# 1.4: Limits at Infinity and Horizontal Asymptotes : End Behavior

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Think Out Loud

What do mean by the end behavior?

Learning Objectives

- Calculate the limit of a function as \(x\) increases or decreases without bound.
- Recognize a horizontal asymptote on the graph of a function.
- Estimate the end behaviour of a function as \(x\) increases or decreases without bound.
- Recognize an oblique asymptote on the graph of a function.

We have learned about \(\displaystyle \lim\limits_{x \to a}f(x) = L\), where \(\displaystyle a\) is a real number. In this section we would like to explore \(\displaystyle a\) to be \(\displaystyle\infty\) or \(\displaystyle -\infty\).

## Limits at Infinity and Horizontal Asymptotes

Recall that \(\displaystyle \lim_{x→a}f(x)=L\) means \(f(x)\) becomes arbitrarily close to \(L\) as long as \(x\) is sufficiently close to \(a\). We can extend this idea to limits at infinity. For example, consider the function \(f(x)=2+\frac{1}{x}\). As can be seen graphically in Figure \(\PageIndex{1}\) and numerically in Table \(\PageIndex{1}\), as the values of \(x\) get larger, the values of \(f(x)\) approach \(2\). We say the limit as \(x\) approaches \(∞\) of \(f(x)\) is \(2\) and write \(\displaystyle \lim_{x→∞}f(x)=2\). Similarly, for \(x<0\), as the values \(|x|\) get larger, the values of \(f(x)\) approaches \(2\). We say the limit as \(x\) approaches \(−∞\) of \(f(x)\) is \(2\) and write \(\displaystyle \lim_{x→a}f(x)=2\).

\(x\) | 10 | 100 | 1,000 | 10,000 |
---|---|---|---|---|

\(2+\frac{1}{x}\) | 2.1 | 2.01 | 2.001 | 2.0001 |

\(x\) | −10 | −100 | −1000 | −10,000 |

\(2+\frac{1}{x}\) | 1.9 | 1.99 | 1.999 | 1.9999 |

More generally, for any function \(f\), we say the limit as \(x→∞\) of \(f(x)\) is \(L\) if \(f(x)\) becomes arbitrarily close to \(L\) as long as \(x\) is sufficiently large. In that case, we write \(\displaystyle \lim_{x→a}f(x)=L\). Similarly, we say the limit as \(x→−∞\) of \(f(x)\) is \(L\) if \(f(x)\) becomes arbitrarily close to \(L\) as long as \(x<0\) and \(|x|\) is sufficiently large. In that case, we write \(\displaystyle \lim_{x→−∞}f(x)=L\). We now look at the definition of a function having a limit at infinity.

Definition: Limit at Infinity (Informal)

If the values of \(f(x)\) become arbitrarily close to \(L\) as \(x\) becomes sufficiently large, we say the function \(f\) has a **limit at infinity** and write

\[\lim_{x→∞}f(x)=L.\]

If the values of \(f(x)\) becomes arbitrarily close to \(L\) for \(x<0\) as \(|x|\) becomes sufficiently large, we say that the function \(f\) has a limit at negative infinity and write

\[\lim_{x→-∞}f(x)=L.\]

If the values \(f(x)\) are getting arbitrarily close to some finite value \(L\) as \(x→∞\) or \(x→−∞\), the graph of \(f\) approaches the line \(y=L\). In that case, the line \(y=L\) is a horizontal asymptote of \(f\) (Figure \(\PageIndex{2}\)). For example, for the function \(f(x)=\dfrac{1}{x}\), since \(\displaystyle \lim_{x→∞}f(x)=0\), the line \(y=0\) is a horizontal asymptote of \(f(x)=\dfrac{1}{x}\).

Definition: Horizontal Asymptote

If \(\displaystyle \lim_{x→∞}f(x)=L\) or \(\displaystyle \lim_{x→−∞}f(x)=L\), we say the line \(y=L\) is a *horizontal asymptote* of \(f\).

A function cannot cross a vertical asymptote because the graph must approach infinity (or \( −∞\)) from at least one direction as \(x\) approaches the vertical asymptote. However, a function may cross a horizontal asymptote. In fact, a function may cross a horizontal asymptote an unlimited number of times. For example, the function \(f(x)=\dfrac{(\cos x)}{x}+1\) shown in Figure \(\PageIndex{3}\) intersects the horizontal asymptote \(y=1\) an infinite number of times as it oscillates around the asymptote with ever-decreasing amplitude.

Example \(\PageIndex{1}\)

Consider \(f(x) = 1/x^2\), as shown in Figure \(\PageIndex{4}\). Note how, as \(x\) approaches 0, \(f(x)\) grows very, very large. It seems appropriate, and descriptive, to state that \[\lim\limits_{x\rightarrow 0} \frac1{x^2}=\infty.\]Also note that as \(x\) gets very large, \(f(x)\) gets very, very small. We could represent this concept with notation such as \[\lim\limits_{x\rightarrow \infty} \frac1{x^2}=0.\]

Figure \(\PageIndex{4}\):: Graphing \(f(x)=1/x^2\) for values of \(x \text{ near }0\).

## Limit Laws

Let \(f(x)=k\). Then \(\displaystyle \lim_{x→∞}f(x)=k\), and \(\displaystyle \lim_{x→-∞}f(x)=k\) as shown in the following graph.

The algebraic limit laws that we introduced in Introduction to Limits also apply to limits at infinity. We illustrate how to use these laws to compute several limits at infinity.

Example \(\PageIndex{2}\): Computing Limits at Infinity

For each of the following functions \(f\), valuate \(\displaystyle \lim_{x→∞}f(x)\) and \(\displaystyle \lim_{x→−∞}f(x)\). Determine the horizontal asymptote(s) for \(f\), where \(f(x)=5−\dfrac{2}{x^2}\)

**Solution**

Using the algebraic limit laws, we have

\[\lim_{x→∞}\left(5−\frac{2}{x^2}\right)=\lim_{x→∞}5−2\left(\lim_{x→∞}\frac{1}{x}\right)\cdot\left(\lim_{x→∞}\frac{1}{x}\right)=5−2⋅0=5.\nonumber\]

Similarly, \(\displaystyle \lim_{x→∞}f(x)=5\). Therefore, \(f(x)=\dfrac{5−2}{x^2}\) has a horizontal asymptote of \(y=5\) and \(f\) approaches this horizontal asymptote as \(x→±∞\) as shown in the following graph.

## Infinite Limits at Infinity

Sometimes the values of a function \(f\) become arbitrarily large as \(x→∞ \)(or as \(x→−∞\)). In this case, we write \(\displaystyle \lim_{x→∞}f(x)=∞\) (or \(\displaystyle \lim_{x→−∞}f(x)=∞\)). On the other hand, if the values of \(f\) are negative but become arbitrarily large in magnitude as \(x→∞\) (or as \(x→−∞\)), we write \(\displaystyle \lim_{x→∞}f(x)=−∞\) (or \(\displaystyle \lim_{x→−∞}f(x)=−∞\)).

For example, consider the function \(f(x)=x^3\). As seen in Table \(\PageIndex{2}\) and Figure \(\PageIndex{8}\), as \(x→∞\) the values \(f(x)\) become arbitrarily large. Therefore, \(\displaystyle \lim_{x→∞}x^3=∞\). On the other hand, as \(x→−∞\), the values of \(f(x)=x^3\) are negative but become arbitrarily large in magnitude. Consequently, \(\displaystyle \lim_{x→−∞}x^3=−∞.\)

Table \(\PageIndex{2}\)

\(x\) |
10 | 20 | 50 | 100 | 1000 |
---|---|---|---|---|---|

\(x^3\) |
1000 | 8000 | 125,000 | 1,000,000 | 1,000,000,000 |

\(x\) |
−10 | −20 | −50 | −100 | −1000 |

\(x^3\) |
−1000 | −8000 | −125,000 | −1,000,000 | −1,000,000,000 |

Values of a power function as \(x→±∞\)

Definition: Infinite Limit at Infinity (Informal)

We say a function \(f\) has an infinite limit at infinity and write

\[\lim_{x→∞}f(x)=∞.\]

if \(f(x)\) becomes arbitrarily large for \(x\) sufficiently large. We say a function has a negative infinite limit at infinity and write

\[\lim_{x→∞}f(x)=−∞.\]

if \(f(x)<0\) and \(|f(x)|\) becomes arbitrarily large for \(x\) sufficiently large. Similarly, we can define infinite limits as \(x→−∞.\)

## End Behavior

The behavior of a function as \(x→±∞\) is called the function’s **end behavior**. At each of the function’s ends, the function could exhibit one of the following types of behavior:

- The function \(f(x)\) approaches a horizontal asymptote \(y=L\).
- The function \(f(x)→∞\) or \(f(x)→−∞.\)
- The function does not approach a finite limit, nor does it approach \(∞\) or \(−∞\). In this case, the function may have some oscillatory behavior.

Let’s consider several classes of functions here and look at the different types of end behaviors for these functions.

## Limit at infinity for Polynomial Functions

Consider the power function \(f(x)=x^n\) where \(n\) is a positive integer. From Figure \(\PageIndex{5}\) and Figure \(\PageIndex{6}\), we see that

\[\lim_{x→∞}x^n=∞;\;n=1,2,3,…\]

and

\[\lim_{x→−∞}x^n=\begin{cases}∞, & n=2,4,6,…\\−∞, & n=1,3,5,….\end{cases}\]

Using these facts, it is not difficult to evaluate \(\displaystyle \lim_{x→∞}cx^n\) and \(\displaystyle \lim_{x→−∞}cx^n\), where \(c\) is any constant and \(n\) is a positive integer. If \(c>0\), the graph of \(y=cx^n\)is a vertical stretch or compression of \(y=x^n,\) and therefore

\(\displaystyle \lim_{x→∞}cx^n=\lim_{x→∞}x^n\) and \(\displaystyle \lim_{x→−∞}cx^n=\lim_{x→−∞}x^n\) if \(c>0\).

If \(c<0,\) the graph of \(y=cx^n\) is a vertical stretch or compression combined with a reflection about the \(x\)-axis, and therefore

\(\displaystyle \lim_{x→∞}cx^n=−\lim_{x→∞}x^n\) and \(\displaystyle \lim_{x→−∞}cx^n=−\lim_{x→−∞}x^n\) if \(c<0.\)

If \(c=0,y=cx^n=0,\) in which case \(\displaystyle \lim_{x→∞}cx^n=0=\lim_{x→−∞}cx^n.\)

Example \(\PageIndex{3}\): Limits at Infinity for Power Functions

For each function \(f\), evaluate \(\displaystyle \lim_{x→∞}f(x)\) and \(\displaystyle \lim_{x→−∞}f(x)\).

- \(f(x)=−5x^3\)
- \(f(x)=2x^4\)

**Solution**

- Since the coefficient of \(x^3\) is \(−5\), the graph of \(f(x)=−5x^3\) involves a vertical stretch and reflection of the graph of \(y=x^3\) about the \(x\)-axis. Therefore, \(\displaystyle \lim_{x→∞}(−5x^3)=−∞\) and \(\displaystyle \lim_{x→−∞}(−5x^3)=∞\).
- Since the coefficient of \(x^4\) is \(2\), the graph of \(f(x)=2x^4\) is a vertical stretch of the graph of \(y=x^4\). Therefore, \(\displaystyle \lim_{x→∞}2x^4=∞\) and \(\displaystyle \lim_{x→−∞}2x^4=∞\).

Exercise \(\PageIndex{3}\)

Let \(f(x)=−3x^4\). Find \(\displaystyle \lim_{x→∞}f(x)\).

**Hint**-
The coefficient \(−3\) is negative.

**Answer**-
\(−∞\)

We now look at how the limits at infinity for power functions can be used to determine \(\displaystyle \lim_{x→±∞}f(x)\) for any polynomial function \(f\). Consider a polynomial function

\[f(x)=a_nx^n+a_{n−1}x^n−1+…+a^1x+a^0\]

of degree \(n≥1\) so that \(a_n≠0.\)Factoring, we see that

\[f(x)=a_nx^n\left(1+\frac{a_{n−1}}{a_n}\frac{1}{x}+…+\frac{a_1}{a_n}\frac{1}{x^{n−1}}+\frac{a_0}{a^n}\right).\]

As \(x→±∞,\) all the terms inside the parentheses approach zero except the first term. We conclude that

\[\lim_{x→±∞}f(x)=\lim_{x→±∞}a_nx^n.\]

For example, the function \(f(x)=5x^3−3x^2+4\) behaves like \(g(x)=5x^3\) as \(x→±∞\) as shown in Figure \(\PageIndex{7}\) and Table \(\PageIndex{3}\).

\(x\) |
10 | 100 | 1000 |
---|---|---|---|

\(f(x)=5x^3−3x^2+4\) |
4704 | 4,970,004 | 4,997,000,004 |

\(g(x)=5x^3\) |
5000 | 5,000,000 | 5,000,000,000 |

\(x\) |
−10 | −100 | −000 |

\(f(x)=5x^3−3x^2+4\) |
−5296 | −5,029,996 | −5,002,999,996 |

\(g(x)=5x^3\) |
−5000 | −5,000,000 | −5,000,000,000 |

## Summary Basic Results

Basic Results

- \(\displaystyle \lim_{x→±∞} k=k\), where \(k\) is a constant.
- \(\displaystyle \lim_{x→∞} x^n=\infty \), for all \(n \in \mathbb{N}\).
- \(\displaystyle \lim_{x→-∞} x^n=\infty \), when \(n\) is even.
- \(\displaystyle \lim_{x→∞} x^n= -\infty \), when \(n\) is odd.
- \(\lim_{x→±∞}a_nx^n+a_{n−1}x^n−1+…+a^1x+a^0=\lim_{x→±∞}a_nx^n.\)
- \(\lim_{x→±∞} \displaystyle \frac{1}{x^n}=0\), for all \(n \in \mathbb{N}\).

## Limit at infinity for Rational Functions

The Limit at infinity for rational functions and functions involving radicals is a little more complicated than for polynomials. In Example, we show that the limits at infinity of a rational function \(f(x)=\dfrac{p(x)}{q(x)}\) depend on the relationship between the degree of the numerator and the degree of the denominator. To evaluate the limits at infinity for a rational function, we divide the numerator and denominator by the highest power of \(x\) appearing in the denominator. This determines which term in the overall expression dominates the behavior of the function at large values of \(x\).

Example \(\PageIndex{5}\): Determining Limit at infinity for Rational Functions

For each of the following functions, determine the limits as \(x→∞\) and \(x→−∞.\) Then, use this information to describe the Limit at infinity of the function.

- \(f(x)=\dfrac{3x−1}{2x+5}\) (Note: The degree of the numerator and the denominator are the same.)
- \(f(x)=\dfrac{3x^2+2x}{4x^3−5x+7}\) (Note: The degree of numerator is less than the degree of the denominator.)
- \(f(x)=\dfrac{3x^2+4x}{x+2}\) in the denominator is \(x\). Therefore, dividing the numerator and denominator by \(x\) and applying the algebraic limit laws, we see that

**Solution**

a. The highest power of \(x\) in the denominator is \(x\). Therefore, dividing the numerator and denominator by \(x\) and applying the algebraic limit laws, we see that

\[ \begin{align*} \lim_{x→±∞}\frac{3x−1}{2x+5} &=\lim_{x→±∞}\frac{3−1/x}{2+5/x} \\[4pt] &=\frac{\lim_{x→±∞}(3−1/x)}{\lim_{x→±∞}(2+5/x)} \\[4pt] &=\frac{\lim_{x→±∞}3−\lim_{x→±∞}1/x}{\lim_{x→±∞}2+\lim_{x→±∞}5/x} \\[4pt] &=\frac{3−0}{2+0}=\frac{3}{2}. \end{align*}\]

Since \(\displaystyle \lim_{x→±∞}f(x)=\frac{3}{2}\), we know that \(y=\frac{3}{2}\) is a horizontal asymptote for this function as shown in the following graph.

b. Since the largest power of \(x\) appearing in the denominator is \(x^3\), divide the numerator and denominator by \(x^3\). After doing so and applying algebraic limit laws, we obtain

\[\lim_{x→±∞}\frac{3x^2+2x}{4x^3−5x+7}=\lim_{x→±∞}\frac{3/x+2/x^2}{4−5/x^2+7/x^3}=\frac{3\cdot 0+2\cdot 0}{4−5\cdot 0+7\cdot 0}=0. \nonumber\]

Therefore \(f\) has a horizontal asymptote of \(y=0\) as shown in the following graph.

c. Dividing the numerator and denominator by \(x\), we have

\[\displaystyle \lim_{x→±∞}\frac{3x^2+4x}{x+2}=\lim_{x→±∞}\frac{3x+4}{1+2/x}. \nonumber\]

As \(x→±∞\), the denominator approaches \(1\). As \(x→∞\), the numerator approaches \(+∞\). As \(x→−∞\), the numerator approaches \(−∞\). Therefore \(\displaystyle \lim_{x→∞}f(x)=∞\), whereas \(\displaystyle \lim_{x→−∞}f(x)=−∞\) as shown in the following figure.

Exercise \(\PageIndex{5}\)

Evaluate \(\displaystyle \lim_{x→±∞}\frac{3x^2+2x−1}{5x^2−4x+7}\) and use these limits to determine the Limit at infinity of \(f(x)=\dfrac{3x^2+2x−2}{5x^2−4x+7}\).

**Hint**-
Divide the numerator and denominator by \(x^2\).

**Answer**-
\(\frac{3}{5}\)

We can summarize the results of Example to make the following conclusion regarding Limit at infinity for rational functions. Consider a rational function

\[f(x)=\frac{p(x)}{q(x)}=\frac{a_nx^n+a_{n−1}x^{n−1}+…+a_1x+a_0}{b_mx^m+b_{m−1}x^{m−1}+…+b_1x+b_0},\nonumber\]

where \(a_n≠0\) and \(b_m≠0.\)

- If the degree of the numerator is the same as the degree of the denominator \((n=m),\) then \(f\) has a horizontal asymptote of \(y=a_n/b_m\) as \(x→±∞.\)
- If the degree of the numerator is less than the degree of the denominator \((n<m),\) then \(f\) has a horizontal asymptote of \(y=0\) as \(x→±∞.\)
- If the degree of the numerator is greater than the degree of the denominator \((n>m),\) then \(f\) does not have a horizontal asymptote. The limits at infinity are either positive or negative infinity, depending on the signs of the leading terms. In addition, using long division, the function can be rewritten as \[f(x)=\frac{p(x)}{q(x)}=g(x)+\frac{r(x)}{q(x)},\] where the degree of \(r(x)\) is less than the degree of \(q(x)\). As a result, \(\displaystyle \lim_{x→±∞}r(x)/q(x)=0\). Therefore, the values of \([f(x)−g(x)]\) approach zero as \(x→±∞\). If the degree of \(p(x)\) is exactly one more than the degree of \(q(x)\) (i.e., \(n=m+1\)), the function \(g(x)\) is a linear function. In this case, we call \(g(x)\) an oblique asymptote.

Now let’s consider the Limit at infinity for functions involving a radical.

Example \(\PageIndex{6}\): Determining Limit at infinity for a Function Involving a Radical

Find the limits as \(x→∞\) and \(x→−∞\) for \(f(x)=\dfrac{3x−2}{\sqrt{4x^2+5}}\).

**Solution**

Let’s use the same strategy as we did for rational functions: divide the numerator and denominator by a power of \(x\). To determine the appropriate power of \(x\), consider the expression \(\sqrt{4x^2+5}\) in the denominator. Since

\[\sqrt{4x^2+5}≈\sqrt{4x^2}=2|x| \nonumber\]

for large values of \(x\) in effect \(x\) appears just to the first power in the denominator. Therefore, we divide the numerator and denominator by \(|x|\). Then, using the fact that \(|x|=x\) for \(x>0, |x|=−x\) for \(x<0\), and \(|x|=\sqrt{x^2}\) for all \(x\), we calculate the limits as follows:

\[\begin{align*} \lim_{x→∞}\frac{3x−2}{\sqrt{4x^2+5}}&=\lim_{x→∞}\frac{(1/|x|)(3x−2)}{(1/|x|)\sqrt{4x^2+5}}\\[4pt]

&=\lim_{x→∞}\frac{(1/x)(3x−2)}{\sqrt{(1/x^2)(4x^2+5)}}\\[4pt]

&=\lim_{x→∞}\frac{3−2/x}{\sqrt{4+5/x^2}}=\frac{3}{\sqrt{4}}=\frac{3}{2} \end{align*}\]

\[\begin{align*} \lim_{x→−∞}\frac{3x−2}{\sqrt{4x^2+5}} &=\lim_{x→−∞}\frac{(1/|x|)(3x−2)}{(1/|x|)\sqrt{4x^2+5}}\\[4pt]

&=\lim_{x→−∞}\frac{(−1/x)(3x−2)}{\sqrt{(1/x^2)(4x^2+5)}}\\[4pt]

&=\lim_{x→−∞}\frac{−3+2/x}{\sqrt{4+5/x^2}}=\frac{−3}{\sqrt{4}}=\frac{−3}{2}. \end{align*}\]

Therefore, \(f(x)\) approaches the horizontal asymptote \(y=\frac{3}{2}\) as \(x→∞\) and the horizontal asymptote \(y=−\frac{3}{2}\) as \(x→−∞\) as shown in the following graph.

Exercise \(\PageIndex{6}\)

Evaluate \(\displaystyle \lim_{x→∞}\frac{\sqrt{3x^2+4}}{x+6}\).

**Hint**-
Divide the numerator and denominator by \(|x|\).

**Answer**-
\(±\sqrt{3}\)

Note

Below we will show two ways of solving limits at the infinity of rational functions.

Example \(\PageIndex{7}\)

Find \(\displaystyle\lim\limits_{x\rightarrow \infty} \dfrac{2x+3}{3x-2}\).

**Solution**

**Method 1: **Divide both numerator and denominator by highest power of \(x\) of the polynomial in the denominator.

\(\displaystyle\lim\limits_{x\rightarrow \infty} \dfrac{2x+3}{3x-2}\)

\(\displaystyle=\lim\limits_{x \rightarrow \infty} \dfrac{(\dfrac{2x+3}{x})}{(\dfrac{3x-2}{x}) }\)

\(\displaystyle=\lim\limits_{x \rightarrow \infty} \dfrac{(\dfrac{2x}{x}+\dfrac{3}{x})}{(\dfrac{3x}{x}-\dfrac{2}{x})}\)

\(\displaystyle=\lim\limits_{x\rightarrow \infty} \dfrac{(2+\dfrac{3}{x})}{(3-\dfrac{2}{x})}\)

\(\displaystyle=\dfrac{2}{3}.\)

**Method 2:**

\(\displaystyle\lim\limits_{x\rightarrow \infty} \dfrac{2x+3}{3x-2}=\lim\limits_{x\rightarrow \infty} \dfrac{2x}{3x} =\dfrac{2}{3}.\)

Exercise \(\PageIndex{7}\)

Find \(\displaystyle\lim\limits_{x\rightarrow -\infty} \dfrac{2x+3}{3x-2}\).

**Answer**-
\(\displaystyle\lim\limits_{x\rightarrow -\infty} \dfrac{2x+3}{3x-2}=\lim\limits_{x\rightarrow -\infty} \dfrac{2x}{3x} =dfrac{2}{3}.\)

Example \(\PageIndex{8}\)

Find \(\displaystyle\lim\limits_{x\rightarrow -\infty} \dfrac{x+9}{\sqrt{4x^2+3x+2}}\).

**Solution**

\(\displaystyle\lim\limits_{x\rightarrow -\infty} \dfrac{x+9}{\sqrt{4x^2+3x+2}}\)

\(\displaystyle =\lim\limits_{x\rightarrow -\infty} \dfrac{x}{\sqrt{4x^2}}\)

\(\displaystyle =\lim\limits_{x\rightarrow -\infty} \dfrac{\dfrac{x}{|x|}}{\dfrac{\sqrt{4x^2}} {|x|} }\)

\(\displaystyle =\lim\limits_{x\rightarrow -\infty} \dfrac{\dfrac{x}{-x}}{\dfrac{\sqrt{4x^2} }{\sqrt{ x^2}}}\), Note that \( \sqrt{ x^2}=|x| \).

\(\displaystyle =\dfrac{-1}{2}.\)

Exercise \(\PageIndex{8}\)

Find \(\displaystyle\lim\limits_{x\rightarrow \infty} \dfrac{x+9}{\sqrt{4x^2+3x+2}}\).

**Answer**-
\(\displaystyle=\dfrac{1}{2}.\)

Example \(\PageIndex{9}\):

Find \(\displaystyle \lim\limits_{x\rightarrow \infty} \sqrt{x^2+3x}-\sqrt{x^2+4x} \).

**Solution**

Exercise \(\PageIndex{9}\)

Find \(\displaystyle \lim\limits_{x\rightarrow -\infty} \sqrt{x^2+3x}-\sqrt{x^2+4x} \)

**Answer**-
\(\displaystyle\dfrac{-1}{2}\).

Example \(\PageIndex{10}\):

Find the horizontal asymptote(s) if any of the function

\(\displaystyle f(x)=\dfrac{x+9}{\sqrt{4x^2+3x+2}}\).

Solution:

\(\displaystyle y=\dfrac{-1}{2}\), and \(y==\dfrac{1}{2}\) are the horizontal asymptotes .

See Example \(\PageIndex{3}\) and Exercise \(\PageIndex{3}\).

## Determining End Behavior for Transcendental Functions

The six basic trigonometric functions are periodic and do not approach a finite limit as \(x→±∞.\) For example, \(\sin x\) oscillates between 1 and −1 (Figure \(\PageIndex{19}\)). The tangent function \(x\) has an infinite number of vertical asymptotes as \(x→±∞\); therefore, it does not approach a finite limit nor does it approach \(±∞\) as \(x→±∞\) as shown in Figure \(\PageIndex{20}\).

Recall that for any base \(b>0,\; b≠1,\) the function \(y=b^x\) is an exponential function with domain \((−∞,∞)\) and range \((0,∞)\). If \(b>1,\;y=b^x\) is increasing over \((−∞,∞)\). If \(0<b<1, \; y=b^x\) is decreasing over \((−∞,∞).\) For the natural exponential function \(f(x)=e^x, \; e≈2.718>1\). Therefore, \(f(x)=e^x\) is increasing on `\((−∞,∞)\) and the range is `\((0,∞)\). The exponential function \(f(x)=e^x\) approaches \(∞\) as \(x→∞\) and approaches \(0\) as \(x→−∞\) as shown in Table \(\PageIndex{4}\) and Figure \(\PageIndex{21}\).

\(x\) |
−5 | −2 | 0 | 2 | 5 |
---|---|---|---|---|---|

\(e^x\) |
0.00674 | 0.135 | 1 | 7.389 | 148.413 |

Recall that the natural logarithm function \(f(x)=\ln(x)\) is the inverse of the natural exponential function \(y=e^x\). Therefore, the domain of \(f(x)=\ln(x)\) is \((0,∞)\) and the range is \((−∞,∞)\). The graph of \(f(x)=\ln(x)\) is the reflection of the graph of \(y=e^x\) about the line \(y=x\). Therefore, \(\ln(x)→−∞\) as \(x→0^+\) and \(\ln(x)→∞\) as \(x→∞\) as shown in Figure \(\PageIndex{22}\) and Table \(\PageIndex{5}\).

\(x\) | 0.01 | 0.1 | 1 | 10 | 100 |
---|---|---|---|---|---|

\(\ln(x)\) | −4.605 | −2.303 | 0 | 2.303 | 4.605 |

Example \(\PageIndex{7}\): Determining End Behavior for a Transcendental Function

Find the limits as \(x→∞\) and \(x→−∞\) for \(f(x)=\dfrac{2+3e^x}{7−5e^x}\) and describe the end behavior of \(f.\)

**Solution**

To find the limit as \(x→∞,\) divide the numerator and denominator by \(e^x\):

\[ \begin{align*} \lim_{x→∞}f(x) &= \lim_{x→∞}\frac{2+3e^x}{7−5e^x} \\[4pt] &=\lim_{x→∞}\frac{(2/e^x)+3}{(7/e^x)−5.} \end{align*}\]

As shown in Figure \(\PageIndex{21}\), \(e^x→∞\) as \(x→∞\). Therefore,

\(\displaystyle \lim_{x→∞}\frac{2}{e^x}=0=\lim_{x→∞}\frac{7}{e^x}\).

We conclude that \(\displaystyle \lim_{x→∞f}(x)=−\frac{3}{5}\), and the graph of \(f\) approaches the horizontal asymptote \(y=−\frac{3}{5}\) as \(x→∞.\) To find the limit as \(x→−∞\), use the fact that \(e^x→0\) as \(x→−∞\) to conclude that \(\displaystyle \lim_{x→∞}f(x)=\frac{2}{7}\), and therefore the graph of approaches the horizontal asymptote \(y=\frac{2}{7}\) as \(x→−∞\).

Exercise \(\PageIndex{7}\)

Find the limits as \(x→∞\) and \(x→−∞\) for \(f(x)=\dfrac{3e^x−4}{5e^x+2}\).

**Hint**-
\(\displaystyle \lim_{x→∞}e^x=∞\) and \(\lim_{x→∞}e^x=0.\)

**Answer**-
\(\displaystyle \lim_{x→∞}f(x)=\frac{3}{5}, \quad\lim_{x→−∞}f(x)=−2\)

## Key Concepts

- The limit of \(f(x)\) is \(L\) as \(x→∞\) (or as \(x→−∞)\) if the values \(f(x)\) become arbitrarily close to \(L\) as \(x\)becomes sufficiently large.
- The limit of \(f(x)\) is \(∞\) as \(x→∞\) if \(f(x)\) becomes arbitrarily large as \(x\) becomes sufficiently large. The limit of \(f(x)\) is \(−∞\) as \(x→∞\) if \(f(x)<0\) and \(|f(x)|\) becomes arbitrarily large as \(x\) becomes sufficiently large. We can define the limit of \(f(x)\) as \(x\) approaches \(−∞\) similarly.
- For a polynomial function \(p(x)=a_nx^n+a_{n−1}x^{n−1}+…+a_1x+a_0,\) where \(a_n≠0\), the Limit at infinity is determined by the leading term \(a_nx^n\). If \(n≠0, p(x)\) approaches \(∞\) or \(−∞\)at each end.
- For a rational function \(f(x)=\frac{p(x)}{q(x),}\) the Limit at infinity is determined by the relationship between the degree of \(p\) and the degree of \(q\). If the degree of \(p\) is less than the degree of \(q\), the line \(y=0\) is a horizontal asymptote for \(f\). If the degree of \(p\) is equal to the degree of \(q\), then the line \(y=\frac{a_n}{b_n}\) is a horizontal asymptote, where \(a_n\) and \(b_n\) are the leading coefficients of \(p\) and \(q\), respectively. If the degree of \(p\) is greater than the degree of \(q\), then \(f\) approaches \(∞\) or \(−∞\) at each end.

## Glossary

**End behaviour**- the behavior of a function as \(x→∞\) and \(x→−∞\)

**horizontal asymptote**- if \(\lim_{x→∞}f(x)=L\) or \(\lim_{x→−∞}f(x)=L\), then \(y=L\) is a horizontal asymptote of \(f\)

**infinite limit at infinity**- a function that becomes arbitrarily large as
*x*becomes large

**limit at infinity**- a function that becomes arbitrarily large as \(x\) becomes large

**oblique asymptote**- the line \(y=mx+b\) if \(f(x)\) approaches it as \(x→∞\) or\( x→−∞\)

## Contributors and Attributions

Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.

Gregory Hartman (Virginia Military Institute). Contributions were made by Troy Siemers and Dimplekumar Chalishajar of VMI and Brian Heinold of Mount Saint Mary's University. This content is copyrighted by a Creative Commons Attribution - Noncommercial (BY-NC) License. http://www.apexcalculus.com/

Pamini Thangarajah (Mount Royal University, Calgary, Alberta, Canada)