Infinite limits

Section 1.17 Infinite limits

Objectives

Provide a rigorous definition of infinite limits at points \(a\in \R\) and at infinity.
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Investigate infinite limits graphically.
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Define vertical asymptotes using language of infinite limits.
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Develop computation techniques for infinite limits.
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As a complement to our limit at infinity discussion, we follow up with a treatment of infinite limits. Whereas previously we introduced the \(\lim\limits_{x\to \pm \infty}f(x)\) notation in order to describe the behavior of a function as the input of the function gets arbitrarily large (in the positive or negative direction), now we wish to introduce notation that indicates that the values of the function get arbitrarily large (in the positive or negative direction).

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Definition 1.17.1. Infinite limits (informal).

Let \(f\) be a function.

Infinite limits at \(a\in \R\).
Fix \(a\in \R\) and assume \(f\) is defined everywhere on an open interval containing \(a\text{,}\) except possibly at \(a\) itself. We say that \(f\) has limit \(\infty\) (resp., limit \(-\infty\)) at \(a\) if the values of \(f(x)\) can be made arbitrarily large and positive (resp. arbitrarily large and negative) provided \(x\) is sufficiently close (but not equal) to \(a\text{.}\) We write \(\lim\limits_{x\to a}f(x)=\infty\) (resp., \(\lim\limits_{x\to a}f(x)=-\infty\)) when this is the case.
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Infinite limit at \(\infty\).
Assume \(f\) is defined on an open interval of the form \((c,\infty)\text{.}\) We say that \(f\) has limit \(\infty\) (resp., limit \(-\infty\)) at \(\infty\) if the values of \(f(x)\) can be made arbitrarily large and positive (resp. arbitrarily large and negative) provided \(x\) is sufficiently large and positive. We write \(\lim\limits_{x\to \infty}f(x)=\infty\) (resp., \(\lim\limits_{x\to \infty}f(x)=-\infty\)) when this is the case.
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Infinite limit at \(-\infty\).
Assume \(f\) is defined on an open interval of the form \((-\infty,c)\text{.}\) We say that \(f\) has limit \(\infty\) (resp., limit \(-\infty\)) at \(-\infty\) if the values of \(f(x)\) can be made arbitrarily large and positive (resp. arbitrarily large and negative) provided \(x\) is sufficiently large and negative. We write \(\lim\limits_{x\to -\infty}f(x)=\infty\) (resp., \(\lim\limits_{x\to -\infty}f(x)=-\infty\)) when this is the case.
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Remark 1.17.2. Infinite limits.

It is important to observe that the various notions of infinite limit defined in Definition 1.17.1 all cover situations where the limit of the function does not exist. The best way to understand an infinite limit statement of the form \(\lim\limits_{x\to a}f(x)=\pm \infty\text{,}\) where \(a\) denotes either a real number, \(\infty\text{,}\) or \(-\infty\text{,}\) is as an assertion that

the limit in question does not exist, and
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its failure to exist is due to values of the function getting arbitrarily large (in positive or negative direction) as \(x\) approaches \(a\text{.}\)
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Accordingly, we must understand this new notation as an extended version of our original limit notation. In particular, we are not treating \(\infty\) or \(-\infty\) here as if they were actual real numbers.

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Remark 1.17.3. One-sided infinite limits.

Yet more variants of infinite limits can be defined for one-sided limits: that is, we can make sense of the following statements for any \(a\in \R\text{:}\)

\begin{align*} \lim\limits_{x\to a^+}f(x) \amp = \infty \amp \lim\limits_{x\to a^+}f(x)\amp =-\infty\\ \lim\limits_{x\to a^-}f(x) \amp = \infty \amp \lim\limits_{x\to a^-}f(x)\amp =-\infty\text{.} \end{align*}

We thought Definition 1.17.1 was long enough as it is.

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Graphically speaking, if \(\lim\limits_{x\to a}f(x)=\pm \infty\) at a value \(a\text{,}\) then the graph of \(f\) will asymptotically approach the vertical line \(x=a\) as \(x\) gets arbitrarily close to \(a\text{.}\) This motivates the following definition.

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Definition 1.17.4. Vertical asymptote.

Fix a constant \(a\in \R\text{.}\) The line \(x=a\) is a vertical asymptote of the graph of a function \(f\) if at least one the of the following conditions holds:

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As with limits at infinity, we would like a starter kit of basic infinite limit formulas that we can build up from. The theorem below is a compendium of limits at infinity and infinite limit formulas for power functions and their reciprocals. Of course, this theorem, like all theorems technically requires proof. In this case, however, the statements of the theorem are essentially just limit notation translations of properties of the graphs of these familiar functions. (See Figure 1.5.1 and Figure 1.5.2.)

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Theorem 1.17.5. Power functions and their reciprocals.

Let \(n\) be a positive integer.

\begin{align*} \lim_{x\to \infty}x^n \amp= \infty.\\ \lim_{x\to -\infty}x^n \amp= \begin{cases} \infty \amp \text{if } n \text{ is even}\\ -\infty \amp \text{ if } n \text{ is odd.} \end{cases}\\ \lim_{x\to 0^+}\frac{1}{x^n} \amp = \infty.\\ \lim_{x\to 0^-}\frac{1}{x^n} \amp =\begin{cases} \infty\amp \text{if } n \text{ is even}\\ -\infty\amp\text{if } n \text{ is odd. } \end{cases} \end{align*}

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The next theorem helps us to compute the limit of functions built from other functions that may have infinite limits. There are a lot of rules listed there, each equipped with a type name that will help you reference it. That said, each rule is the consequence of very elementary reasoning about the size of various quantities. Instead of trying to commit all of Theorem 1.17.6 to memory, we encourage you instead to internalize this reasoning. For example, the fact that if \(\lim\limits_{x\to a}f(x)=2\) and \(\lim\limits_{x\to a}g(x)=\infty \text{,}\) then \(\lim\limits_{x\to a}\frac{f(x)}{g(x)}=0\text{,}\) follows from the simple fact that if the numerator \(f(x)\) is approaching some finite number \(c\) as \(x\to a\text{,}\) while the denominator \(g(x)\) gets arbitrarily large, then the quotient \(f(x)/g(x)\) is the result of taking a number very close to \(2\) and dividing it by a very large positive number, which yields a very small positive number. Thus the limit is equal to 0: i.e., \(\lim\limits_{x\to a}f(x)/g(x)=0\text{.}\)

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All of the arguments behind the results of Theorem 1.17.6 are of a similar elementary nature. The type descriptions of each result (e.g., \(\infty+\infty\text{,}\) \(\infty\cdot c\text{,}\) etc.) should be thought of as helpful shorthand for the simple principles at work. You should use these in the parenthetical justifications of steps in an infinite limit computation.

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Theorem 1.17.6. Infinite limit rules.

For some arithmetic combinations of functions with infinite limits, we can easily determine the resulting limit using a simple argument about size of quantities. We collect some of these results below. In what follows the limit point \(a\) can be either a real number or \(\pm\infty\text{.}\)

\begin{align*} \lim\limits_{ x\to a}f(x)=\infty, \lim\limits_{ x\to a}g(x)=c \amp \implies \lim\limits_{ x\to a}f(x)+g(x)=\infty \amp (\text{type } \pm\infty+c) \\ \lim\limits_{ x\to a}f(x)=\lim\limits_{ x\to a}g(x)=\infty \amp \implies \lim\limits_{ x\to a}f(x)+g(x)=\infty \amp (\text{type } \infty+\infty) \\ \lim\limits_{ x\to a}f(x)=\lim\limits_{ x\to a}g(x)=-\infty \amp \implies \lim\limits_{ x\to a}f(x)+g(x)=-\infty \amp (\text{type } -\infty -\infty) \\ \lim\limits_{ x\to a}f(x)=\lim\limits_{ x\to a}g(x)=\infty \amp \implies \lim\limits_{ x\to a}f(x)g(x)=\infty \amp (\text{type } \infty\cdot\infty) \\ \lim\limits_{ x\to a}f(x)=\lim\limits_{ x\to a}g(x)=-\infty \amp \implies \lim\limits_{ x\to a}f(x)g(x)=\infty \amp (\text{type } (-\infty)\cdot(-\infty))\\ \lim\limits_{ x\to a}f(x)=\infty, \lim\limits_{x\to a}g(x)=-\infty \amp \implies \lim\limits_{x\to a}f(x)g(x)=-\infty \amp (\text{type } (\infty)\cdot(-\infty))\\ \lim\limits_{x\to a}f(x)=\infty \amp \implies \lim\limits_{x\to a}\sqrt[n]{f(x)}=\infty \amp (\text{type } \sqrt[n]{\infty}) \\ \lim\limits_{x\to a}f(x)=-\infty \text{ and } n \text{ odd} \amp \implies \lim\limits_{x\to a}\sqrt[n]{f(x)}=-\infty \amp (\text{type } \sqrt[n]{-\infty}) \\ \lim\limits_{x\to a}f(x)=c, \lim\limits_{x\to a}g(x)=\pm \infty \amp \implies \lim\limits_{x\to a}\frac{f(x)}{g(x)}=0 \amp (\text{type } \tfrac{c}{\pm\infty}) \\ \lim\limits_{x\to a}f(x)=\pm\infty, \lim\limits_{x\to a}g(x)=c\ne 0 \amp \implies \lim\limits_{x\to a}f(x)g(x)=\pm \infty \amp (\text{type } \infty\cdot c)\\ \lim\limits_{x\to a}f(x)=c\ne 0, \lim\limits_{x\to a}g(x)=0 \amp \implies \lim\limits_{x\to a}\frac{f(x)}{g(x)}=\pm \infty \amp (\text{type } \tfrac{c}{0}) \text{.} \end{align*}

In the type \(\infty\cdot c\) rule, the sign of the limit depends on the sign of \(c\) and whether the limit of \(f\) is \(\infty\) or \(-\infty\text{.}\) Similarly, in the type \(c/0\) rule, the sign of the limit depends on the sign of \(c\) and the sign of \(g(x)\) as \(x\) approaches \(a\text{.}\)

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Let’s see how to write up our work when making use of the results of Theorem 1.17.6. The explanations in these situations tend to be slightly less streamlined than usual. This is in large part a result of the fact that in these situations we cannot make use of our usual limit rules (e.g., sum, product, quotient, etc.); and this is so precisely because those rules require that the limits involved exist!

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Example 1.17.7. Infinite limit.

Compute the following limits. Your answer should be either a real number, \(\infty\text{,}\) or \(-\infty\text{.}\)

\(\displaystyle \lim\limits_{x\to 1^-}\sqrt[5]{\frac{3}{x-1}}\)
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\(\displaystyle \lim\limits_{x\to -\infty}x^3+7x^2\)
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Solution.

In this case a quick appraisal tells us that the limit in question will fall under type \(\sqrt[5]{-\infty}\text{,}\) and thus should be equal to \(-\infty\text{,}\) the basic principle being the following: the \(5\)-th root of a very large negative number is a very large negative number. Here is how we can formally write this up. In this case we come right out and state the answer, indicating that this limit is of one of our known types:

\begin{align*} \lim\limits_{x\to 1^-}\sqrt[5]{\frac{3}{x-1}} \amp = -\infty \amp (\text{type } \sqrt[5]{-\infty}) \text{.} \end{align*}

We then provide justification for the claim that the limit is of the stated type. This can be done via a series of limit observations about component functions as follows:
- \(\lim\limits_{x\to 1^-}x-1=0\text{;}\)
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- \(\lim\limits_{x\to 1^-}\tfrac{3}{x-1}= -\infty\) (type \(3/0\)), since \(\lim\limits_{x\to 1^-} x-1=0\) and \(x< 1\implies x-1< 0\text{.}\)
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Note first that \(\lim\limits_{x\to -\infty}x^3=-\infty\) (1.17.5), and \(\lim\limits_{x\to -\infty}7x^2=\infty\) (type \(\infty\cdot c\)). Unfortunately, we do not have any infinite limit principles with descriptive type \(\infty-\infty\text{,}\) so we cannot use any of the results of Theorem 1.17.6 directly. Instead, we first do some algebra, using our intuition that the \(x^3\) term “dominates” the \(7x^2\) term. We make this intuition precise by forcibly factoring out \(x^3\) from the expression, in a way that allows us to recognize the type of the limit:

\begin{align*} \lim\limits_{x\to -\infty}x^3+7x^2 \amp = \lim\limits_{x\to -\infty}x^3(1+7/x) \\ \amp = -\infty \amp (\text{type } -\infty\cdot 1)\text{,} \end{align*}

since \(\lim\limits_{x\to -\infty}x^3=-\infty\) and \(\lim\limits_{x\to -\infty}1+\frac{7}{x}=1+0=1\text{.}\)

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Remark 1.17.8. Determinate versus indeterminate forms.

In part (b) of Example 1.17.7, we saw that the limit expression in its initial form was of type form \(\infty-\infty\text{.}\) In contrast to the limit types of Theorem 1.17.6, the description \(\infty-\infty\) alone is not sufficient for determining the limit. Why? In a limit of form \(\infty-\infty\text{,}\) we have one term that is getting arbitrarily large and positive, and the other getting arbitrarily large and negative. That information alone, does not tell us what the limit behavior is. Indeed, in general the answer will depend on how quickly each term grows.

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To make this more explicit consider the following three limits:

\begin{align*} \lim\limits_{x\to \infty}x-x \amp = \lim\limits_{x\to \infty}0=0 \\ \lim\limits_{x\to \infty} x^2-x\amp =\lim\limits_{x\to \infty}x^2(1-1/x)=\infty \\ \lim\limits_{x\to \infty}x-2x \amp= \lim\limits_{x\to \infty}-x=-\infty \text{.} \end{align*}

All three limits are of type \(\infty-\infty\text{,}\) and yet we get three different answers. For this reason, limit types like \(\infty-\infty\) are called indeterminate forms, because the information they convey is not sufficient for determining the limit. In contrast, the limit types appearing in Theorem 1.17.6 are called determinate forms.

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Below is a list of some common indeterminate forms to watch out for:

\begin{align*} \infty-\infty \amp \amp \frac{0}{0} \amp \amp \frac{\infty}{\infty} \amp\amp 0\cdot \infty \text{.} \end{align*}

It is important to realize that if in the course of your analysis you encounter an indeterminate form, this does not necessarily mean that the limit in question does not exist. Instead, it means that your analysis thus far simply isn’t detailed enough to provide an answer: i.e., your investigation so far is inconclusive, and more work is required. For the time being that additional work will be algebraic in nature.

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The technique used for the computations in Example 1.17.7 can be summarized by the following rough procedure.

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Procedure 1.17.9. Infinite limit computation.

To compute an infinite limit \(\lim\limits_{x\to a}f(x)\text{,}\) with justification, follow the steps below.

If you cannot immediately identify the limit as one of the types of Theorem 1.17.6, do some algebraic manipulation. It may help to forcibly factor out terms that dominate the behavior of the function as \(x\to a\text{.}\)
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Once you have identified the type of the infinite limit, compute the limit using Theorem 1.17.6 and cite the limit type.
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Below your computed limit, provide the necessary justification showing that the limit is indeed of the stated type. This typically involves computing limits of simpler functions that are used to construct \(f\text{.}\) You may need to cite limit types and Theorem 1.17.6 for some of these simpler limit computations as well.
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Let’s see this procedure in action with another example.

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Example 1.17.10. Infinite limits.

Compute the following limit. Your answer should be a real number or \(\pm \infty\text{.}\)

\begin{equation*} \lim\limits_{x\to \infty}\sqrt{\frac{2x^3-5x+1}{x^2+x-2}}\text{.} \end{equation*}

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Solution.

We have

\begin{align*} \lim\limits_{x\to \infty}\sqrt{\frac{2x^3-5x+1}{x^2+x-2}} \amp = \lim\limits_{x\to \infty}\sqrt{\frac{x^3(2-5/x^2+1/x^3)}{x^2(1+1/x-2/x^2)}} \\ \amp = \lim\limits_{x\to \infty}\sqrt{x\cdot \frac{2-5/x^2+1/x^3}{1+1/x-2/x^2}} \\ \amp = \infty \amp (\text{type } \sqrt{\infty}) \text{,} \end{align*}

since

\begin{align*} \lim\limits_{x\to \infty}x \amp =\infty\\ \lim\limits_{x\to \infty}\frac{2-5/x^2+1/x^3}{1+1/x-2/x^2} \amp =\frac{2}{1}=2 \end{align*}

and thus

\begin{align*} \lim\limits_{x\to \infty}x\cdot \frac{2-5/x^2+1/x^3}{1+1/x-2/x^2} \amp= \infty \amp (\text{type }\infty\cdot 2) \text{.} \end{align*}

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Remark 1.17.11. End behavior of polynomials.

The technique in part (b) of Example 1.17.7 can be generalized to analyze the behavior of an arbitrary polynomial.

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Let \(f(x)=\anpoly\text{,}\) where \(a_n\ne 0\text{.}\) Our intuition tells us that the leading term \(a_nx^n\) should “dominate” the other terms as \(x\to\pm\infty\text{.}\) We make precise sense of this algebraically by forcibly factoring out \(x^n\text{.}\) Namely, we write

\begin{equation*} f(x)=x^n\underset{p(x)}{\underbrace{\left(a_n+\frac{a_{n-1}}{x}+\frac{a_{n-2}}{x^2}+\cdots \frac{a_1}{x^{n-1}}+\frac{a_0}{x^n}\right)}}\text{.} \end{equation*}

Next, we compute

\begin{align*} \lim\limits_{x\to \infty}p(x) \amp = \lim\limits_{x\to \infty} (a_n+a_{n-1}/x+a_{n-2}/x^2+\cdots +a_1/x^{n-1}+a_0/x^n) \\ \amp = (\lim\limits_{x\to \infty}a_n+\lim\limits_{x\to \infty}a_{n-1}/x++\cdots+\lim\limits_{x\to \infty} a_1/x^{n-1}+\lim\limits_{x\to \infty}a_0/x^n) \amp \\ \amp =a_n+0+\cdots +0 \amp \\ \amp =a_n\text{.} \end{align*}

We conclude that \(\lim\limits_{x\to \infty}f(x)\) is of type \(\infty\cdot a_n\text{.}\) It follows from Theorem 1.17.6 that

\begin{equation*} \lim\limits_{x\to \pm \infty}\anpoly=\pm\infty\text{.} \end{equation*}

However, the precise sign depends on the sign of \(a_n\) and the sign of \(x^n\) as \(x\to \pm \infty\text{.}\) For example, we have

\begin{equation*} \lim\limits_{x\to -\infty}-3x^5+x^4+x^2+1=\infty\text{,} \end{equation*}

since \(-3x^5> 0\) as \(x\to -\infty\text{.}\) As a shorthand, we write

\begin{equation*} \lim\limits_{x\to \pm\infty}\anpoly=\lim\limits_{x\to \pm\infty}a_nx^n\text{,} \end{equation*}

and leave it to you to determine whether the last limit is \(\infty\) or \(-\infty\text{.}\)

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The argument above can be extended even further to compute limits at infinity of an arbitrary rational function.

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Theorem 1.17.12. Limit at infinity: rational functions.

Assume we are given polynomials \(f(x)=ax^m+a_{m-1}x^{m-1}+\cdots +a_1x+a_0\) and \(g(x)=bx^n+b_{n-1}x^{n-1}+\cdots +b_1x+b_0\text{,}\) where \(a\) and \(b\) are nonzero constants.

Equal degree.

If \(m=n\text{,}\) then

\begin{align*} \lim\limits_{x\to \infty}\frac{f(x)}{g(x)} \amp=\lim\limits_{x\to -\infty}\frac{f(x)}{g(x)}=\frac{a}{b} \text{.} \end{align*}

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Denominator degree bigger.

If \(m< n\text{,}\) then

\begin{align*} \lim\limits_{x\to \infty}\frac{f(x)}{g(x)} \amp =\lim\limits_{x\to -\infty}\frac{f(x)}{g(x)}=0\text{.} \end{align*}

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Numerator degree bigger.

If \(m> n\text{,}\) then

\begin{align*} \lim\limits_{x\to \infty}\frac{f(x)}{g(x)} \amp =\lim\limits_{x\to \infty}\frac{a}{b}x^{m-n}\\ \lim\limits_{x\to -\infty}\frac{f(x)}{g(x)} \amp =\lim\limits_{x\to -\infty}\frac{a}{b}x^{m-n}\text{.} \end{align*}

Furthermore, both these limits are infinite. However, their sign depends on whether \(a/b\) is positive or negative, and whether \(m-n\) is even or odd. Alternatively, the sign of the limit depends on the limits of \(f\) and \(g\) at \(\infty\) and \(-\infty\text{,}\) respectively.

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Since both limits at infinity do not exist, \(f\) does not have any horizontal asymptotes.
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The next example will have us put Theorem 1.17.12 to good use in the context of computing horizontal and vertical asymptotes of a rational function.

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Example 1.17.13. Asymptotes: rational function.

Let \(\displaystyle f(x)=\frac{x^5-2x^4}{-x^2+3x-2}\text{.}\) Find all and any horizontal and vertical asymptotes of \(f\text{.}\) For any vertical asymptotes, compute both one-sided limits.

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Solution.

For our horizontal asymptote investigation we compute the limits at infinity of \(f\text{.}\) At \(\infty\) we have

\begin{align*} \lim\limits_{x\to \infty}f(x) \amp= \lim\limits_{x\to \infty}\frac{x^5-2x^4}{-x^2+3x-2} \\ \amp =\lim\limits_{x\to \infty}-x^{3} \amp (\knowl{./knowl/xref/th_rational_function.html}{\text{Theorem 1.17.12}}) \\ \amp = -\infty \amp (\infty\cdot c) \end{align*}

since \(\lim\limits_{x\to \infty}x^3=\infty\text{,}\) and \(-x^3\) is negative eventually as \(x\to \infty\text{.}\)

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At \(-\infty\) we have

\begin{align*} \lim\limits_{x\to -\infty}f(x) \amp= \lim\limits_{x\to -\infty}\frac{x^5-2x^4}{-x^2+3x-2} \\ \amp =\lim\limits_{x\to -\infty}-x^{3} \amp (\knowl{./knowl/xref/th_rational_function.html}{\text{Theorem 1.17.12}}) \\ \amp = \infty \amp (\text{type } \infty\cdot c) \end{align*}

since \(\lim\limits_{x\to -\infty}x^3=-\infty\text{,}\) and \(-x^3\) is positive eventually as \(x\to -\infty\text{.}\)

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Since \(f\) is algebraic, the possible vertical asymptotes are of the form \(x=a\text{,}\) where \(a\) is not in the domain of \(f\text{.}\) We have

\begin{equation*} f(x)=\frac{x^5-2x^4}{-x^2+3x-2}=\frac{x^4(x-2)}{-(x-2)(x-1)}\text{.} \end{equation*}

From this we see that the only candidates for vertical asymptotes are the lines \(x=2\) and \(x=1\text{.}\) We investigate the limits at these points:

\begin{align*} \lim\limits_{x\to 2 }f(x)\amp=\lim\limits_{x\to 2}\frac{x^4(x-2)}{-(x-2)(x-1)} \\ \amp = \lim\limits_{x\to 2 }\frac{x^4}{-(x-1)} \amp \text{(repl.)}\\ \amp = -16 \amp \text{(poly. eval.)}\text{.} \end{align*}

Since the limit exists at \(2\text{,}\) the line \(x=2\) is not a vertical asymptote.

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We now compute the one-sided limits at \(x=1\text{.}\) We have

\begin{align*} \lim\limits_{x\to 1^-}f(x) \amp = \lim\limits_{x\to 1^-}\frac{-x^4}{x-1} \amp \text{(repl.)}\\ \amp = \infty \amp (\text{type } c/0)\text{,} \end{align*}

since \(\lim\limits_{x\to 1^-}x-1=0\text{,}\) \(\lim\limits_{x\to 1^-}-x^4=-1\text{,}\) and \(-x^4/(x-1)\) is positive for all \(x\) close to and less than \(1\text{.}\) Since once of the one-sided limits is infinite, we conclude that \(x=1\) is a vertical asymptote of the graph of \(f\text{.}\)

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We are asked to compute the other one-sided limit. The computation is similar:

\begin{align*} \lim\limits_{x\to 1^+}f(x) \amp = \lim\limits_{x\to 1^+}\frac{-x^4}{x-1} \amp \text{(repl.)}\\ \amp = -\infty \amp (\text{type } c/0)\text{,} \end{align*}

since \(\lim\limits_{x\to 1^-}x-1=0\text{,}\) \(\lim\limits_{x\to 1^-}-x^4=-1\text{,}\) and \(-x^4/(x-1)\) is negative for all \(x\) close to and greater than \(1\text{.}\)

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Warning 1.17.14. End behavior of rational functions.

Note well the important restriction imposed in Theorem 1.17.12: namely, this can only be applied to limits at infinity of rational functions. Do not assume that it can be applied to any quotient of functions!

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For example, we cannot apply Theorem 1.17.12 to the limit

\begin{equation*} \lim\limits_{x\to \infty}\frac{\sqrt{\frac{1}{x}+x^2}}{\frac{1}{x^3}+7x}\text{,} \end{equation*}

since this function is not rational. However, the same approach of forcibly factoring out a dominating term can be used:

\begin{align*} \lim\limits_{x\to \infty} \frac{\sqrt{\frac{1}{x}+x^2}}{\frac{1}{x^3}+7x} \amp =\lim\limits_{x\to \infty}\frac{\sqrt{x^2(\frac{1}{x^3}+1)}}{x(\frac{1}{x^4}+7)} \\ \amp =\lim\limits_{x\to \infty}\frac{\abs{x}\sqrt{1/x^3+1}}{x(1/x^4+7)} \\ \amp = \lim\limits_{x\to \infty}\frac{x\sqrt{1/x^3+1}}{x(1/x^4+7)} \amp (x> 0)\\ \amp =\lim\limits_{x\to \infty}\frac{\sqrt{1/x^3+1}}{1/x^4+7}\\ \amp = \frac{\sqrt{\lim\limits_{x\to \infty}1/x^3+1}}{\lim\limits_{x\to \infty} 1/x^4+7}\\ \amp =\frac{\sqrt{0+1}}{0+7}\\ \amp =\frac{1}{7}\text{.} \end{align*}

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