical asymptote.

Factor the numerator:

x^2-9=(x-3)(x+3)

. Then

h(x)=\frac{(x-3)(x+3)}{x-3}=x+3

for

x\neq 3

. The common factor cancels, so the limit is finite:

\lim_{x\to 3}h(x)=3+3=6

. Since the limit exists and is finite, no infinite limit occurs, so

x=3

is a hole, not a vertical asymptote.

The function

f(x)=\dfrac{2x}{x-5}

has

\lim_{x\to 5^-}f(x)=

which of the following?

$+\infty$
$-\infty$
$2$
Does not exist and is not infinite

Answer: $-\infty$

Near

x=5

the numerator

2x\approx 10>0

. Approaching from the left,

x-5

is a small negative number. A positive divided by a small negative yields a large negative value, so the limit is

-\infty

FAQ

Is an infinite limit the same as saying the limit does not exist?: Yes, formally. Since $\infty$ is not a real number, a limit equal to $+\infty$ or $-\infty$ technically does not exist. We still write $=\pm\infty$ because it communicates the specific way the function diverges, which is more informative than just saying 'DNE'.
How do I know whether a denominator zero gives an asymptote or a hole?: Factor both numerator and denominator and cancel common factors. If a factor $(x-a)$ remains in the denominator after canceling, you get a vertical asymptote at $x=a$ . If the factor cancels completely, you get a removable hole and the limit is finite.
How do I decide between +∞ and −∞ for a one-sided limit?: Determine the sign of the numerator near the point, then determine the sign of the denominator as you approach from the chosen side. Positive over tiny positive is $+\infty$ ; positive over tiny negative is $-\infty$ , and so on. Watch even versus odd powers of the vanishing factor.
Can non-rational functions have vertical asymptotes?: Absolutely. For example $\ln x$ has a vertical asymptote at $x=0$ because $\lim_{x\to 0^+}\ln x=-\infty$ , and $\tan x$ has vertical asymptotes wherever $\cos x=0$ . Any function where values grow unbounded at a finite input qualifies.

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