# Horizontal Asymptotes: AP Calculus Crash Course Review

AP Calculus is not easy, and it is hard to score a five on the test. Only about 25% people who take the test can earn that top score on the AP Calculus exam. Because there are so many concepts and theorems to remember, this AP Calculus review will cover horizontal asymptotes and graphs of a function when $x$ approaches infinity. If you want to ace the exam, you should continue reading this blog post about horizontal asymptotes for AP Calculus practice—horizontal asymptotes calculus is likely to appear on the AP Calculus exam.

To understand horizontal asymptotes, we first need to comprehend what asymptotes are in general. In short, asymptotes are based on graphs of the different limits of a function. There are three types of asymptotes: horizontal, vertical, and oblique/slanted. Take a look at the graphs below for examples of each. This AP Calculus review blog post will focus on finding horizontal asymptotes.

## What are Horizontal Asymptotes Calculus?

Now that we know what asymptotes are, what exactly are horizontal asymptotes? The mathematical definition of horizontal asymptotes is as follows:

If a function $f\left( x \right)$ becomes arbitrarily close to a finite number $L$ for all sufficiently large and positive $x$, then we can write

$\lim _{ x\rightarrow +\infty }{ f\left( x \right) } =L$

In other words, the limit of $f\left( x \right)$ as $x$ approaches infinity is $L$. In this case, the line $y=L$ is a horizontal asymptote of $f$The limit at negative infinity,

$\lim _{ x\rightarrow -\infty }{ f\left( x \right) } =M$

is defined similarly. When this limit exists $y=M$ is also a horizontal asymptote.

To illustrate this abstract definition, let’s look at a basic problem to explain the concept further.

Explain the meaning of

$\lim _{ x\rightarrow \infty }{ f\left( x \right) } =9$

This question straight forwardly draws from the above definition. This expression means that the value of the limit of the function $f\left( x \right)$ when $x$ approaches positive infinity is 9.

Here is another problem for the other case:

Explain the meaning of

$\lim _{ x\rightarrow -\infty }{ f\left( x \right) } =-1$

The answer here is a little different. This equation means that the value of the limit for the function $f\left( x \right)$ when $x$ approaches negative infinity is -1.

These two examples represent the definition of horizontal asymptotes. This definition is the key for calculating horizontal asymptotes.

## Why do Horizontal Asymptotes Matter?

Limits will appear on both AP Calculus AB and BC exams. As for horizontal asymptotes, the AP exam can ask you about definitions or definition-related problems, like the examples above. However, the most popular problems on the exam about limits at infinity include finding the limits and whether they are vertical, slanted, or horizontal. Horizontal asymptotes are finite limits that are rampant on the AP exams.

For rational functions, it is simple to evaluate the limits and find horizontal. We follow The Limit Laws to find the limits:

### Limit Laws:

Assume

$\lim _{ x\rightarrow \infty }{ f\left( x \right) }$ and $\lim _{ x\rightarrow \infty }{ g\left( x \right) }$ exist.

The following properties hold, where $a$ is a real number, and $m>0$ and $n>0$:

1. Sum: $\lim _{ x\rightarrow a }{ \left( f\left( x \right) +g\left( x \right) \right) } =\lim _{ x\rightarrow a }{ f\left( x \right) } +\lim _{ x\rightarrow a }{ g\left( x \right) }$
2. Difference: $\lim _{ x\rightarrow a }{ \left( f\left( x \right) -g\left( x \right) \right) } =\lim _{ x\rightarrow a }{ f\left( x \right) } -\lim _{ x\rightarrow a }{ g\left( x \right) }$
3. Constant multiple: $\lim _{ x\rightarrow a }{ \left( cf\left( x \right) \right) } =c\lim _{ x\rightarrow a }{ f\left( x \right) }$
4. Product: $\lim _{ x\rightarrow a }{ \left( f\left( x \right) g\left( x \right) \right) } =\lim _{ x\rightarrow a }{ f\left( x \right) \times } \lim _{ x\rightarrow a }{ g\left( x \right) }$
5. Quotient: $\lim _{ x\rightarrow a }{ \left( f\left( x \right) /g\left( x \right) \right) } =\lim _{ x\rightarrow a }{ f\left( x \right) } /\lim _{ x\rightarrow a }{ g\left( x \right) }$ provided that $\lim _{ x\rightarrow a }{ g\left( x \right) \neq } 0$
6. Power: $\lim _{ x\rightarrow a }{ { \left( f\left( x \right) \right) }^{ n } } ={ \left( \lim _{ x\rightarrow a }{ f\left( x \right) } \right) }^{ n }$
7. Fractional power: $\lim _{ x\rightarrow a }{ { \left( f\left( x \right) \right) }^{ \frac { n }{ m } } } ={ \left( \lim _{ x\rightarrow a }{ f\left( x \right) } \right) }^{ \frac { n }{ m } }$

Before tackling an example from an AP Calculus practice test, let’s first warm up with some easier, similar problems.

Here’s one:

1. Evaluate the limit: $\lim _{ n\rightarrow -\infty }{ \left( 2+\frac { 10 }{ { x }^{ 2 } } \right) }$

Solution:

We use the sum limit law to solve this problem:

$\lim _{ n\rightarrow a }{ \left( f\left( x \right) +g\left( x \right) \right) } =\lim _{ n\rightarrow a }{ f\left( x \right) } +\lim _{ n\rightarrow a }{ g\left( x \right) }$

Although $x$ goes to negative infinity, ${ x }^{ 2 }$ goes to positive infinity. In turn,

$\frac { 10 }{ { x }^{ 2 } }$

goes to 0. By the limit laws, we have

$\lim _{ n\rightarrow -\infty }{ \left( 2+\frac { 10 }{ { x }^{ 2 } } \right) } =\lim _{ n\rightarrow -\infty }{ 2 } +\lim _{ n\rightarrow -\infty }{ \frac { 10 }{ { x }^{ 2 } } } =2+0=2$

Thus, the graph of $y=2+\frac { 10 }{ { x }^{ 2 } }$ approaches the horizontal asymptote $y=2$ as $x\rightarrow -\infty$.

To solve these horizontal asymptotes problems, you should have an in-depth understanding of related theorems about limits, especially for those approaching infinity. For rational functions, the limits act in certain, predictable ways. However, this is not the only way to find the horizontal asymptotes for rational functions.

The following trick is crucial to help you find the horizontal asymptotes of any rational function.

A rational function can be written as

$f\left( x \right) =\frac { p\left( x \right) }{ q\left( x \right) }$

where $p\left( x \right)$ and $q\left( x \right)$ are both polynomials. Even without using limits for rational functions, you can find its horizontal asymptotes (HA) based on the degree of the function—that is, the highest power of the variable that appears in your polynomial. For instance, in the function ${ x }^{ 2 }+2x+3,{ x }^{ 2 }$ is the highest power of $x$, so the degree is 2. There are three possible cases:

1. If the degree of $p\left( x \right)$ < degree of $q\left( x \right)$, then the HA is $y=0$ (the x-axis).
2. If the degree of $p\left( x \right)$degree of $q\left( x \right)$, then the HA is $y=\frac { leading\enspace term\enspace on\enspace top }{ leading\enspace term\enspace on\enspace bottom }$. Remember to SIMPLIFY the equation. This should result in the expression $y=a$, where $a$ is a constant.
3. If the degree of $p\left( x \right)$degree of $q\left( x \right)$, then the function has no HA.

This theorem implies that a rational function can have at most one horizontal asymptote, and whenever there is a horizontal asymptote,

$\lim _{ n\rightarrow \infty }{ \frac { p\left( x \right) }{ q\left( x \right) } } =\lim _{ n\rightarrow -\infty }{ \frac { p\left( x \right) }{ q\left( x \right) } }$.

That is, horizontal asymptotes for rational functions are singular.

There are other rules for finding horizontal asymptotes. Whenever more than one rule comes into play, the problem becomes a little more complicated. For instance, rational functions are different from trigonometric functions. For such functions, we find the horizontal asymptotes differently. We use the same concept, but tweak the definitions and rules to suit the function.

Here is an AP Calculus practice problem for finding the horizontal asymptotes of trigonometric functions:

Evaluate the limits:

$\lim _{ n\rightarrow -\infty }{ 5+\frac { \sin { x } }{ \sqrt { x } } }$

We use the Squeeze Theorem to solve this problem:

Suppose the functions $f$, $g$, and $h$ satisfy

$f\left( x \right)\le g\left( x \right)\le h\left( x \right)$

for all values of $x$ near $a$, except at $a$. If

$\lim _{ n\rightarrow a }{ f\left( x \right) } =\lim _{ n\rightarrow a }{ g\left( x \right)=L }$, then $\lim _{ n\rightarrow a }{ g\left( x \right)=L }$

We will also use the Sum Limit Law:

$\lim _{ n\rightarrow a }{ \left( f\left( x \right) +g\left( x \right) \right) } =\lim _{ n\rightarrow a }{ f\left( x \right) } +\lim _{ n\rightarrow a }{ g\left( x \right) }$

The numerator of $\frac { \sin { x } }{ \sqrt { x } }$ is bounded within -1 and 1/ For every $x>0$,

$\frac { -1 }{ \sqrt { x } } \le \frac { \sin { x } }{ \sqrt { x } } \le \frac { 1 }{ \sqrt { x } }$

As $x\rightarrow \infty$, $\sqrt { x } \rightarrow \infty$. Consequently,

$\lim _{ n\rightarrow -\infty }{ \frac { -1 }{ \sqrt { x } } } =\lim _{ n\rightarrow -\infty }{ \frac { 1 }{ \sqrt { x } } } =0$

By using the Squeeze Theorem,

$\lim _{ n\rightarrow -\infty }{ \frac { \sin { x } }{ \sqrt { x } } } =0$.

Also,

$\lim _{ n\rightarrow -\infty }{ 5 } =5$.

Applying the limit laws stated above, we can now evaluate the limits and find the horizontal asymptotes:

$\lim _{ n\rightarrow -\infty }{ 5+\frac { \sin { x } }{ \sqrt { x } } } =\lim _{ n\rightarrow -\infty }{ 5 } +\lim _{ n\rightarrow -\infty }{ \frac { \sin { x } }{ \sqrt { x } } } =5+0=0$

Hence, the graph $y=5+\frac { \sin { x } }{ \sqrt { x } }$ approaches the horizontal asymptote $y=5$ as $x$ goes to infinty.

On the AP exams, the questions will usually include non-rational functions. The test often asks the same questions,but for some more complicated cases, which require you to modify the original functions or to apply different rules, such as the Squeeze Theorem or other cases of the Limit Laws. They might ask you to apply the quotient rule, for instance. Here is an example from the CollegeBoard’s AP Calculus sample questions:

2. Evaluate the limit: $\lim _{ x\rightarrow \infty }{ \frac { \sqrt { 9{ x }^{ 2 }+1 } }{ { x }^{ 2 }-3x+5 } }$

(A) 1

(B) 3

(C) 9

(D) Does not exist

Before tackling the question, step back and think about how to solve this problem. It is not a normal rational The numerator is a polynomial, but it is under a square root.

We cannot directly find the limits of the function. Notice that you can only evaluate the limits of the denominator and the numerator separately. Therefore, you apply the Quotient Limit Law to solve this:

$\lim _{ x\rightarrow a }{ \left( f\left( x \right) /g\left( x \right) \right) } =\lim _{ x\rightarrow a }{ f\left( x \right) /\lim _{ x\rightarrow a }{ g\left( x \right) } }$

$\lim _{ x\rightarrow a }{ g\left( x \right) } \neq 0$

Apparently, this does work, as $\lim _{ x\rightarrow a }{ { x }^{ 2 }-3x+5 }$ does not exist. It goes to infinity as $x$ approaches infinity.

Oops—but don’t worry!Here is how we can solve the problem:

$\lim _{ x\rightarrow \infty }{ \frac { \sqrt { 9{ x }^{ 4 }+1 } }{ { x }^{ 2 }-3x+5 } } =\lim _{ x\rightarrow \infty }{ \frac { \sqrt { \frac { 9{ x }^{ 4 } }{ { x }^{ 4 } } +\frac { 1 }{ { x }^{ 4 } } } }{ \frac { { x }^{ 2 } }{ { x }^{ 2 } } -\frac { 3x }{ { x }^{ 2 } } +\frac { 5 }{ { x }^{ 2 } } } }$

Divide by $\sqrt { { x }^{ 2 } } ={ x }^{ 2 }$

$=\lim _{ x\rightarrow \infty }{ \frac { \sqrt { 9+\frac { 1 }{ { x }^{ 4 } } } }{ 1-\frac { 3 }{ x } +\frac { 5 }{ { x }^{ 2 } } } }$

Simplify,

as $x\rightarrow \infty$, $\frac { 1 }{ { x }^{ 4 } }$, $\frac { 3 }{ x }$, and $\frac { 5 }{ { x }^{ 2 } } \rightarrow 0$, so the horizontal asymptotes are:

$=\frac { \sqrt { 9 } }{ 1 } =3$

The answer is (B).

If you have mastered these problems, you should be an expert at calculating horizontal asymptotes. If you applied this technique but did not arrive at an answer, try one more important rule for finding limits: the L’Hopital’s Rule. (Check out our post on L’Hopital’s Rule for more details).

If $\lim _{ x\rightarrow a }{ f\left( x \right) } =\lim _{ x\rightarrow a }{ g\left( x \right) } =0$ or $\lim _{ x\rightarrow a }{ f\left( x \right) } =\lim _{ x\rightarrow a }{ g\left( x \right) } =\infty$, then:

$\lim _{ x\rightarrow a }{ \frac { f\left( x \right) }{ g\left( x \right) } } =\lim _{ x\rightarrow a }{ \frac { f^{ \prime }\left( x \right) }{ g^{ \prime }\left( x \right) } }$

The rule also applies if $x\rightarrow a$ is replaced with $x\rightarrow \pm \infty$, $x\rightarrow { a }^{ - }$$x\rightarrow { a }^{ + }$.

L’Hopital’s rule is life-saving in many ways. Whenever there is a limit that you have no idea how to solve after trying again and again, apply L’Hopital’s Rule—this strategy will often lead you in the right direction. Here is an example using L’Hopital’s rule from CollegeBoard’s AP Calculus sample questions:

1. $\lim _{ x\rightarrow \pi }{ \frac { \cos { x } +\sin { 2x+1 } }{ { x }^{ 2 }-{ \pi }^{ 2 } } }$ is:

(A) $\frac { 1 }{ 2\pi }$

(B) $\frac { 1 }{ \pi }$

(C) $1$

(D) None existent

This limit has the form of, so we apply L’Hopital’s rule. Both the denominator and the numerator approach infinity as $x$ goes to infinity.

Here is how to set up the problem using L’Hopital’s rule

$\lim _{ x\rightarrow \pi }{ \frac { \cos { x } +\sin { 2x+1 } }{ { x }^{ 2 }-{ \pi }^{ 2 } } } =\lim _{ x\rightarrow \pi }{ \frac { -\sin { x+2\cos { \left( 2x \right) } } }{ 2x } =\frac { 1 }{ \pi } }$

The answer is B.

This blog post reviewed all that you need to know about horizontal asymptotes calculus for the AP Calculus exam. Follow all the rules and laws, improvise with the techniques, and you will ace any horizontal asymptote-related questions on the exam. Have fun and good luck on the test!

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