2.5: Limits Involving Infinity; Asymptotes of Graphs

Learning Objectives

• Calculate the limit of a function as x increases or decreases without bound.
• Using correct notation, describe an infinite limit.
• Define a vertical asymptote.
• Recognize a horizontal asymptote on the graph of a function.
• Estimate the end behavior of a function as x increases or decreases without bound.
• Recognize an oblique asymptote on the graph of a function.

Infinite Limits

Evaluating the limit of a function at a point or evaluating the limit of a function from the right and left at a point helps us to characterize the behavior of a function around a given value. As we shall see, we can also describe the behavior of functions that do not have finite limits.

We now turn our attention to \(h(x)=1/(x−2)^2\), the third and final function introduced at the beginning of this section (see Figure \(\PageIndex{1}\)(c)). From its graph we see that as the values of \(x\) approach \(2\), the values of \(h(x)=1/(x−2)^2\) become larger and larger and, in fact, become infinite. Mathematically, we say that the limit of \(h(x)\) as \(x\) approaches \(2\) is positive infinity. Symbolically, we express this idea as

\[\lim_{x \to 2}h(x)=+∞. \nonumber \]

More generally, we define infinite limits as follows:

Definitions: Infinite Limits

We define three types of infinite limits.

Infinite limits from the left: Let \(f(x)\) be a function defined at all values in an open interval of the form \((b,a)\).

i. If the values of \(f(x)\) increase without bound as the values of \(x\) (where \(x<a\)) approach the number \(a\), then we say that the limit as \(x\) approaches \(a\) from the left is positive infinity and we write \[\lim_{x \to a^−}f(x)=+∞. \nonumber \]

ii. If the values of \(f(x)\) decrease without bound as the values of \(x\) (where \(x<a\)) approach the number \(a\), then we say that the limit as \(x\) approaches \(a\) from the left is negative infinity and we write \[\lim_{x \to a^−}f(x)=−∞. \nonumber \]

Infinite limits from the right: Let \(f(x)\) be a function defined at all values in an open interval of the form \((a,c)\).

i. If the values of \(f(x)\) increase without bound as the values of \(x\) (where \(x>a\)) approach the number \(a\), then we say that the limit as \(x\) approaches \(a\) from the right is positive infinity and we write \[\lim_{x \to a^+}f(x)=+∞. \nonumber \]

ii. If the values of \(f(x)\) decrease without bound as the values of \(x\) (where \(x>a\)) approach the number \(a\), then we say that the limit as \(x\) approaches \(a\) from the right is negative infinity and we write \[\lim_{x \to a^+}f(x)=−∞. \nonumber \]

Two-sided infinite limit: Let \(f(x)\) be defined for all \(x≠a\) in an open interval containing \(a\)

i. If the values of \(f(x)\) increase without bound as the values of \(x\) (where \(x≠a\)) approach the number \(a\), then we say that the limit as \(x\) approaches \(a\) is positive infinity and we write \[\lim_{x \to a} f(x)=+∞. \nonumber \]

ii. If the values of \(f(x)\) decrease without bound as the values of \(x\) (where \(x≠a\)) approach the number \(a\), then we say that the limit as \(x\) approaches \(a\) is negative infinity and we write \[\lim_{x \to a}f(x)=−∞. \nonumber \]

It is important to understand that when we write statements such as \(\displaystyle \lim_{x \to a}f(x)=+∞\) or \(\displaystyle \lim_{x \to a}f(x)=−∞\) we are describing the behavior of the function, as we have just defined it. We are not asserting that a limit exists. For the limit of a function \(f(x)\) to exist at \(a\), it must approach a real number \(L\) as \(x\) approaches \(a\). That said, if, for example, \(\displaystyle \lim_{x \to a}f(x)=+∞\), we always write \(\displaystyle \lim_{x \to a}f(x)=+∞\) rather than \(\displaystyle \lim_{x \to a}f(x)\) DNE.

Example \(\PageIndex{1}\): Recognizing an Infinite Limit

Evaluate each of the following limits, if possible. Use a table of functional values and graph \(f(x)=1/x\) to confirm your conclusion.

1. \(\displaystyle \lim_{x \to 0^−} \frac{1}{x}\)
2. \(\displaystyle \lim_{x \to 0^+} \frac{1}{x}\)
3. \( \displaystyle \lim_{x \to 0}\frac{1}{x}\)

Solution

Begin by constructing a table of functional values.

Table \(\PageIndex{7}\)

\(x\)	\(\dfrac{1}{x}\)	\(x\)	\(\dfrac{1}{x}\)
-0.1	-10	0.1	10
-0.01	-100	0.01	100
-0.001	-1000	0.001	1000
-0.0001	-10,000	0.0001	10,000
-0.00001	-100,000	0.00001	100,000
-0.000001	-1,000,000	0.000001	1,000,000

a. The values of \(1/x\) decrease without bound as \(x\) approaches \(0\) from the left. We conclude that

\[\lim_{x \to 0^−}\frac{1}{x}=−∞.\nonumber \]

b. The values of \(1/x\) increase without bound as \(x\) approaches \(0\) from the right. We conclude that

\[\lim_{x \to 0^+}\frac{1}{x}=+∞. \nonumber \]

c. Since \(\displaystyle \lim_{x \to 0^−}\frac{1}{x}=−∞\) and \(\displaystyle \lim_{x \to 0^+}\frac{1}{x}=+∞\) have different values, we conclude that

\[\lim_{x \to 0}\frac{1}{x}\quad\text{DNE.} \nonumber \]

The graph of \(f(x)=1/x\) in Figure \(\PageIndex{8}\) confirms these conclusions.

Figure \(\PageIndex{8}\): The graph of \(f(x)=1/x\) confirms that the limit as \(x\) approaches \(0\) does not exist.

Exercise \(\PageIndex{1}\):

Evaluate each of the following limits, if possible. Use a table of functional values and graph \(f(x)=1/x^2\) to confirm your conclusion.

1. \(\displaystyle \lim_{x \to 0^−}\frac{1}{x^2}\)
2. \(\displaystyle \lim_{x \to 0^+}\frac{1}{x^2}\)
3. \(\displaystyle \lim_{x \to 0}\frac{1}{x^2}\)

Hint

Follow the procedures from Example \(\PageIndex{5}\).

Answer

a. \(\displaystyle \lim_{x \to 0^−}\frac{1}{x^2}=+∞\);

b. \(\displaystyle \lim_{x \to 0^+}\frac{1}{x^2}=+∞\);

c. \(\displaystyle \lim_{x \to 0}\frac{1}{x^2}=+∞\)

It is useful to point out that functions of the form \(f(x)=1/(x−a)^n\), where n is a positive integer, have infinite limits as \(x\) approaches \(a\) from either the left or right (Figure \(\PageIndex{9}\)). These limits are summarized in the above definitions.

Figure \(\PageIndex{9}\): The function \(f(x)=1/(x−a)^n\) has infinite limits at \(a\).

Infinite Limits from Positive Integers

If \(n\) is a positive even integer, then

\[\lim_{x \to a}\frac{1}{(x−a)^n}=+∞.\label{infLim1} \]

If \(n\) is a positive odd integer, then

\[\lim_{x \to a^+}\frac{1}{(x−a)^n}=+∞\label{infLim2} \]

and

\[\lim_{x \to a^−}\frac{1}{(x−a)^n}=−∞.\label{infLim3} \]

We should also point out that in the graphs of \(f(x)=1/(x−a)^n\), points on the graph having \(x\)-coordinates very near to \(a\) are very close to the vertical line \(x=a\). That is, as \(x\) approaches \(a\), the points on the graph of \(f(x)\) are closer to the line \(x=a\). The line \(x=a\) is called a vertical asymptote of the graph. We formally define a vertical asymptote as follows:

Definition: Vertical Asymptotes

Let \(f(x)\) be a function. If any of the following conditions hold, then the line \(x=a\) is a vertical asymptote of \(f(x)\).

\[\lim_{x \to a^−}f(x)=+∞ \nonumber \]

\[\lim_{x \to a^−}f(x)=−∞ \nonumber \]

\[\lim_{x \to a^+}f(x)=+∞ \nonumber \]

\[\lim_{x \to a^+}f(x)=−∞ \nonumber \]

\[\lim_{x \to a}f(x)=+∞ \nonumber \]

\[\lim_{x \to a}f(x)=−∞ \nonumber \]

Example \(\PageIndex{2}\): Finding a Vertical Asymptote

Evaluate each of the following limits using Equations \ref{infLim1}, \ref{infLim2}, and \ref{infLim3} above. Identify any vertical asymptotes of the function \(f(x)=1/(x+3)^4.\)

1. \(\displaystyle \lim_{x \to −3^−}\frac{1}{(x+3)^4}\)
2. \(\displaystyle \lim_{x \to −3^+}\frac{1}{(x+3)^4}\)
3. \(\displaystyle \lim_{x \to −3}\frac{1}{(x+3)^4}\)

Solution

We can use the above equations directly.

1. \(\displaystyle \lim_{x \to −3^−}\frac{1}{(x+3)^4}=+∞\)
2. \(\displaystyle \lim_{x \to −3^+}\frac{1}{(x+3)^4}=+∞\)
3. \(\displaystyle \lim_{x \to −3}\frac{1}{(x+3)^4}=+∞\)

The function \(f(x)=1/(x+3)^4\) has a vertical asymptote of \(x=−3\).

Exercise \(\PageIndex{2}\):

Evaluate each of the following limits. Identify any vertical asymptotes of the function \(f(x)=\dfrac{1}{(x−2)^3}\).

1. \(\displaystyle \lim_{x→2^−}\frac{1}{(x−2)^3}\)
2. \(\displaystyle \lim_{x→2^+}\frac{1}{(x−2)^3}\)
3. \(\displaystyle \lim_{x→2}\frac{1}{(x−2)^3}\)

Answer a

\(\displaystyle \lim_{x→2^−}\frac{1}{(x−2)^3}=−∞\)

Answer b

\(\displaystyle \lim_{x→2^+}\frac{1}{(x−2)^3}=+∞\)

Answer c

\(\displaystyle \lim_{x→2}\frac{1}{(x−2)^3}\) DNE. The line \(x=2\) is the vertical asymptote of \(f(x)=1/(x−2)^3.\)

In the next example we put our knowledge of various types of limits to use to analyze the behavior of a function at several different points.

Example \(\PageIndex{3}\): Behavior of a Function at Different Points

Use the graph of \(f(x)\) in Figure \(\PageIndex{10}\) to determine each of the following values:

1. \(\displaystyle \lim_{x \to −4^−}f(x)\); \(\displaystyle \lim_{x \to −4^+}f(x)\); \(\displaystyle \lim_{x→−4}f(x);\;f(−4)\)
2. \(\displaystyle \lim_{x \to −2^−}f(x\)); \(\displaystyle \lim_{x \to −2^+}f(x)\); \(\displaystyle \lim_{x→−2}f(x);\;f(−2)\)
3. \( \displaystyle \lim_{x \to 1^−}f(x)\); \(\displaystyle \lim_{x \to 1^+}f(x)\); \(\displaystyle \lim_{x \to 1}f(x);\;f(1)\)
4. \( \displaystyle \lim_{x \to 3^−}f(x)\); \(\displaystyle \lim_{x \to 3^+}f(x)\); \(\displaystyle \lim_{x \to 3}f(x);\;f(3)\)

Figure \(\PageIndex{10}\): The graph shows \(f(x)\).

Solution

Using the definitions above and the graph for reference, we arrive at the following values:

1. \(\displaystyle \lim_{x \to −4^−}f(x)=0\); \(\displaystyle \lim_{x \to −4^+}f(x)=0\); \(\displaystyle \lim_{x \to −4}f(x)=0;\;f(−4)=0\)
2. \(\displaystyle \lim_{x \to −2^−}f(x)=3\); \(\displaystyle \lim_{x \to −2^+}f(x)=3\); \(\displaystyle \lim_{x \to −2}f(x)=3;\;f(−2)\) is undefined
3. \(\displaystyle \lim_{x \to 1^−}f(x)=6\); \(\displaystyle \lim_{x \to 1^+}f(x)=3\); \(\displaystyle \lim_{x \to 1}f(x)\) DNE; \(f(1)=6\)
4. \(\displaystyle \lim_{x \to 3^−}f(x)=−∞\); \(\displaystyle \lim_{x \to 3^+}f(x)=−∞\); \(\displaystyle \lim_{x \to 3}f(x)=−∞\); \(f(3)\) is undefined

Exercise \(\PageIndex{3}\):

Evaluate \(\displaystyle\lim_{x \to 1}f(x)\) for \(f(x)\) shown here:

Figure \(\PageIndex{11}\). The graph of a piecewise function \(f\).

Hint

Compare the limit from the right with the limit from the left.

Answer

\(\displaystyle\lim_{x \to 1}f(x)\) does not exist

Vertical Asymptotes

If the limit of \(f(x)\) as \(x\) approaches \(c\) from either the left or right (or both) is \(\infty\) or \(-\infty\), we say the function has a vertical asymptote at \(c\).

Example \(\PageIndex{4}\): Finding vertical asymptotes

Find the vertical asymptotes of \(f(x)=\dfrac{3x}{x^2-4}\).

\(\text{FIGURE 1.33}\): Graphing \(f(x) = \frac{3x}{x^2-4}\).

Vertical asymptotes occur where the function grows without bound; this can occur at values of \(c\) where the denominator is 0. When \(x\) is near \(c\), the denominator is small, which in turn can make the function take on large values. In the case of the given function, the denominator is 0 at \(x=\pm 2\). Substituting in values of \(x\) close to \(2\) and \(-2\) seems to indicate that the function tends toward \(\infty\) or \(-\infty\) at those points. We can graphically confirm this by looking at Figure 1.33. Thus the vertical asymptotes are at \(x=\pm2\).

When a rational function has a vertical asymptote at \(x=c\), we can conclude that the denominator is 0 at \(x=c\). However, just because the denominator is 0 at a certain point does not mean there is a vertical asymptote there. For instance, \(f(x)=(x^2-1)/(x-1)\) does not have a vertical asymptote at \(x=1\), as shown in Figure 1.34. While the denominator does get small near \(x=1\), the numerator gets small too, matching the denominator step for step. In fact, factoring the numerator, we get\[f(x)=\frac{(x-1)(x+1)}{x-1}.\]

Canceling the common term, we get that \(f(x)=x+1\) for \(x\not=1\). So there is clearly no asymptote, rather a hole exists in the graph at \(x=1\).

\(\text{FIGURE 1.34}\): Graphically showing that \(f(x)=\frac{x^2-1}{x-1}\) does not have an asymptote at \(x=1\).

The above example may seem a little contrived. Another example demonstrating this important concept is \(f(x)= (\sin x)/x\). We have considered this function several times in the previous sections. We found that \( \lim\limits_{x\to0}\frac{\sin x}{x}=1\); i.e., there is no vertical asymptote. No simple algebraic cancellation makes this fact obvious; we used the Squeeze Theorem in Section 1.3 to prove this.

If the denominator is 0 at a certain point but the numerator is not, then there will usually be a vertical asymptote at that point. On the other hand, if the numerator and denominator are both zero at that point, then there may or may not be a vertical asymptote at that point. This case where the numerator and denominator are both zero returns us to an important topic.

Indeterminate Forms

We have seen how the limits

\[\lim\limits_{x\rightarrow 0}\frac{\sin x}{x}\quad \text{and}\quad \lim\limits_{x\to1}\frac{x^2-1}{x-1}\]each return the indeterminate form "\(0/0\)'' when we blindly plug in \(x=0\) and \(x=1\), respectively. However, \(0/0\) is not a valid arithmetical expression. It gives no indication that the respective limits are 1 and 2.

With a little cleverness, one can come up \(0/0\) expressions which have a limit of \(\infty\), 0, or any other real number. That is why this expression is called indeterminate.

A key concept to understand is that such limits do not really return \(0/0\). Rather, keep in mind that we are taking limits. What is really happening is that the numerator is shrinking to 0 while the denominator is also shrinking to 0. The respective rates at which they do this are very important and determine the actual value of the limit.

An indeterminate form indicates that one needs to do more work in order to compute the limit. That work may be algebraic (such as factoring and canceling) or it may require a tool such as the Squeeze Theorem. In a later section we will learn a technique called l'Hospital's Rule that provides another way to handle indeterminate forms.

Some other common indeterminate forms are \(\infty-\infty\), \(\infty\cdot 0\), \(\infty/\infty\), \(0^0\), \(\infty^0\) and \(1^{\infty}\). Again, keep in mind that these are the "blind'' results of evaluating a limit, and each, in and of itself, has no meaning. The expression \(\infty-\infty\) does not really mean "subtract infinity from infinity.'' Rather, it means "One quantity is subtracted from the other, but both are growing without bound.'' What is the result? It is possible to get every value between \(-\infty\) and \(\infty\)

Note that \(1/0\) and \(\infty/0\) are not indeterminate forms, though they are not exactly valid mathematical expressions, either. In each, the function is growing without bound, indicating that the limit will be \(\infty\), \(-\infty\), or simply not exist if the left- and right-hand limits do not match.

Limits at Infinity and Horizontal Asymptotes

At the beginning of this section we briefly considered what happens to \(f(x) = 1/x^2\) as \(x\) grew very large. Graphically, it concerns the behavior of the function to the "far right'' of the graph. We make this notion more explicit in the following definition.

Definition: Limits at Infinity and Horizontal Asymptote

1. We say \(\lim\limits_{x\rightarrow\infty} f(x)=L\) if for every \(\epsilon>0\) there exists \(M>0\) such that if \(x\geq M\), then \(|f(x)-L|<\epsilon\).
2. We say \(\lim\limits_{x\rightarrow-\infty} f(x)=L\) if for every \(\epsilon>0\) there exists \(M<0\) such that if \(x\leq M\), then \(|f(x)-L|<\epsilon\).
3. If \(\lim\limits_{x\rightarrow\infty} f(x)=L\) or \(\lim\limits_{x\rightarrow-\infty} f(x)=L\), we say that \(y=L\) is a horizontal asymptote of \(f\).

We can also define limits such as \(\lim\limits_{x\rightarrow\infty}f(x)=\infty\) by combining this definition with Definition 5.

Example \(\PageIndex{5}\): Approximating horizontal asymptotes

Approximate the horizontal asymptote(s) of \( f(x)=\frac{x^2}{x^2+4}\).

Solution

We will approximate the horizontal asymptotes by approximating the limits \[\lim\limits_{x\to-\infty} \frac{x^2}{x^2+4}\quad \text{and}\quad \lim\limits_{x\to\infty} \frac{x^2}{x^2+4}.\]Figure 1.35(a) shows a sketch of \(f\), and part (b) gives values of \(f(x)\) for large magnitude values of \(x\). It seems reasonable to conclude from both of these sources that \(f\) has a horizontal asymptote at \(y=1\).

\(\text{FIGURE 1.35}\): Using a graph and a table to approximate a horizontal asymptote in Example 29.

Later, we will show how to determine this analytically.

Horizontal asymptotes can take on a variety of forms. Figure 1.36(a) shows that \(f(x) = x/(x^2+1)\) has a horizontal asymptote of \(y=0\), where 0 is approached from both above and below.

Figure 1.36(b) shows that \(f(x) =x/\sqrt{x^2+1}\) has two horizontal asymptotes; one at \(y=1\) and the other at \(y=-1\).

Figure 1.36(c) shows that \(f(x) = (\sin x)/x\) has even more interesting behavior than at just \(x=0\); as \(x\) approaches \(\pm\infty\), \(f(x)\) approaches 0, but oscillates as it does this.

\(\text{FIGURE 1.36}\): Considering different types of horizontal asymptotes.

We can analytically evaluate limits at infinity for rational functions once we understand \(\lim\limits_{x\rightarrow\infty} 1/x\). As \(x\) gets larger and larger, the \(1/x\) gets smaller and smaller, approaching 0. We can, in fact, make \(1/x\) as small as we want by choosing a large enough value of \(x\). Given \(\epsilon\), we can make \(1/x<\epsilon\) by choosing \(x>1/\epsilon\). Thus we have \(\lim\limits_{x\rightarrow\infty} 1/x=0\).

It is now not much of a jump to conclude the following:

\[\lim\limits_{x\rightarrow\infty}\frac1{x^n}=0\quad \text{and}\quad \lim\limits_{x\rightarrow-\infty}\frac1{x^n}=0\]

Now suppose we need to compute the following limit:

\[\lim\limits_{x\rightarrow\infty}\frac{x^3+2x+1}{4x^3-2x^2+9}.\]

A good way of approaching this is to divide through the numerator and denominator by \(x^3\) (hence dividing by 1), which is the largest power of \(x\) to appear in the function. Doing this, we get

\[\begin{align*}\lim\limits_{x\rightarrow\infty}\frac{x^3+2x+1}{4x^3-2x^2+9} &=\lim\limits_{x\rightarrow\infty}\frac{1/x^3}{1/x^3}\cdot\frac{x^3+2x+1}{4x^3-2x^2+9}\\ &=\lim\limits_{x\rightarrow\infty}\frac{x^3/x^3+2x/x^3+1/x^3}{4x^3/x^3-2x^2/x^3+9/x^3}\\ &= \lim\limits_{x\rightarrow\infty}\frac{1+2/x^2+1/x^3}{4-2/x+9/x^3}.\end{align*}\]

Then using the rules for limits (which also hold for limits at infinity), as well as the fact about limits of \(1/x^n\), we see that the limit becomes\[\frac{1+0+0}{4-0+0}=\frac14.\]

This procedure works for any rational function. In fact, it gives us the following theorem.

Theorem: Limits of Rational Functions at Infinity

Let \(f(x)\) be a rational function of the following form:

\[f(x)=\frac{a_nx^n + a_{n-1}x^{n-1}+\dots + a_1x + a_0}{b_mx^m + b_{m-1}x^{m-1} + \dots + b_1x + b_0},\]

where any of the coefficients may be 0 except for \(a_n\) and \(b_m\).

1. If \(n=m\), then \(\lim\limits_{x\rightarrow\infty} f(x) = \lim\limits_{x\rightarrow-\infty} f(x) = \frac{a_n}{b_m}\).
2. If \(n<m\), then \(\lim\limits_{x\rightarrow\infty} f(x) = \lim\limits_{x\rightarrow-\infty} f(x) = 0\).
3. If \(n>m\), then \(\lim\limits_{x\rightarrow\infty} f(x)\) and \(\lim\limits_{x\rightarrow-\infty} f(x)\) are both infinite.

We can see why this is true. If the highest power of \(x\) is the same in both the numerator and denominator (i.e. \(n=m\)), we will be in a situation like the example above, where we will divide by \(x^n\) and in the limit all the terms will approach 0 except for \(a_nx^n/x^n\) and \(b_mx^m/x^n\). Since \(n=m\), this will leave us with the limit \(a_n/b_m\). If \(n<m\), then after dividing through by \(x^m\), all the terms in the numerator will approach 0 in the limit, leaving us with \(0/b_m\) or 0. If \(n>m\), and we try dividing through by \(x^n\), we end up with all the terms in the denominator tending toward 0, while the \(x^n\) term in the numerator does not approach 0. This is indicative of some sort of infinite limit.

Intuitively, as \(x\) gets very large, all the terms in the numerator are small in comparison to \(a_nx^n\), and likewise all the terms in the denominator are small compared to \(b_nx^m\). If \(n=m\), looking only at these two important terms, we have \((a_nx^n)/(b_nx^m)\). This reduces to \(a_n/b_m\). If \(n<m\), the function behaves like \(a_n/(b_mx^{m-n})\), which tends toward 0. If \(n>m\), the function behaves like \(a_nx^{n-m}/b_m\), which will tend to either \(\infty\) or \(-\infty\) depending on the values of \(n\), \(m\), \(a_n\), \(b_m\) and whether you are looking for \(\lim\limits_{x\rightarrow\infty} f(x)\) or \(\lim\limits_{x\rightarrow-\infty} f(x)\).

With care, we can quickly evaluate limits at infinity for a large number of functions by considering the largest powers of \(x\). For instance, consider again \(\lim\limits_{x\to\pm\infty}\frac{x}{\sqrt{x^2+1}},\) graphed in Figure \ref{fig:hzasy}(b). When \(x\) is very large, \(x^2+1 \approx x^2\). Thus \[\sqrt{x^2+1}\approx \sqrt{x^2} = |x|,\quad \text{and}\quad \frac{x}{\sqrt{x^2+1}} \approx \frac{x}{|x|}.\]This expression is 1 when \(x\) is positive and \(-1\) when \(x\) is negative. Hence we get asymptotes of \(y=1\) and \(y=-1\), respectively.

Example \(\PageIndex{6}\): Finding a limit of a rational function

Confirm analytically that \(y=1\) is the horizontal asymptote of \( f(x) = \frac{x^2}{x^2+4}\), as approximated in Example 29.

Solution

Before using Theorem 11, let's use the technique of evaluating limits at infinity of rational functions that led to that theorem. The largest power of \(x\) in \(f\) is 2, so divide the numerator and denominator of \(f\) by \(x^2\), then take limits.

\[\begin{align*}\lim\limits_{x\to\infty}\frac{x^2}{x^2+4} &= \lim\limits_{x\to\infty}\frac{x^2/x^2}{x^2/x^2+4/x^2}\\ &=\lim\limits_{x\to\infty}\frac{1}{1+4/x^2}\\ &=\frac{1}{1+0}\\ &= 1. \end{align*}\]

We can also use Theorem 11 directly; in this case \(n=m\) so the limit is the ratio of the leading coefficients of the numerator and denominator, i.e., 1/1 = 1.

Example \(\PageIndex{6}\): Finding limits of rational functions

Use Theorem 11 to evaluate each of the following limits.

\(\begin{align}&1.\,\,\lim\limits_{x\rightarrow-\infty}\frac{x^2+2x-1}{x^3+1} \qquad\qquad &&3.\,\,\lim\limits_{x\rightarrow\infty}\frac{x^2-1}{3-x} \\ &2.\,\,\lim\limits_{x\rightarrow\infty}\frac{x^2+2x-1}{1-x-3x^2} && \\ \end{align}\)

\(\text{FIGURE 1.37}\): Visualizing the functions in Example 31.

Solution

1. The highest power of \(x\) is in the denominator. Therefore, the limit is 0; see Figure 1.37(a).
2. The highest power of \(x\) is \(x^2\), which occurs in both the numerator and denominator. The limit is therefore the ratio of the coefficients of \(x^2\), which is \(-1/3\). See Figure 1.37(b).
3. The highest power of \(x\) is in the numerator so the limit will be \(\infty\) or \(-\infty\). To see which, consider only the dominant terms from the numerator and denominator, which are \(x^2\) and \(-x\). The expression in the limit will behave like \(x^2/(-x) = -x\) for large values of \(x\). Therefore, the limit is \(-\infty\). See Figure 1.37(c).