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## Section3.5Limits at Infinity, Infinite Limits and Asymptotes

### Subsection3.5.1Limits at Infinity

We occasionally want to know what happens to some quantity when a variable gets very large or “goes to infinity”.

###### Example3.18. Limit at Infinity.

What happens to the function $\ds \cos(1/x)$ as $x$ goes to infinity? It seems clear that as $x$ gets larger and larger, $1/x$ gets closer and closer to zero, so $\cos(1/x)$ should be getting closer and closer to $\cos(0)=1\text{.}$

As with ordinary limits, this concept of “limit at infinity” can be made precise. Roughly, we want $\ds \lim_{x\to \infty}f(x)=L$ to mean that we can make $f(x)$ as close as we want to $L$ by making $x$ large enough.

###### Definition3.19. Limit at Infinity.

In general, we write

\begin{equation*} \lim_{x \to \infty} f(x) = L \end{equation*}

if $f(x)$ can be made arbitrarily close to $L$ by taking $x$ large enough. If this limits exists, we say that the function $f$ has the limit $L$ as $x$ increases without bound.

Similarly, we write

\begin{equation*} \lim_{x \to -\infty} f(x) = M \end{equation*}

if $f(x)$ can be made arbitrarily close to $M$ by taking $x$ to be negative and sufficiently large in absolute value. If this limit exists, we say that the function $f$ has the limit $L$ as $x$ decreases without bound.

###### Example3.20. Limit at Infinity.

Let $f$ and $g$ be the functions

\begin{equation*} f(x) = \begin{cases} -2 \amp \text{ if } x \lt 0 \\ 2 \amp \text{ if } x \geq 0 \end{cases} \text{ and } g(x) = \dfrac{1}{x^{2}} \end{equation*}

Evaluate:

1. $\lim\limits_{x \to \infty} f(x)$ and $\lim\limits_{x \to -\infty} f(x)\text{.}$

2. $\lim\limits_{x \to \infty} g(x)$ and $\lim\limits_{x \to -\infty} g(x)\text{.}$

Solution

Referring to the graphs of $f(x)$ and $g(x)$ shown below, we see that

1. $\lim\limits_{x \to \infty} f(x) = 2$ and $\lim\limits_{x \to -\infty} f(x) = -2\text{.}$

2. $\lim\limits_{x \to \infty} g(x) = 0$ and $\lim\limits_{x \to -\infty} g(x) = 0\text{.}$  ###### Definition3.21. Limit at Infinity (Formal Definition).

If $f$ is a function, we say that $\ds \lim_{x\to \infty}f(x)=L$ if for every $\epsilon>0$ there is an $N > 0$ so that whenever $x>N\text{,}$ $|f(x)-L|\lt \epsilon\text{.}$ We may similarly define $\ds \lim_{x\to-\infty}f(x)=L\text{.}$

We include this definition for completeness, but we will not explore it in detail. Suffice it to say that such limits behave in much the same way that ordinary limits do; in particular there is a direct analog of Theorem 3.9.

###### Example3.22. Limit at Infinity.

Compute $\ds\lim_{x\to \infty}{2x^2-3x+7\over x^2+47x+1}\text{.}$

Solution

As $x$ goes to infinity both the numerator and denominator go to infinity. We divide the numerator and denominator by $\ds x^2\text{:}$

\begin{equation*} \lim_{x\to \infty}{2x^2-3x+7\over x^2+47x+1}= \lim_{x\to \infty}{2-\ds{3\over x}+\ds{7\over x^2}\over 1+\ds{47\over x}+\ds{1\over x^2}} \end{equation*}

Now as $x$ approaches infinity, all the quotients with some power of $x$ in the denominator approach zero, leaving 2 in the numerator and 1 in the denominator, so the limit again is 2,

\begin{equation*} \lim\limits_{x\to\infty} {2x^2-3x+7\over x^2+47x+1} = 2\text{.} \end{equation*}

In the previous example, we divided by the highest power of $x$ that occurs in the denominator in order to evaluate the limit. We illustrate another technique similar to this.

###### Example3.23. More Limits at Infinity.

Compute the following limit:

\begin{equation*} \lim_{x\to\infty}\frac{2x^2+3}{5x^2+x}\text{.} \end{equation*}
Solution

As $x$ becomes large, both the numerator and denominator become large, so it isn't clear what happens to their ratio. The highest power of $x$ in the denominator is $x^2\text{,}$ therefore we will divide every term in both the numerator and denominator by $x^2$ as follows:

\begin{equation*} \lim_{x\to\infty}\frac{2x^2+3}{5x^2+x}=\lim_{x\to\infty}\frac{2+3/x^2}{5+1/x} \end{equation*}

We can also apply limit laws to infinite limits instead of arguing as we did in Example 3.20.

\begin{equation*} \lim_{x\to\infty}\frac{2+3/x^2}{5+1/x}=\frac{\ds{\lim_{x\to\infty}2+3\lim_{x\to\infty}\frac{1}{x^2}}}{\ds{\lim_{x\to\infty}5+\lim_{x\to\infty}\frac{1}{x}}}=\frac{2+3(0)}{5+0}=\frac{2}{5} \end{equation*}

Note that we used the theorem above to get that $\ds{\lim_{x\to\infty}\frac{1}{x}=0}$ and $\ds{\lim_{x\to\infty}\frac{1}{x^2}=0}\text{.}$

A shortcut technique is to analyze only the leading terms of the numerator and denominator. A leading term is a term that has the highest power of $x\text{.}$ If there are multiple terms with the same exponent, you must include all of them.

Top: The leading term is $2x^2\text{.}$

Bottom: The leading term is $5x^2\text{.}$

Now only looking at leading terms and ignoring the other terms we get:

\begin{equation*} \lim_{x\to\infty}\frac{2x^2+3}{5x^2+x}=\lim_{x\to\infty}\frac{2x^2}{5x^2}=\frac{2}{5} \end{equation*}
###### Example3.24. Application of Limits at Infinity.

A certain manufacturer makes a line of luxurious chairs. It is estimated that the total cost of making $x$ luxurious chairs is

\begin{equation*} C(x) = 350x + 200,000 \end{equation*}

dollars per year. Thus, the average cost of making $x$ chairs is given by

\begin{equation*} \overline{C(x)} = \dfrac{C(x)}{x} = \dfrac{350x + 200,000}{x} = 350 + \dfrac{200,000}{x} \end{equation*}

dollars per chair. Evaluate $\lim\limits_{x\to\infty} \overline{C(x)}$ and interpret your results.

Solution
\begin{equation*} \begin{split} \lim_{x\to\infty} \overline{C(x)} \amp = \lim_{x\to\infty} \left(350 + \dfrac{200,000}{x} \right) \\ \amp = \lim\limits_{x\to\infty} 350 + \lim\limits_{x\to\infty} \dfrac{200,000}{x} \\ \amp = 350 \end{split} \end{equation*}

A sketch of the graph of the function $\overline{C(x)}$ is shown below. The result we obtained is fully expected if we consider its economic implications. Note that as the level of production increases, the fixed cost per chair produced, represented by the term $\dfrac{200,000}{x}\text{,}$ drops steadily. The average cost should approach a constant unit cost of production — 350 in this case (see below). Interactive Demonstration. Use the slider below to investigate the limit $\ds\lim_{x\to\infty} \overline{C(x)} \text{,}$ where $\overline{C(x)} = 350 + \dfrac{200,000}{x}$ as found in Example 3.24. ### Subsection3.5.2Infinite Limits We next look at functions whose limit at $x = a$ does not exist, but whose values increase or decrease without bound as $x$ approaches $a$ from the left or right. ###### Definition3.25. Infinite Limit (Useable Definition). In general, we write \begin{equation*} \lim_{x\to a}f(x)=\infty \end{equation*} if we can make the value of $f(x)$ arbitrarily large by taking $x$ to be sufficiently close to $a$ (on either side of $a$) but not equal to $a\text{.}$ Similarly, we write \begin{equation*} \lim_{x\to a}f(x)=-\infty \end{equation*} if we can make the value of $f(x)$ arbitrarily large and \blue{negative} by taking $x$ to be sufficiently close to $a$ (on either side of $a$) but not equal to $a\text{.}$ Note: 1. We want to emphasize that by the proper definition of limits, the above limits do not exist, since they are not real numbers. However, writing $\pm \infty$ provides us with more information than simply writing DNE. 2. This definition can be modified for one-sided limits as well as limits with $x\to a$ replaced by $x\to\infty$ or $x\to-\infty\text{.}$ ###### Example3.26. Simple Infinite Limit. Compute the following limit: $\ds\lim\limits_{x\to 0} \frac{1}{x^2}\text{.}$ Solution We refer to the graph below. Let's first look at the limit as $x \to 0^+\text{,}$ and notice that $\frac{1}{x^2}$ increases without bound. Therefore, \begin{equation*} \lim_{x\to 0^+} \frac{1}{x^2} = +\infty\text{.} \end{equation*} As $x \to 0^-\text{,}$ we again see that $\frac{1}{x^2}$ increases without bound: \begin{equation*} \lim_{x\to 0^-} \frac{1}{x^2} = +\infty\text{.} \end{equation*} We conclude that \begin{equation*} \lim_{x\to 0} \frac{1}{x^2} = \infty\text{.} \end{equation*} ###### Example3.27. Infinite Limit. Compute the following limit: $\ds\lim_{x\to\infty}(x^3-x)\text{.}$ Solution One might be tempted to write: \begin{equation*} \lim_{x\to\infty}x^3-\lim_{x\to\infty}x=\infty-\infty\text{,} \end{equation*} however, we do not know what $\infty-\infty$ is, as $\infty$ is not a real number and so cannot be treated like one. Incidentally, the expression $\infty - \infty$ is another indeterminate form. We instead write: \begin{equation*} \lim_{x\to\infty}(x^3-x)=\lim_{x\to\infty}x(x^2-1)\text{.} \end{equation*} As $x$ becomes arbitrarily large, then both $x$ and $x^2-1$ become arbitrarily large, and hence their product $x(x^2-1)$ will also become arbitrarily large. Thus we see that \begin{equation*} \lim_{x\to\infty}(x^3-x)=\infty\text{.} \end{equation*} ###### Example3.28. More Infinite Limit. Let \begin{equation*} f(x) = \dfrac{5x^{3}-3x^{2}+1}{x^{2}+2x+4}\text{.} \end{equation*} Evaluate 1. $\lim\limits_{x\to\infty} f(x)$ 2. $\lim\limits_{x\to -\infty} f(x)$ Solution 1. Dividing the numerator and the denominator of the rational expression by $x^{2}\text{,}$ we obtain \begin{equation*} \lim_{x\to\infty} \dfrac{5x^{3}-3x^{2}+1}{x^{2}+2x+4} = \lim_{x\to\infty} \dfrac{5x-3+\frac{1}{x^{2}}}{1+\frac{2}{x}+\frac{4}{x^{2}}} \end{equation*} Since the numerator becomes arbitrarily large whereas the denominator approaches $1$ as $x$ tends to infinity, we see that the quotient $f(x)$ gets larger and larger as $x$ approaches infinity. In other words, the limit does not exist. We indicate this by writing \begin{equation*} \lim_{x\to\infty} \dfrac{5x^{3}-3x^{2}+1}{x^{2}+2x+4} = \infty\text{.} \end{equation*} 2. Once again, dividing both the numerator and the denominator by $x^{2}\text{,}$ we obtain \begin{equation*} \lim_{x\to -\infty} \dfrac{5x^{3}-3x^{2}+1}{x^{2}+2x+4} = \lim_{x\to -\infty} \dfrac{5x-3+\frac{1}{x^{2}}}{1+\frac{2}{x}+\frac{4}{x^{2}}} \end{equation*} In this case, the numerator becomes arbitrarily large in magnitude, bu negative in sign, whereas the denominator approaches $1$ as $x$ appraoches negative infinity. Therefore, the quotient $f(x)$ decreases without bound, and the limit does not exists. We indicate this by writing \begin{equation*} \lim_{x\to -\infty} \dfrac{5x^{3}-3x^{2}+1}{x^{2}+2x+4} = -\infty\text{.} \end{equation*} ###### Example3.29. Limit at Infinity, Infinite Limit and Basic Functions. Find the following limits by observing the behaviour of the graph of each function. 1. $\ds\lim_{x\to \infty} \frac{6}{\sqrt{x^3}}$ 2. $\ds\lim_{x\to -\infty} \left(x-x^2\right)$ 3. $\ds\lim_{x\to \infty}\left(x^3+x\right)$ 4. $\ds\lim_{x\to \infty} \cos(x)$ 5. $\ds\lim_{x\to \infty} e^x$ 6. $\ds\lim_{x\to -\infty} e^x$ 7. $\ds\lim_{x\to 0^+} \ln x$ 8. $\ds\lim_{x\to 0} \cos(1/x)$ Solution We can easily evaluate the following limits by observation:        Note: Often, the shorthand notation $\ds\frac{1}{0^+}=+\infty$ and $\ds\frac{1}{0^-}=-\infty$ is used to represent the following two limits respectively: \begin{equation*} \lim_{x\to 0^+}\frac{1}{x}=+\infty, \text{ and} \end{equation*} \begin{equation*} \lim_{x\to 0^-}\frac{1}{x}=-\infty\text{.} \end{equation*} Making use of the results from Example 3.29 we can compute the following limits. ###### Example3.30. More Limit at Infinity, Infinite Limit and Basic Functions. Compute $\ds{\lim_{x\to 0^+} e^{1/x}}\text{,}$ $\ds{\lim_{x\to 0^-} e^{1/x}}$ and $\ds{\lim_{x\to 0} e^{1/x}}\text{.}$ Solution We have: \begin{equation*} \lim_{x\to 0^+} e^{\frac{1}{x}}=e^{\frac{1}{0^+}}=e^{+\infty}=\infty\text{.} \end{equation*} \begin{equation*} \lim_{x\to 0^-} e^{\frac{1}{x}}=e^{\frac{1}{0^-}}=e^{-\infty}=0\text{.} \end{equation*} Thus, as left-hand limit $\neq$ right-hand limit, \begin{equation*} \lim_{x\to 0} e^{\frac{1}{x}}=\mbox{DNE}\text{.} \end{equation*} ### Subsection3.5.3Vertical Asymptotes ###### Definition3.31. Vertical Asymptote. The line $x=a$ is called a vertical asymptote of $f(x)$ if at least one of the following is true:      Interactive Demonstration. Use the slider below to investigate the limit $\ds \lim_{x\to 10^{-}} f(x) \text{.}$ ###### Example3.32. Vertical Asymptotes. Find the vertical asymptotes of $\ds f(x)=\frac{2x}{x-4}\text{.}$ Solution In the definition of vertical asymptotes we need a certain limit to be $\pm\infty\text{.}$ Candidates would be to consider values not in the domain of $f(x)\text{,}$ such as $a=4\text{.}$ As $x$ approaches $4$ but is larger than $4$ then $x-4$ is a small positive number and $2x$ is close to $8\text{,}$ so the quotient $2x/(x-4)$ is a large positive number. Thus we see that \begin{equation*} \lim_{x\to 4^+}\frac{2x}{x-4}=\infty\text{.} \end{equation*} Thus, at least one of the conditions in the definition above is satisfied. Therefore $x=4$ is a vertical asymptote, as shown below. ### Subsection3.5.4Horizontal Asymptotes ###### Definition3.33. Horizontal Asymptote. The line $y=L$ is a horizontal asymptote of $f(x)$ if either \begin{equation*} \lim_{x\to\infty}f(x)=L\qquad\mbox{or} \qquad\lim_{x\to-\infty}f(x)=L\text{.} \end{equation*} ###### Example3.34. Horizontal Asymptotes. Find the horizontal asymptotes of $\ds f(x)=\frac{|x|}{x}\text{.}$ Solution We must compute two infinite limits. First, \begin{equation*} \lim_{x\to\infty}\frac{|x|}{x}\text{.} \end{equation*} Notice that for $x$ arbitrarily large that $x>0\text{,}$ so that $|x|=x\text{.}$ In particular, for $x$ in the interval $(0,\infty)$ we have \begin{equation*} \lim_{x\to\infty}\frac{|x|}{x}=\lim_{x\to\infty}\frac{x}{x}=1\text{.} \end{equation*} Second, we must compute \begin{equation*} \lim_{x\to-\infty}\frac{|x|}{x}\text{.} \end{equation*} Notice that for $x$ arbitrarily large negative that $x\lt 0\text{,}$ so that $|x|=-x\text{.}$ In particular, for $x$ in the interval $(-\infty,0)$ we have \begin{equation*} \lim_{x\to -\infty}\frac{|x|}{x}=\lim_{x\to -\infty}\frac{-x}{x}=-1\text{.} \end{equation*} Therefore there are two horizontal asymptotes, namely, $y=1$ and $y=-1\text{,}$ as shown to the right.  ### Subsection3.5.5Slant Asymptotes Some functions may have slant (or oblique) asymptotes, which are neither vertical nor horizontal. ###### Definition3.35. Slant Asymptote. The line $y=mx+b$ is a slant asymptote of $f(x)$ if either \begin{equation*} \ds\lim_{x\to\infty}\left[f(x)-(mx+b)\right]=0 \ \ \ \text{ or } \end{equation*} \begin{equation*} \ds\lim_{x\to -\infty}\left[f(x)-(mx+b)\right]=0\text{,} \end{equation*} as shown in Figure 3.1. Visually, the vertical distance between $f(x)$ and $y=mx+b$ is decreasing towards 0 and the curves do not intersect or cross at any point as $x$ approaches infinity. ###### Example3.36. Slant Asymptote in a Rational Function. Find the slant asymptotes of $\ds f(x)=\frac{-3x^2+4}{x-1}\text{.}$ Solution Note that this function has no horizontal asymptotes since $f(x)\to-\infty$ as $x\to\infty$ and $f(x)\to\infty$ as $x\to-\infty\text{.}$ In rational functions, slant asymptotes occur when the degree in the numerator is one greater than in the denominator. We use long division to rearrange the function: \begin{equation*} \ds\frac{-3x^2+4}{x-1}=-3x-3+\frac{1}{x-1}\text{.} \end{equation*} The part we're interested in is the resulting polynomial $-3x-3\text{.}$ This is the line $y=mx+b$ we were seeking, where $m=-3$ and $b=-3\text{.}$ Notice that \begin{equation*} \ds\lim_{x\to\infty}\frac{-3x^2+4}{x-1}-(-3x-3)=\lim_{x\to\infty}\frac{1}{x-1}=0 \end{equation*} and \begin{equation*} \ds\lim_{x\to-\infty}\frac{-3x^2+4}{x-1}-(-3x-3)=\lim_{x\to-\infty}\frac{1}{x-1}=0\text{.} \end{equation*} Thus, $y=-3x-3$ is a slant asymptote of $f(x)\text{,}$ as shown in Figure 3.1 below. Figure 3.1. $y=f(x)=\frac{-3x^2+4}{x-1}$ has slant asymptote $y=-3x-3\text{.}$ Although rational functions are the most common type of function we encounter with slant asymptotes, there are other types of functions we can consider that present an interesting challenge. ###### Example3.37. Slant Asymptote. Show that $y=2x+4$ is a slant asymptote of $f(x)=2x-3^x+4\text{.}$ Solution This is because \begin{equation*} \lim_{x\to -\infty}[f(x)-(2x+4)]=\lim_{x\to -\infty}(-3^x)=0\text{.} \end{equation*} We note that $\lim_{x\to \infty}[f(x)-(2x+4)]=\lim_{x\to \infty}(-3^x)=-\infty\text{.}$ So, the vertical distance between $y=f(x)$ and the line $y=2x+4$ decreases toward 0 only when $x\to -\infty$ and not when $x\to \infty\text{.}$ The graph of $f$ approaches the slant asymptote $y=2x+4$ only at the far left and not at the far right. One might ask if $y=f(x)$ approaches a slant asymptote when $x\to \infty\text{.}$ The answer turns out to be no, but we will need to know something about the relative growth rates of the exponential functions and linear functions in order to prove this. Specifically, one can prove that when the base is greater than 1 the exponential functions grows faster than any power function as $x\to \infty\text{.}$ This can be phrased like this: For any $a>1$ and any $n>0\text{,}$ \begin{equation*} \lim_{x\to \infty}\frac{a^x}{x^n}=\infty \text{ and } \lim_{x\to \infty}\frac{x^n}{a^x}=0\text{.} \end{equation*} These facts are most easily proved with the aim of something called the L'Hôpital's Rule. ### Subsection3.5.6End Behaviour and Comparative Growth Rates Let us now look at the last two subsections and go deeper. In the last two subsections we looked at horizontal and slant asymptotes. Both are special cases of the end behaviour of functions, and both concern situations where the graph of a function approaches a straight line as $x\to \infty$ or $-\infty\text{.}$ But not all functions have this kind of end behaviour. For example, $f(x)=x^2$ and $f(x)=x^3$ do not approach a straight line as $x\to \infty$ or $-\infty\text{.}$ The best we can say with the notion of limit developed at this stage are that \begin{align*} \lim\limits_{x\to \infty}x^2 =\infty, \amp \lim\limits_{x\to -\infty}x^2=\infty,\\ \lim\limits_{x\to \infty}x^3 =\infty, \amp \lim\limits_{x\to -\infty}x^3=-\infty\text{.} \end{align*} Similarly, we can describe the end behaviour of transcendental functions such as $f(x)=e^x$ using limits, and in this case, the graph approaches a line as $x\to -\infty$ but not as $x\to \infty\text{.}$ \begin{equation*} \lim_{x\to -\infty}e^x=0,\lim_{x\to \infty}e^x=\infty\text{.} \end{equation*} People have found it useful to make a finer distinction between these end behaviours all thus far captured by the symbols $\infty$ and $-\infty\text{.}$ Specifically, we will see that the above functions have different growth rates at infinity. Some increases to infinity faster than others. Specifically, ###### Definition3.38. Comparative Growth Rates. Suppose that $f$ and $g$ are two functions such that $\lim\limits_{x\to \infty }f\left( x\right) =\infty$ and $\lim\limits_{x\to \infty }g\left( x\right) =\infty \text{.}$ We say that $f\left( x\right)$ grows faster than $g\left( x\right)$ as $x\to \infty$ if the following holds: \begin{equation*} \lim_{x\to \infty }\frac{f\left( x\right) }{g\left( x\right) }=\infty\text{,} \end{equation*} or equivalently, \begin{equation*} \lim_{x\to \infty }\frac{g\left( x\right) }{f\left( x\right) }=0\text{.} \end{equation*} Here are a few obvious examples: ###### Example3.39. Growth Rate in Polynomial Functions. Show that if $m>n$ are two positive integers, then $f(x)=x^m$ grows faster than $g(x)=x^n$ as $x\to \infty\text{.}$ Solution Since $m>n\text{,}$ $m-n$ is a positive integer. Therefore, \begin{equation*} \lim_{x\to \infty}\frac{f(x)}{g(x)}=\lim_{x\to \infty}\frac{x^m}{x^n}=\lim_{x\to \infty}x^{m-n}=\infty\text{.} \end{equation*} ###### Example3.40. Growth Rate in Polynomial Functions. Show that if $m>n$ are two positive integers, then any monic polynomial $P_{m}(x)$ of degree $m$ grows faster than any monic polynomial $P_{n}(x)$ of degree $n$ as $x\to \infty\text{.}$ [Recall that a polynomial is monic if its leading coefficient is 1.] Solution By assumption, $P_{m}(x)=x^m+$ terms of degrees less than $m=x^{m}+a_{m-1}x^{m-1}+\ldots\text{,}$ and $P_{n}(x)=x^n+$ terms of degrees less than $n=x^{n}+b_{n-1}x^{n-1}+\ldots\text{.}$ Dividing the numerator and denominator by $x^n\text{,}$ we get \begin{equation*} \begin{split} \lim_{x\to \infty}\frac{f(x)}{g(x)}=\amp \lim_{x\to \infty}\frac{x^{m-n}+a_{m-1}x^{m-n-1}+\ldots}{1+\frac{b_{n-1}}{x}+\ldots}\\ =\amp \lim_{x\to \infty}x^{m-n}\left(\frac{1+\frac{a_{m-1}}{x}+\ldots}{1+\frac{b_{n-1}}{x}+\ldots}\right)\\ \amp=\infty \end{split}\text{,} \end{equation*} since the limit of the bracketed fraction is 1 and the limit of $x^{m-n}$ is $\infty\text{,}$ as we showed in Example 3.39. ###### Example3.41. Growth Rate in Polynomial Functions. Show that a polynomial grows exactly as fast as its highest degree term as $x\to \infty$ or $-\infty\text{.}$ That is, if $P(x)$ is any polynomial and $Q(x)$ is its highest degree term, then both limits \begin{equation*} \lim_{x\to \infty}\frac{P(x)}{Q(x)}\text{ and } \lim_{x\to -\infty}\frac{P(x)}{Q(x)} \end{equation*} are finite and nonzero. Solution Suppose that $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\ldots+a_{1}x+a_{0}\text{,}$ where $a_{n}\neq 0\text{.}$ Then the highest degree term is $Q\left( x\right) =a_{n}x^{n}\text{.}$ So, \begin{equation*} \lim_{x\to \infty }\frac{P\left( x\right) }{Q\left( x\right) }=\lim_{x\to \infty }\left( a_{n}+\frac{a_{n-1}}{x}+...+\frac{a_{1}}{ x^{n-1}}+\frac{a_{0}}{x^{n}}\right) =a_{n}\neq 0\text{.} \end{equation*} Let's state a theorem we mentioned when we discussed the last example in the last subsection: The easiest way to prove this is to use the L'Hôpital's Rule, which we will introduce in a later chapter. For now, one can plot and compare the graphs of an exponential function and a power function. Here is a comparison between $f(x)=x^2$ and $g(x)=2^x\text{:}$ Notice also that as $x\to -\infty\text{,}$ $x^n$ grows in size but $e^x$ does not. More specifically, $x^n\to \infty$ or $-\infty$ according as $n$ is even or odd, while $e^x\to 0\text{.}$ So, it is meaningless to compare their “growth” rates, although we can still calculate the limit \begin{equation*} \lim_{x\to -\infty}\frac{e^x}{x^n}=0\text{.} \end{equation*} Let's see an application of our theorem. ###### Example3.43. Horizontal Asymptotes. Find the horizontal asymptote(s) of $f(x)=\dfrac{x^3+2e^x}{e^x-4x^2}\text{.}$ Solution To find horizontal asymptotes, we calculate the limits of $f(x)$ as $x\to \infty$ and $x\to -\infty\text{.}$ For $x\to \infty\text{,}$ we divide the numerator and the denominator by $e^x\text{,}$ and then we take limit to get \begin{equation*} \lim_{x\to \infty}\dfrac{x^3+2e^x}{e^x-4x^2}=\lim_{x\to \infty}\dfrac{\frac{x^3}{e^x}+2}{1-4\frac{x^2}{e^x}}=\frac{0+2}{1-4(0)}=2\text{.} \end{equation*} For $x\to -\infty\text{,}$ we divide the numerator and the denominator by $x^2$ to get \begin{equation*} \lim_{x\to -\infty}\dfrac{x^3+2e^x}{e^x-4x^2}=\lim_{x\to -\infty}\dfrac{x+2\frac{e^x}{x^2}}{\frac{e^x}{x^2}-4}\text{.} \end{equation*} The denominator now approaches $0-4=-4\text{.}$ The numerator has limit $-\infty\text{.}$ So, the quotient has limit $\infty\text{:}$ \begin{equation*} \lim_{x\to -\infty}\dfrac{x+2\frac{e^x}{x^2}}{\frac{e^x}{x^2}-4}=\infty\text{.} \end{equation*} So, $y=2$ is a horizontal asymptote. The function $y=f(x)$ approaches the line $y=2$ as $x\to \infty\text{.}$ And this is the only horizontal asymptote, since the function $y=f(x)$ does not approach any horizontal line as $x\to -\infty\text{.}$ Since the growth rate of a polynomial is the same as that of its leading term, the following is obvious: ###### Example3.44. Growth Rate in Polynomial and Exponential Functions. If $P(x)$ is any polynomial, then \begin{equation*} \lim_{x\to \infty}\frac{P(x)}{e^x}=0\text{.} \end{equation*} Also, if $r$ is any real number, then we can place it between two consecutive integers $n$ and $n+1\text{.}$ For example, $\sqrt{3}$ is between 1 and 2, $e$ is between 2 and 3, and $\pi$ is between 3 and 4. Then the following is totally within our expectation: ###### Example3.45. Growth Rate in Exponential Functions. Prove that if $a>1$ is any basis and $r>0$ is any exponent, then $f(x)=a^x$ grows faster than $g(x)=x^r$ as $x\to \infty\text{.}$ Solution Let $r$ be between consecutive integers $n$ and $n+1\text{.}$ Then for all $x>1\text{,}$ $x^{n}\leq x^{r}\leq x^{n+1}\text{.}$ Dividing by $a^{x}\text{,}$ we get \begin{equation*} \frac{x^n}{a^x}\leq \frac{x^r}{a^x}\leq \frac{x^{n+1}}{a^x}\text{.} \end{equation*} Since \begin{equation*} \lim_{x\to \infty}\frac{x^n}{a^x}=0\text{.} \end{equation*} What about exponential functions with different bases? We recall from the graphs of the exponential functions that for any base $a>1\text{,}$ \begin{equation*} \lim_{x\to \infty}a^x=\infty\text{.} \end{equation*} So, the exponential functions with bases greater than 1 all grow to infinity as $x\to \infty\text{.}$ How do their growth rates compare? Since $a\lt b\text{,}$ we have $\frac{b}{a}>1\text{.}$ So, \begin{equation*} \lim_{x\to \infty}\frac{b^x}{a^x}=\lim_{x\to \infty}\left(\frac{b}{a}\right)^x=\infty\text{.} \end{equation*} We further have that $\frac{a}{b} \lt 1\text{.}$ So, \begin{equation*} \lim_{x\to \infty}\frac{a^x}{b^x}=\lim_{x\to \infty}\left(\frac{a}{b}\right)^x=0\text{.} \end{equation*} Another function that grows to infinity as $x\to \infty$ is $g(x)=\ln x\text{.}$ Recall that the natural logarithmic function is the inverse of the exponential function $y=e^x\text{.}$ Since $e^x$ grows very fast as $x$ increases, we should expect $\ln x$ to grow very slowly as $x$ increases. The same applies to logarithmic functions with any basis $a>1\text{.}$ This is the content of the next theorem. 1. We use a change of variable. Letting $t=\ln x\text{,}$ then $x=e^t\text{.}$ So, $x\to \infty$ if and only if $t\to \infty\text{,}$ and \begin{equation*} \lim_{x\to \infty}\frac{\ln x}{x^r}=\lim_{t\to \infty}\frac{t}{(e^t)^r}=\lim_{t\to \infty}\frac{t}{(e^r)^t}\text{.} \end{equation*} Now, since $r>0\text{,}$ $a=e^r>1\text{.}$ So, $a^t$ grows as $t$ increases, and it grows faster than $t$ as $t\to \infty\text{.}$ Therefore, \begin{equation*} \lim_{x\to \infty}\frac{\ln x}{x^{r}}=\lim_{t\to \infty}\frac{t}{(e^r)^t}=\lim_{t\to \infty}\frac{t}{a^t}=0\text{.} \end{equation*} 2. The change of base identity $\log_{a}x=\dfrac{\ln x}{\ln a}$ implies that $\log_{a}x$ is simply a constant multple of $\ln x\text{.}$ The result now follows from (a). ##### Exercises for Section 3.5. Compute the following limits. 1. $\ds\lim_{x\to \infty} \left(\sqrt{x^2+x}-\sqrt{x^2-x}\right)$ Answer $1$ Solution We note that this is of the indeterminate form $\infty \pm \infty\text{.}$ Let's try to simplify the expression by multiplying by \begin{equation*} \dfrac{\sqrt{x^{2}+x} + \sqrt{x^{2} - x}}{\sqrt{x^{2}+x} + \sqrt{x^{2} - x}}\text{.} \end{equation*} We get that \begin{equation*} \begin{split} \lim\limits_{x \to \infty} \left(\sqrt{x^{2}+x} - \sqrt{x^{2} - x}\right)\amp = \lim\limits_{x \to \infty} \left(\sqrt{x^{2}+x} - \sqrt{x^{2} - x}\right)\left(\dfrac{\sqrt{x^{2}+x} + \sqrt{x^{2} - x}}{\sqrt{x^{2}+x} + \sqrt{x^{2} - x}}\right)\\ \amp = \lim\limits_{x \to \infty} \dfrac{2x}{\sqrt{x^{2}+x} + \sqrt{x^{2} - x}} \\ \amp = \lim\limits_{x \to \infty} \dfrac{2}{\sqrt{1+\frac{1}{x}} + \sqrt{1 - \frac{1}{x}}} \\ \amp = \dfrac{2}{1+1} \\ \amp = 1. \end{split} \end{equation*} 2. $\ds\lim_{x\to\infty} {e^x + e^{-x}\over e^x -e^{-x}}$ Answer $1$ Solution Let $w = e^{x}\text{.}$ Then as $x \to \infty\text{,}$ $w \to \infty\text{.}$ The limit becomes \begin{equation*} \begin{split} \lim_{w\to\infty} \dfrac{w + \frac{1}{w}}{w - \frac{1}{w}} \amp = \lim_{w\to\infty} \dfrac{1 + \frac{1}{w^{2}}}{1 - \frac{1}{w^{2}}} \\ \amp = 1. \end{split} \end{equation*} 3. $\ds\lim_{t\to1^+}{(1/t)-1\over t^2-2t+1}$ Answer $-\infty$ Solution First, we factorize the denominator to get \begin{equation*} \lim\limits_{t\to 1^{+}} \dfrac{\frac{1}{t} - 1}{t^{2}-2t+1} = \lim\limits_{t\to 1^{+}} \dfrac{\frac{1}{t} - 1}{(t-1)^{2}}\text{.} \end{equation*} It is now clear that this limit is of the indeterminate form $\frac{0}{0}\text{.}$ Now we can multiply by $\frac{t}{t}$ obtaining \begin{equation*} \begin{split} \lim\limits_{t\to 1^{+}} \dfrac{\frac{1}{t} - 1}{(t-1)^{2}} \amp = \lim\limits_{t\to 1^{+}} \dfrac{1 - t}{t(t-1)^{2}} \\ \amp = \lim\limits_{t\to 1^{+}} \dfrac{-1}{t(t-1)}\\ \amp = -\infty. \end{split} \end{equation*} 4. $\ds\lim_{t\to\infty}{t+5-2/t-1/t^3\over 3t+12-1/t^2}$ Answer $1/3$ Solution \begin{equation*} \lim\limits_{t \to \infty} \dfrac{t + 5 - \frac{2}{t} - \frac{1}{t^3}}{3t + 12 - \frac{1}{t^2}} = \lim\limits_{t \to \infty} \dfrac{1 + \frac{5}{t} - \frac{2}{t^2} - \frac{1}{t^4}}{3 + \frac{12}{t} - \frac{1}{t^3}} = \dfrac{1}{3}\text{.} \end{equation*} 5. $\ds\lim_{y\to\infty}{\sqrt{y+1}+\sqrt{y-1}\over y}$ Answer $0$ Solution $\lim\limits_{y\to \infty} \dfrac{\sqrt{y+1}+\sqrt{y-1}}{y} = \lim\limits_{y\to \infty} \sqrt{\dfrac{1}{y}+\dfrac{1}{y^2}} + \sqrt{\dfrac{1}{y} - \dfrac{1}{y^2}} = 0\text{.}$ 6. $\ds\lim_{x\to 0^+}{3+x^{-1/2}+x^{-1}\over 2+4x^{-1/2}}$ Answer $\infty$ Solution $\lim\limits_{x \to 0^+} \dfrac{3+\frac{1}{\sqrt{x}} + \frac{1}{\sqrt{x}}}{2+\frac{4}{\sqrt{x}}} = \lim\limits_{x \to 0^+} \dfrac{3x+ \sqrt{x} + 1}{2x+4\sqrt{x}} = \lim\limits_{x \to 0^+} \dfrac{1}{2x+4\sqrt{x}} = \infty\text{.}$ 7. $\ds\lim_{x\to\infty}{x+x^{1/2}+x^{1/3}\over x^{2/3}+x^{1/4}}$ Answer $\infty$ Solution $\lim\limits_{x \to \infty} \dfrac{x + x^{1/2} + x^{1/3}}{x^{2/3} + x^{1/4}} = \lim\limits_{x \to \infty} \dfrac{1 + x^{-1/2} + x^{-2/3}}{x^{-1/3} + x^{-3/4}} = \lim\limits_{x \to \infty} \dfrac{1}{x^{-1/3} + x^{-3/4}} = \infty\text{.}$ 8. $\ds\lim_{t\to\infty}{1-\sqrt{t\over t+1}\over 2-\sqrt{4t+1\over t+2}}$ Answer $2/7$ Solution $\begin{array}{rl} \lim\limits_{t \to \infty} \dfrac{1 - \sqrt{\frac{t}{t+1}}}{2-\sqrt{\frac{4t+1}{t+2}}} \amp = \lim\limits_{t \to \infty} \dfrac{1 - \sqrt{\frac{t}{t+1}}}{2-\sqrt{\frac{4t+1}{t+2}}} \cdot \dfrac{2+ \sqrt{\frac{4t+1}{t+2}}}{2+\sqrt{\frac{4t+1}{t+2}}} \\ \amp = \lim\limits_{t \to \infty} \dfrac{1}{7} \cdot (t+2)\left( 1- \sqrt{\frac{t}{t+1}}\right) \left(2+\sqrt{\frac{4t+1}{t+2}}\right) \\ \amp = \dfrac{1}{7} \lim\limits_{t \to \infty} (t+2) \left(2+\sqrt{\frac{4t+1}{t+2}}\right) \left( 1- \sqrt{\frac{t}{t+1}}\right) \cdot \dfrac{\left( 1+ \sqrt{\frac{t}{t+1}}\right)}{\left( 1+ \sqrt{\frac{t}{t+1}}\right)} \\ \amp = \dfrac{1}{7} \lim\limits_{t \to \infty} \dfrac{t+2}{t+1} \cdot \lim\limits_{t \to \infty} \dfrac{\left(2+\sqrt{\frac{4t+1}{t+2}}\right)}{\left( 1+ \sqrt{\frac{t}{t+1}}\right)} \\ \amp = \dfrac{1}{7} \cdot 1 \cdot \lim\limits_{t \to \infty} \dfrac{2 + 2 \sqrt{\frac{t+\frac{1}{4}}{t+2}}}{1 + \sqrt{\frac{t}{t+1}}} \\ \amp = \dfrac{1}{7} \cdot 1 \cdot \dfrac{4}{2} \\ \amp = \dfrac{2}{7}. \end{array}$ 9. $\ds\lim_{t\to\infty}{1-{t\over t-1}\over 1-\sqrt{t\over t-1}}$ Answer $2$ Solution $\begin{array}{rl} \lim\limits_{t \to \infty} \dfrac{1 - \frac{t}{t-1}}{1 - \sqrt{\frac{t}{t-1}}} \amp = \lim\limits_{t \to \infty} \dfrac{1 - \frac{t}{t-1}}{1 - \sqrt{\frac{t}{t-1}}} \cdot \dfrac{1 + \sqrt{\frac{t}{t-1}}}{1 + \sqrt{\frac{t}{t-1}}}\\ \amp = \lim\limits_{t \to \infty} -(t-1)\left( 1-\frac{t}{t-1}\right) \left( 1+ \sqrt{\frac{t}{t-1}}\right) \\ \amp = 1 +1 = 2. \end{array}$ 10. $\ds\lim_{x\to-\infty}{x+x^{-1}\over 1+\sqrt{1-x}}$ Answer $-\infty$ Solution $\begin{array}{rl} \lim\limits_{x \to -\infty} \dfrac{1 + \frac{1}{x}}{1 + \sqrt{1-x}} \amp = \lim\limits_{x \to -\infty} \dfrac{(x+\frac{1}{x})(1-\sqrt{1-x})}{x} \\ \amp = \lim\limits_{x \to -\infty} \left(1 + \frac{1}{x^2} - \frac{\sqrt{1-x}}{x^2}-\sqrt{1-x}\right) \\ \amp = \lim\limits_{x \to -\infty} (-\sqrt{1-x}) \\ \amp = -\infty. \end{array}$ 11. $\ds\lim_{x\to1^+}{\sqrt{x}\over x-1}$ Answer $\infty$ Solution $\lim\limits_{x \to 1^+} \dfrac{\sqrt{x}}{1-x} = \lim\limits_{x \to 1^+} \dfrac{\sqrt{\frac{1}{x}}}{\frac{1}{x}-1} = -\infty\text{.}$ 12. $\ds\lim_{x\to\infty}{x^{-1}+x^{-1/2}\over x+x^{-1/2}}$ Answer $0$ Solution $\lim\limits_{x \to \infty} \dfrac{\frac{1}{x} + \frac{1}{\sqrt{x}}}{x+\frac{1}{\sqrt{x}}} = \lim\limits_{x \to \infty} \left(\dfrac{1}{x^2 + \sqrt{x}}\right) + \lim\limits_{x \to \infty} \left(\dfrac{1}{x\sqrt{x}+1}\right) = 0+0 = 0\text{.}$ 13. $\ds\lim_{x\to\infty}{x+x^{-2}\over 2x+x^{-2}}$ Answer $1/2$ Solution $\lim\limits_{x \to \infty} \dfrac{x + \frac{1}{x^2}}{2x+\frac{1}{x^2}} = \lim\limits_{x \to \infty} \dfrac{1 + \frac{1}{x^3}}{2+\frac{1}{x^3}} = \dfrac{1}{2}\text{.}$ 14. $\ds\lim_{x\to\infty}{5+x^{-1}\over 1+2x^{-1}}$ Answer $5$ Solution $\lim\limits_{x \to \infty} \dfrac{5 + \frac{1}{x}}{1+\frac{2}{x}} = \dfrac{5}{1} = 5\text{.}$ 15. $\ds\lim_{x\to\infty}{4x\over\sqrt{2x^2+1}}$ Answer $\ds 2\sqrt2$ Solution $\lim\limits_{x \to \infty} \dfrac{4x}{\sqrt{2x^2+1}} = \lim\limits_{x \to \infty} \dfrac{\frac{4x}{x}}{\frac{\sqrt{2x^2+1}}{\sqrt{x^2}}} = \lim\limits_{x \to \infty} \dfrac{4}{\sqrt{2+\frac{1}{x^2}}} = \dfrac{4}{\sqrt{2}}\text{.}$ 16. $\ds\lim_{x\to\infty}{(x+5)\left({1\over 2x}+{1\over x+2}\right)}$ Answer $3/2$ Solution \begin{equation*} \begin{split}\lim\limits_{x \to \infty} (x+5)\left(\dfrac{1}{2x}+\dfrac{1}{x+2}\right) \amp= \lim\limits_{x \to \infty} \dfrac{(x+5)}{2x} + \lim\limits_{x \to \infty} \dfrac{(x+5)}{(x+2)}\\ \amp = \lim\limits_{x \to \infty} \dfrac{1+\frac{5}{x}}{2} + \lim\limits_{x \to \infty} \dfrac{1+\frac{5}{x}}{1+\frac{2}{x}}\\ \amp = \dfrac{1}{2} + 1 = \dfrac{3}{2}.\end{split} \end{equation*} 17. $\ds\lim_{x\to0^+}{(x+5)\left({1\over 2x}+{1\over x+2}\right)}$ Answer $\infty$ Solution \begin{equation*} \begin{split}\lim\limits_{x \to 0^+} (x+5) \left(\dfrac{1}{2x} + \dfrac{1}{x+2}\right) \amp= \lim\limits_{x \to 0^+}\dfrac{(x+5)}{2x} + \lim\limits_{x \to 0^+} \dfrac{(x+5)}{(x+2)}\\ \amp = \lim\limits_{x \to 0^+} \dfrac{1+\frac{5}{x}}{2} + \dfrac{5}{2} = +\infty.\end{split} \end{equation*} 18. $\ds\lim_{x\to2}{x^3-6x-2\over x^3-4x}$ Answer DNE Solution $\lim\limits_{x \to 2} \dfrac{x^3-6x-2}{x^3-4x} = \dfrac{x^3-6x-2}{x(x+2)(x-2)}\text{.}$ Since $\lim\limits_{x \to 2} (x^3-6x-2)=-6$ and $\lim\limits_{x \to 2^+} \dfrac{1}{x(x+2)(x-2)} = \infty\text{,}$ $\lim\limits_{x \to 2^-} \dfrac{1}{x(x+2)(x-2)} = -\infty\text{,}$ we see that $\lim\limits_{x \to 2} \dfrac{x^3-6x-2}{x^3-4x}$ DNE. 19. $\lim\limits_{x \to -\infty} \dfrac{3x^{3}+x^{2}+1}{x^{3}+1}$ Answer $3$ Solution $\lim\limits_{x \to -\infty} \dfrac{3x^3+x^2+1}{x^3+1} = \lim\limits_{x \to -\infty} \dfrac{3+\frac{1}{x} + \frac{1}{x^3}}{1+\frac{1}{x^3}} = 3\text{.}$ 20. $\lim\limits_{x \to -\infty} \dfrac{x^{4}+1}{x^{3}-1}$ Answer $-\infty$ Solution $\lim\limits_{x \to -\infty} \dfrac{x^4+1}{x^3-1} = \lim\limits_{x \to -\infty} \dfrac{x+\frac{1}{x^3}}{1-\frac{1}{x^3}} = -\infty\text{.}$ 21. $\lim\limits_{x \to \infty} \dfrac{x^{5}-x^{3}+x-1}{x^{6} + 2x^{2} + 1}$ Answer $0$ Solution \begin{equation*} \begin{split}\lim\limits_{x \to \infty} \dfrac{x^5-x^3+x-1}{x^6+2x^2+1} \amp= \lim\limits_{x \to \infty} \dfrac{\frac{1}{x} - \frac{1}{x^3}+\frac{1}{x^5}-\frac{1}{x^6}}{1+\frac{2}{x^4}+\frac{1}{x^6}}\\ \amp = \dfrac{0}{1} = 0.\end{split} \end{equation*} The function $\ds f(x) = {x\over\sqrt{x^2+1}}$ has two horizontal asymptotes. Find them and give a rough sketch of $f$ with its horizontal asymptotes. Answer $y=1$ and $y=-1$ Solution To find the horizontal asymptotes of $f\text{,}$ we compute the limits \begin{equation*} \begin{split} \lim_{x \to\infty} f(x) \amp = \lim_{x \to\infty} \frac{x}{\sqrt{x^2+1}} \\ \amp = \lim_{x \to\infty}\frac{x/x}{\sqrt{\frac{x^2+1}{x^2}}} \\ \amp = \lim_{x \to\infty}\frac{1}{\sqrt{1+1/x^2}} = 1, \end{split} \end{equation*} and \begin{equation*} \begin{split} \lim_{x \to-\infty} f(x) \amp = \lim_{x \to-\infty} \frac{x}{\sqrt{x^2+1}} \\ \amp = \lim_{x \to-\infty}\frac{x/(-x)}{\sqrt{\frac{x^2+1}{x^2}}} \\ \amp = \lim_{x \to-\infty}\frac{-1}{\sqrt{1+1/x^2}} = -1. \end{split} \end{equation*} Thus, $f$ has two horizontal asymptotes: at $y=1$ and $y=-1\text{.}$ The graph of $f$ is provided below. Find the vertical asymptotes of $\ds f(x)=\frac{\ln x}{x-2}\text{.}$ Answer $x=0$ and $x=2\text{.}$ Solution Observe that $f(x) = \dfrac{\ln x}{x-2}$ is undefined at $x=0$ and $x=2\text{.}$ To find any vertical asymptotes, we thus compute the limits \begin{equation*} \lim_{x\to 2^+} \frac{\ln x}{x-2} = +\infty\text{,} \end{equation*} \begin{equation*} \lim_{x\to 2^-} \frac{\ln x}{x-2} = -\infty\text{,} \end{equation*} \begin{equation*} \lim_{x\to 0+} \frac{\ln x}{x-2} = -\infty\text{.} \end{equation*} Therefore, $f$ has two vertical asymptotes: at $x=2$ and $x=0\text{.}$ Suppose that a falling object reaches velocity $v(t)=50(1-e^{-t/5})$ at time $t\text{,}$ where distance is measured in $m$ and time $s\text{.}$ What is the object's terminal velocity, i.e. the value of $v(t)$ as $t$ goes to infinity? Answer 50 m/s Solution $\lim\limits_{t \to \infty} v(t) = \lim\limits_{t \to \infty} 50(1- e^{-t/5}) = 50 (1- \lim\limits_{t \to \infty} e^{-t/5}) = 50(1-0) = 50\text{.}$ Hence the object's terminal velocity is $50$ m/s. Find the slant asymptote of $f(x)=\dfrac{x^2+x+6}{x-3}\text{.}$ Answer $y=x+4$ Solution We first see that $f(x)$ has no horizontal asymptotes since \begin{equation*} \lim_{x\to \infty} f(x) = \lim_{x\to\infty}\frac{x^2+x+6}{x-3} = \infty, \ \ \text{ and } \ \ \lim_{x\to-\infty}\frac{x^2+x+6}{x-3} = -\infty\text{.} \end{equation*} The degree of the polynomial in the numerator of $f$ is higher than the degree of the polynomial in the denominator, and so we look instead for a slant asymptote. Notice that $(x-3)(x+4) = x^2+x-12\text{.}$ Thus, \begin{equation*} f(x) = \frac{x^2+x+6}{x-3} = \frac{(x-3)(x+4)+18}{x-3} = (x+4) + \frac{18}{x-3} \cdot \end{equation*} We now take the limits \begin{equation*} \lim_{x\to\infty} \left(f(x)-(x+4)\right) = \lim_{x\to\infty} \frac{18}{x-3} = 0, \ \ \text{ and } \end{equation*} \begin{equation*} \lim_{x\to-\infty} \left(f(x)-(x+4)\right) = \lim_{x\to-\infty} \frac{18}{x-3} = 0\text{,} \end{equation*} to confirm that the line $y=x+4$ is a slant asymptote of $f\text{.}$ The graph of $f$ is shown below. Compute the following limits. 1. $\ds\lim_{x\to -\infty}(2x^3-x)$ Answer $-\infty$ Solution $\lim\limits_{x\to -\infty} \left(2x^{3}-x\right) = \lim\limits_{x\to -\infty} x\left(2x^{2}-1\right) = -\infty\text{,}$ since $\lim\limits_{x\to-\infty} \left(2x^{2} -1\right) = \infty\text{.}$ 2. $\ds\lim_{x\to \infty}\tan^{-1}(e^x)$ Answer $\pi/2$ Solution $\lim\limits_{x\to\infty} \tan^{-1}\left(e^{x}\right) = \lim\limits_{w\to\infty} \tan^{-1}\left(w\right) = \frac{\pi}{2}\text{.}$ 3. $\ds\lim_{x\to -\infty}\tan^{-1}(e^x)$ Answer $-\pi/2$ Solution $\lim\limits_{x\to -\infty} \tan^{-1}\left(e^{x}\right) = \lim\limits_{w\to -\infty} \tan^{-1}\left(w\right) = -\frac{\pi}{2}\text{.}$ 4. $\ds\lim_{x\to \infty}\dfrac{e^x+x^4}{x^3+5\ln x}$ Answer $\infty$ Solution $\lim\limits_{x\to\infty} \dfrac{e^x+x^4}{x^3+5\ln x} = \lim\limits_{x\to\infty} \dfrac{\frac{e^x+x^4}{x^3}}{\frac{x^3+5\ln x}{x^3}} = \lim\limits_{x\to\infty} \dfrac{\frac{e^x}{x^3}+x}{1+\frac{5\ln x}{x^3}} = +\infty\text{,}$ since $\lim\limits_{x\to\infty} \dfrac{\ln x}{x^3} = 0$ and $\lim\limits_{x\to\infty} \dfrac{e^x}{x^3} = \infty\text{.}$ 5. $\ds\lim_{x\to \infty}\dfrac{2^x+5(3^x)}{3(2^x)-3^x}$ Answer $0$ Solution $\displaystyle{\lim\limits_{x\to\infty} \frac{2^x+5(3^x)}{3(2^x)-3^x} = \lim\limits_{x\to\infty} \frac{\frac{2^x+5(3^x)}{3^x}}{\frac{3(2^x)-3^x}{3^x}} = \lim\limits_{x\to\infty} \frac{\left(\frac{2}{3}\right)^x+5}{3\left(\frac{2}{3}\right)^x-1} = -5}\text{,}$ since $\lim\limits_{x\to\infty} \left(\frac{2}{3}\right)^x = 0\text{.}$ 6. $\ds\lim_{x\to -\infty}\dfrac{2^x+5(3^x)}{3(2^x)-3^x}$ Answer 1/3 Solution Now we want to use the fact that $\lim\limits_{x \to -\infty} \left(\dfrac{3}{2}\right)^x = \lim\limits_{x \to \infty} \left(\dfrac{2}{3}\right)^x=0\text{.}$ So we take $\displaystyle{\lim\limits_{x\to-\infty} \frac{2^x+5(3^x)}{3(2^x)-3^x} = \lim\limits_{x\to-\infty} \frac{\frac{2^x+5(3^x)}{2^x}}{\frac{3(2^x)-3^x}{2^x}} = \lim\limits_{x\to-\infty} \frac{1+5\left(\frac{3}{2}\right)^x}{3-\left(\frac{3}{2}\right)^x} = \frac{1}{3}}\text{.}$ Due to the mining of some oil sands, a river flowing nearby is found to be contaminated with bitumen, a dense and extremely viscous form of petroleum. Since this river is part of the water supply to a large farm region downstream from it, a proposal submitted to the farm council indicates that the cost, measured in millions of dollars, of removing x% of the toxic pollutant is given by \begin{equation*} C(x) = \dfrac{0.5x}{100-x} \ \ \ (0 \lt x \lt 100) \end{equation*} 1. Find the cost of removing $50$%, $60$%, $70$%, $80$%, $90$% and $95$% of the pollutant. Answer$500,000\text{,}$ $$750,000\text{,}$$$1,166,667\text{,}$ $$2$ million,$$4.5$ million, $$9.5$ million. Solution $C(50) = \dfrac{0.5(50)}{100-50} = 0.5\text{.}$ So the cost of removing $50\%$ of the contaminant is estimated to be $\ 500,000\text{.}$ Similarly, the cost to remove $60\%\text{,}$ $70\%\text{,}$ $80\%\text{,}$ $90\%$ and $95\%$ are calculated: \begin{equation*} \begin{array}{cc} \% \;\mbox{removed} \amp \mbox{cost estimate} \\ \hline 50 \amp \$ 500,000\\ 60 \amp \$750,000\\ 70 \amp \$ 1,166,667 \\ 80 \amp \$2 \;\mbox{million} \\ 90 \amp \$ 4.5 \;\mbox{million} \\ 95 \amp \$9.5 \;\mbox{million} \end{array} \end{equation*} 2. Evaluate \begin{equation*} \lim_{x \to 100} \dfrac{0.5x}{100-x} \end{equation*} and interpret your results. Answer \begin{equation*} \lim\limits_{x \to 100^-} \dfrac{0.5x}{100-x} = \lim\limits_{x \to 100^-} \dfrac{0.5}{\frac{100}{x}-1} = \infty \end{equation*} Therefore, as the percent of pollutant to be removes approaches $100\text{,}$ the cost becomes astronomical. The total sales from a Pulitzer-prized book is approximated by the function \begin{equation*} T(x) = \dfrac{120x^{2}}{x^{2}+4} \end{equation*} where $T(x)$ is measured in millions of dollars and $x$ is the number of months since the book's release. 1. What are the sales after the first month? The second month? the third month? Answer$$24$ million, $$60$ million,$$83.1$ million.
Solution

The sales after one month is approximated by $T(1) = \frac{120}{5} = 24\text{,}$ or $$24$ million. Similarly, after the second and third months, we find that the total sales are approximately$$60$ and $$83.1$ million, respectively. 2. What will the book gross in the long run? Answer$$120$ million.
Solution

To estimate the long-run gross of the book, we find

\begin{equation*} \lim\limits_{x \to \infty} T(x) = \lim_{x \to \infty} \dfrac{120x^{2}}{x^{2}+4} = 120\text{,} \end{equation*}

or \$$120$ million.