How do I quickly find a horizontal asymptote of a rational function?

Compare degrees. If the bottom degree is bigger, the asymptote is y=0. If degrees are equal, it is y= the ratio of leading coefficients. If the top degree is bigger, there is no horizontal asymptote (though there may be a slant one).

Can a function cross its horizontal asymptote?

Yes. A horizontal asymptote only describes end behavior as x→±∞. The graph can cross it any number of times for finite x; the curve just settles toward the line eventually.

Why does sqrt(x^2) become -x when x is negative?

Because sqrt(x^2)=|x| by definition, and |x|=-x whenever x<0. This sign flip is why limits as x→-∞ involving radicals can differ from the x→+∞ case, producing two different horizontal asymptotes.

Do exponential functions have horizontal asymptotes?

Often, but only in one direction. For example, e^x→0 as x→-∞ (asymptote y=0) but →∞ as x→+∞ (no asymptote). Exponentials also dominate any polynomial, so limits like x^n/e^x equal 0.

AP-CALCBC-1.15

U1.15 Limits at Infinity and Horizontal Asymptotes

Q: Do exponential functions have horizontal asymptotes?

Often, but only in one direction. For example, e^x→0 as x→-∞ (asymptote y=0) but →∞ as x→+∞ (no asymptote). Exponentials also dominate any polynomial, so limits like x^n/e^x equal 0.

Master limits at infinity for AP Calculus BC: use the degree comparison rule and dominant-term technique to find horizontal asymptotes fast.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on U1.15 Limits at Infinity and Horizontal Asymptotes, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

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What this lesson covers

When you zoom out on a graph and ask what happens as

x

races toward positive or negative infinity, you are computing a limit at infinity. These limits answer a practical question: does the function settle toward a fixed height, blow up, or oscillate? In this lesson you will learn to evaluate

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

, connect the answer directly to horizontal asymptotes, and use two reliable shortcuts — the degree comparison rule for rational functions and the dominant-term technique for messier expressions. These skills show up constantly on the AP exam, both in standalone multiple-choice items and buried inside FRQs about long-run behavior.

What a Limit at Infinity Means

A limit at infinity describes the end behavior of a function: the value

f(x)

approaches as

x

grows without bound in the positive direction (

x\to+\infty

) or the negative direction (

x\to-\infty

). Unlike the infinite limits of U1.14 (which describe vertical asymptotes as

x

approaches a finite number), here the input is unbounded and we watch the output.

Three outcomes are possible. First, the limit equals a finite number

L

, meaning the graph flattens toward the horizontal line

y=L

. Second, the limit is

+\infty

-\infty

, meaning the function grows or falls without bound. Third, the limit may not exist because the function oscillates, as with

\sin x

, whose values keep cycling between

-1

and

1

.

Notation matters on the exam. Writing

\lim_{x\to\infty} f(x)=2

is a precise claim that outputs get arbitrarily close to

2

for large enough

x

. A common misconception is treating

\infty

as a number you can plug in — you cannot. Instead you reason about which parts of the expression dominate. Keep in mind the two directions can give different answers: a function can approach one asymptote on the right and a different one on the left.

The Degree Comparison Rule for Rational Functions

For a rational function

f(x)=\frac{p(x)}{q(x)}

, the limit at infinity depends only on the leading terms of the numerator and denominator. Let

n

be the degree of

p

and

m

the degree of

q

, with leading coefficients

a

and

b

Case	$\lim_{x\to\pm\infty}f(x)$	Horizontal asymptote
$n<m$ (bottom heavy)	$0$	$y=0$
$n=m$ (equal)	$\frac{a}{b}$	$y=\frac{a}{b}$
$n>m$ (top heavy)	$\pm\infty$	none

For example,

\lim_{x\to\infty}\frac{3x^2+5}{7x^2-x}=\frac{3}{7}

because both are degree 2. And

\lim_{x\to\infty}\frac{4x}{x^3+1}=0

because the denominator wins. When the top degree exceeds the bottom by exactly one, the graph has a slant (oblique) asymptote rather than a horizontal one, found by polynomial long division — but the limit itself is still

\pm\infty

.

The rule is a shortcut, not magic. It follows from dividing every term by the highest power of

x

in the denominator, which is the technique the next section makes explicit.

The Dominant-Term Technique

The dominant-term technique justifies the degree rule and extends to expressions the rule alone cannot handle. The idea: for large

|x|

, the term with the fastest growth swamps the others, so divide numerator and denominator by the highest power of

x

present, then send the leftover terms to

0

.

Consider

\lim_{x\to\infty}\frac{2x^2-x+4}{5x^2+3}

. Divide top and bottom by

x^2

\frac{2-\frac{1}{x}+\frac{4}{x^2}}{5+\frac{3}{x^2}}

. As

x\to\infty

, each

\frac{1}{x}

-type term vanishes, leaving

\frac{2}{5}

.

The technique also handles radicals, but watch signs. For

\lim_{x\to-\infty}\frac{\sqrt{x^2+1}}{x}

, note

\sqrt{x^2}=|x|=-x

when

x<0

. Dividing by

x

introduces a sign flip, giving

-1

, whereas the

x\to+\infty

limit is

+1

. Forgetting

\sqrt{x^2}=|x|

is one of the most tested traps in this topic.

Exponentials dominate polynomials. In

\lim_{x\to\infty}\frac{x^{100}}{e^x}=0

, the exponential grows faster than any power, so the limit is

0

. This growth hierarchy — logarithms slower than powers slower than exponentials — is worth memorizing for BC problems.

Connecting Limits to Horizontal Asymptotes

A horizontal asymptote is a direct consequence of a finite limit at infinity. The line

y=L

is a horizontal asymptote of

f

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

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

. Because the two directions are evaluated separately, a function may have zero, one, or two horizontal asymptotes.

A classic two-asymptote example is

f(x)=\frac{x}{\sqrt{x^2+1}}

, which approaches

1

x\to+\infty

and

-1

x\to-\infty

, giving the horizontal asymptotes

y=1

and

y=-1

. Another familiar case,

\arctan x

, has

y=\frac{\pi}{2}

and

y=-\frac{\pi}{2}

.

A frequent misconception is believing a graph can never cross its horizontal asymptote. It can — repeatedly — near the middle of the domain; the asymptote only governs long-run behavior. On the AP exam, expect to translate between a limit statement, a table of large-input values, and a graph. If a table shows

f(1000)\approx 2.99

and

f(10000)\approx 2.999

, you should recognize

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

and report the asymptote

y=3

. This multi-representational fluency is exactly what the objective targets.

Key terms

Limit at infinity.: The value $f(x)$ approaches as $x$ increases or decreases without bound, written $\lim_{x\to\pm\infty}f(x)$ .
Horizontal asymptote.: A line $y=L$ such that $\lim_{x\to+\infty}f(x)=L$ or $\lim_{x\to-\infty}f(x)=L$ ; describes end behavior.
Degree comparison rule.: For a rational function, the infinity limit is $0$ if the numerator degree is smaller, the ratio of leading coefficients if degrees match, and $\pm\infty$ if the numerator degree is larger.
Dominant term.: The term that grows fastest for large $|x|$ and therefore controls a function's end behavior.
Dominant-term technique.: Dividing numerator and denominator by the highest power of $x$ so that smaller terms vanish, revealing the limit.
Slant (oblique) asymptote.: A non-horizontal line the graph approaches when the numerator degree exceeds the denominator degree by exactly one, found by long division.
Growth hierarchy.: The ordering that logarithms grow slower than powers, which grow slower than exponentials, used to evaluate limits like $\frac{x^n}{e^x}$ .

Worked example

Find all horizontal asymptotes of

f(x)=\dfrac{6x^2-x}{\sqrt{4x^4+1}}

by evaluating

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

and

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

The highest power inside the radical is

x^4

, whose square root behaves like

x^2

, so the dominant power overall is

x^2

. Divide numerator and denominator by

x^2

.

For the denominator, move the

x^2

inside the square root as

x^4

\frac{\sqrt{4x^4+1}}{x^2}=\sqrt{\frac{4x^4+1}{x^4}}=\sqrt{4+\frac{1}{x^4}}

. This works for both directions because

x^2>0

regardless of sign.

The numerator becomes

\frac{6x^2-x}{x^2}=6-\frac{1}{x}

.

So

f(x)=\dfrac{6-\frac{1}{x}}{\sqrt{4+\frac{1}{x^4}}}

. As

x\to+\infty

, the small terms

\frac{1}{x}

and

\frac{1}{x^4}

go to

0

, leaving

\frac{6}{\sqrt{4}}=\frac{6}{2}=3

.

As

x\to-\infty

, the same terms still vanish (odd and even powers of

\frac{1}{x}

both approach

0

), and since we divided by

x^2>0

there is no sign flip. The limit is again

\frac{6}{2}=3

.

Both directions give

3

, so the only horizontal asymptote is

y=3

. Note the sign flip trap did not appear here because the dominant denominator power was even; always check whether you divided by an odd or even power.

Practice questions

What is

\lim_{x\to\infty}\dfrac{5x^3+2x}{2x^3-7x^2+1}

$0$
$\frac{5}{2}$
$\frac{2}{5}$
$\infty$

Answer: $\frac{5}{2}$

Numerator and denominator both have degree

3

, so by the degree comparison rule the limit equals the ratio of leading coefficients,

\frac{5}{2}

. Dividing every term by

x^3

confirms this: the lower-order terms all vanish, leaving

\frac{5+0}{2-0+0}=\frac{5}{2}

Evaluate

\lim_{x\to-\infty}\dfrac{\sqrt{9x^2+4}}{2x-1}

and explain the sign carefully.

Answer: $-\frac{3}{2}$

Divide top and bottom by

x

. For the numerator,

\frac{\sqrt{9x^2+4}}{x}

; since

x<0

\sqrt{x^2}=|x|=-x

, so this equals

-\sqrt{9+\frac{4}{x^2}}

, approaching

-3

. The denominator

\frac{2x-1}{x}=2-\frac{1}{x}

approaches

2

. The limit is

\frac{-3}{2}=-\frac{3}{2}

. The key is recognizing the absolute value forces a negative sign because

x\to-\infty

A function satisfies

f(100)=4.2

f(1000)=4.02

f(10000)=4.002

. What horizontal asymptote does this evidence suggest, and does it prove the function never equals

5

Answer: $y=4$

The outputs approach

4

x

grows, so the data strongly suggests

\lim_{x\to\infty}f(x)=4

and the horizontal asymptote

y=4

. This says nothing about whether

f

equals

5

at some finite input — a function can cross or exceed values far from its asymptote in the middle of its domain. The asymptote only constrains long-run behavior, not every point.

FAQ

How do I quickly find a horizontal asymptote of a rational function?: Compare degrees. If the bottom degree is bigger, the asymptote is $y=0$ . If degrees are equal, it is $y=$ the ratio of leading coefficients. If the top degree is bigger, there is no horizontal asymptote (though there may be a slant one).
Can a function cross its horizontal asymptote?: Yes. A horizontal asymptote only describes end behavior as $x\to\pm\infty$ . The graph can cross it any number of times for finite $x$ ; the curve just settles toward the line eventually.
Why does $\sqrt{x^2}$ become $-x$ when $x$ is negative?: Because $\sqrt{x^2}=|x|$ by definition, and $|x|=-x$ whenever $x<0$ . This sign flip is why limits as $x\to-\infty$ involving radicals can differ from the $x\to+\infty$ case, producing two different horizontal asymptotes.
Do exponential functions have horizontal asymptotes?: Often, but only in one direction. For example, $e^x\to0$ as $x\to-\infty$ (asymptote $y=0$ ) but $\to\infty$ as $x\to+\infty$ (no asymptote). Exponentials also dominate any polynomial, so limits like $\frac{x^n}{e^x}$ equal $0$ .

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