APEX Calculus

Section1.6Limits Involving Infinity

In Definition 1.2.2 we stated that in the equation $$\lim_{x\to c}f(x) = L\text{,}$$ both $$c$$ and $$L$$ were numbers. In this section we relax that definition a bit by considering situations when it makes sense to let $$c$$ and/or $$L$$ be “infinity.”
As a motivating example, consider $$f(x) = 1/x^2\text{,}$$ as shown in Figure 1.6.1. Note how, as $$x$$ approaches 0, $$f(x)$$ grows very, very large—in fact, it grows without bound. It seems appropriate, and descriptive, to state that
\begin{equation*} \lim_{x\to 0} \frac1{x^2}=\infty\text{.} \end{equation*}
Also note that as $$x$$ gets very large, $$f(x)$$ gets very, very small. We could represent this concept with notation such as
\begin{equation*} \lim_{x\to \infty} \frac1{x^2}=0\text{.} \end{equation*}
We explore both types of use of $$\infty$$ in turn.

Definition1.6.2.Limit of Infinity, $$\infty$$.

Let $$I$$ be an open interval containing $$c\text{,}$$ and let $$f$$ be a function defined on $$I\text{,}$$ except possibly at $$c\text{.}$$
• The limit of $$f(x)\text{,}$$ as $$x$$ approaches $$c\text{,}$$ is infinity, denoted by
\begin{equation*} \lim_{x\rightarrow c} f(x) = \infty\text{,} \end{equation*}
if given any $$N \gt 0\text{,}$$ there exists $$\delta \gt 0$$ such that for all $$x$$ in $$I\text{,}$$ where $$x\neq c\text{,}$$ if $$\abs{x - c} \lt \delta\text{,}$$ then $$f(x) \gt N\text{.}$$
• The limit of $$f(x)\text{,}$$ as $$x$$ approaches $$c\text{,}$$ is negative infinity, denoted by
\begin{equation*} \lim_{x\rightarrow c} f(x) = -\infty\text{,} \end{equation*}
if given any $$N \lt 0\text{,}$$ there exists $$\delta \gt 0$$ such that for all $$x$$ in $$I\text{,}$$ where $$x\neq c\text{,}$$ if $$\abs{x - c} \lt \delta\text{,}$$ then $$f(x) \lt N\text{.}$$
The first definition is similar to the $$\varepsilon$$-$$\delta$$ definition in Definition 1.2.2 from Section 1.2. In that definition, given any (small) value $$\varepsilon\text{,}$$ if we let $$x$$ get close enough to $$c$$ (within $$\delta$$ units of $$c$$) then $$f(x)$$ is guaranteed to be within $$\varepsilon$$ of $$L\text{.}$$ Here, given any (large) value $$N\text{,}$$ if we let $$x$$ get close enough to $$c$$ (within $$\delta$$ units of $$c$$), then $$f(x)$$ will be at least as large as $$N\text{.}$$ In other words, if we get close enough to $$c\text{,}$$ then we can make $$f(x)$$ as large as we want.
It is important to note that by saying $$\lim_{x\to c}f(x) = \infty$$ we are implicitly stating that the limit of $$f(x)\text{,}$$ as $$x$$ approaches $$c\text{,}$$ does not exist. A limit only exists when $$f(x)$$ approaches an actual numeric value. We use the concept of limits that approach infinity because it is helpful and descriptive. It is one specific way in which a limit can fail to exist.
We define one-sided limits that approach infinity in a similar way.

Definition1.6.4.One-Sided Limits of Infinity.

• Let $$f$$ be a function defined on $$(a,c)$$ for some $$a\lt c\text{.}$$ We say the limit of $$f(x)\text{,}$$ as $$x$$ approaches $$c$$ from the left, is infinity, or, the left-hand limit of $$f$$ at $$c$$ is infinity, denoted by
\begin{equation*} \lim_{x\rightarrow c^-} f(x) = \infty\text{,} \end{equation*}
if given any $$N \gt 0\text{,}$$ there exists $$\delta \gt 0$$ such that for all $$a\lt x\lt c\text{,}$$ if $$\abs{x - c} \lt \delta\text{,}$$ then $$f(x) \gt N\text{.}$$
• Let $$f$$ be a function defined on $$(c,b)$$ for some $$b \gt c\text{.}$$ We say the limit of $$f(x)\text{,}$$ as $$x$$ approaches $$c$$ from the right, is infinity, or, the right-hand limit of $$f$$ at $$c$$ is infinity, denoted by
\begin{equation*} \lim_{x\rightarrow c^+} f(x) = \infty\text{,} \end{equation*}
if given any $$N \gt 0\text{,}$$ there exists $$\delta \gt 0$$ such that for all $$c\lt x\lt b\text{,}$$ if $$\abs{x - c} \lt \delta\text{,}$$ then $$f(x) \gt N\text{.}$$
• The term left- (or, right-) hand limit of $$f$$ at c is negative infinity is defined in a manner similar to Definition 1.6.2.

Example1.6.5.Evaluating limits involving infinity.

Find $$\lim\limits_{x\to 1}\frac1{(x-1)^2}$$ as shown in Figure 1.6.6.
Solution 1. Video solution
Solution 2.
In Example 1.1.18 of Section 1.1, by inspecting values of $$x$$ close to $$1$$ we concluded that this limit does not exist. That is, it cannot equal any real number. But the limit could be infinite. And in fact, we see that the function does appear to be growing larger and larger, as $$f(0.99)=10^4\text{,}$$ $$f(0.999)=10^6\text{,}$$ $$f(0.9999)=10^8\text{.}$$ A similar thing happens on the other side of $$1\text{.}$$ From the graph and the numeric information, we could state $$\lim_{x \to 1}1/(x-1)^2=\infty\text{.}$$ We can prove this by using Definition 1.6.2
In general, let a “large” value $$N$$ be given. Let $$\delta=1/\sqrt{N}\text{.}$$ If $$x$$ is within $$\delta$$ of $$1\text{,}$$ i.e., if $$\abs{x-1}\lt 1/\sqrt{N}\text{,}$$ then:
\begin{align*} \abs{x-1} \amp \lt \frac{1}{\sqrt{N}}\\ (x-1)^2 \amp \lt \frac{1}{N}\\ \frac{1}{(x-1)^2} \amp \gt N\text{,} \end{align*}
which is what we wanted to show. So we may say $$\lim_{x\to 1}1/{(x-1)^2}=\infty\text{.}$$

Example1.6.7.Evaluating limits involving infinity.

Find $$\lim\limits_{x\to 0}\frac1x\text{,}$$ as shown in Figure 1.6.8.
Solution 1. Video solution
Solution 2.
It is easy to see that the function grows without bound near $$0\text{,}$$ but it does so in different ways on different sides of $$0\text{.}$$ Since its behavior is not consistent, we cannot say that $$\lim_{x\to 0}\frac{1}{x}=\infty\text{.}$$ Instead, we will say $$\lim_{x\to 0}\frac{1}{x}$$ does not exist. However, we can make a statement about one-sided limits. We can state that $$\lim_{x\to 0^+}\frac1x=\infty$$ and $$\lim_{x\to 0^-}\frac1x=-\infty\text{.}$$

Subsection1.6.1Vertical asymptotes

The graphs in the two previous examples demonstrate that if a function $$f$$ has a limit (or, left- or right-hand limit) of infinity at $$x=c\text{,}$$ then the graph of $$f$$ looks similar to a vertical line near $$x=c\text{.}$$ This observation leads to a definition.

Definition1.6.9.Vertical Asymptote.

Let $$I$$ be an interval that either contains $$c$$ or has $$c$$ as an endpoint, and let $$f$$ be a function defined on $$I\text{,}$$ except possibly at $$c\text{.}$$
If the limit of $$f(x)$$ as $$x$$ approaches $$c$$ from either the left or right (or both) is $$\infty$$ or $$-\infty\text{,}$$ then the line $$x=c$$ is a vertical asymptote of $$f\text{.}$$

Example1.6.11.Finding vertical asymptotes.

Find the vertical asymptotes of $$f(x)=\frac{3x}{x^2-4}\text{.}$$
Solution 1. Video solution
Solution 2.
Vertical asymptotes occur where the function grows without bound; this can occur at values of $$c$$ where the denominator is $$0\text{.}$$ When $$x$$ is near $$c\text{,}$$ 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 $$x=\pm 2\text{.}$$ 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 1.6.12. Thus the vertical asymptotes are at $$x=\pm2\text{.}$$
When a rational function has a vertical asymptote at $$x=c\text{,}$$ we can conclude that the denominator is $$0$$ at $$x=c\text{.}$$ 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 $$x=1\text{,}$$ as shown in Figure 1.6.13. While the denominator does get small near $$x=1\text{,}$$ the numerator gets small too, matching the denominator step for step. In fact, factoring the numerator, we get
\begin{equation*} f(x)=\frac{(x-1)(x+1)}{x-1}\text{.} \end{equation*}
Canceling the common term, we get that $$f(x)=x+1$$ for $$x\not=1\text{.}$$ So there is clearly no asymptote; rather, a hole exists in the graph at $$x=1\text{.}$$
The above example may seem a little contrived. Another example demonstrating this important concept is $$f(x)= (\sin(x) )/x\text{.}$$ We have considered this function several times in the previous sections. We found that $$\lim_{x\to0}\frac{\sin(x) }{x}=1\text{;}$$ i.e., there is no vertical asymptote. No simple algebraic cancellation makes this fact obvious; we used the Theorem 1.3.10 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.

Subsection1.6.2Indeterminate Forms

We have seen how the limits $$\lim_{x\to 0}\frac{\sin(x) }{x}$$ and $$\lim_{x\to1}\frac{x^2-1}{x-1}$$ each return the indeterminate form $$0/0$$ when we blindly plug in $$x=0$$ and $$x=1\text{,}$$ respectively. However, $$0/0$$ is not a valid arithmetical expression. It gives no indication that the respective limits are $$1$$ and $$2\text{.}$$
With a little cleverness, one can come up with $$0/0$$ expressions which have a limit of $$\infty\text{,}$$ 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 $$0/0\text{.}$$ 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 $$0\text{.}$$ 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), it may involve using trigonometric identities or logarithm rules, or it may require a tool such as the Squeeze Theorem. In Section 6.7 we will learn yet another technique called L'Hospital's Rule that provides another way to handle indeterminate forms.
Some other common indeterminate forms are $$\infty-\infty\text{,}$$ $$\infty\cdot 0\text{,}$$ $$\infty/\infty\text{,}$$ $$0^0\text{,}$$ $$\infty^0$$ and $$1^{\infty}\text{.}$$ Again, keep in mind that these are the “blind” results of directly substituting $$c$$ into the expression, 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 $$\infty\text{.}$$
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 $$\infty\text{,}$$ $$-\infty\text{,}$$ or simply not exist if the left- and right-hand limits do not match.

Subsection1.6.3Limits 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.

Definition1.6.14.Limits at Infinity and Horizontal Asymptotes.

Let $$L$$ be a real number.
1. Let $$f$$ be a function defined on $$(a,\infty)$$ for some number $$a\text{.}$$ The limit of $$f$$ at infinity is $$L\text{,}$$ denoted $$\lim_{x\to\infty} f(x)=L\text{,}$$ if for every $$\epsilon \gt 0$$ there exists $$M \gt a$$ such that if $$x \gt M\text{,}$$ then $$\abs{f(x)-L}\lt \epsilon\text{.}$$
2. Let $$f$$ be a function defined on $$(-\infty,b)$$ for some number $$b\text{.}$$ The limit of $$f$$ at negative infinity is $$L\text{,}$$ denoted $$\lim_{x\to-\infty} f(x)=L\text{,}$$ if for every $$\epsilon \gt 0$$ there exists $$M\lt b$$ such that if $$x \lt M\text{,}$$ then $$\abs{f(x)-L}\lt \epsilon\text{.}$$
3. If $$\lim_{x\rightarrow\infty} f(x)=L$$ or $$\lim_{x\rightarrow-\infty} f(x)=L\text{,}$$ we say the line $$y=L$$ is a horizontal asymptote of $$f\text{.}$$
We can also define limits such as $$\lim_{x\to\infty}f(x)=\infty$$ by combining this definition with Definition 1.6.2.

Example1.6.16.Approximating horizontal asymptotes.

Approximate the horizontal asymptote(s) of $$f(x)=\frac{x^2}{x^2+4}\text{.}$$
Solution.
We will approximate the horizontal asymptotes by approximating the limits $$\lim_{x\to-\infty} \frac{x^2}{x^2+4}$$ and $$\lim_{x\to\infty} \frac{x^2}{x^2+4}\text{.}$$ (A rational function can have at most one horizontal asymptote. So we could get away with only taking $$x \to \infty$$).
Figure 1.6.17.(a) shows a sketch of $$f\text{,}$$ and the table in Figure 1.6.17.(b) gives values of $$f(x)$$ for large magnitude values of $$x\text{.}$$ It seems reasonable to conclude from both of these sources that $$f$$ has a horizontal asymptote at $$y=1\text{.}$$
Later, we will show how to determine this analytically.
The video in Figure 1.6.18 shows how to prove the result from Example 1.6.16 using the limit definition.
Horizontal asymptotes can take on a variety of forms. Figure 1.6.19.(a) shows that $$f(x) = x/(x^2+1)$$ has a horizontal asymptote of $$y=0\text{,}$$ where $$0$$ is approached from both above and below.
Figure 1.6.19.(b) shows that $$f(x) =x/\sqrt{x^2+1}$$ has two horizontal asymptotes; one at $$y=1$$ and the other at $$y=-1\text{.}$$
Figure 1.6.19.(c) shows that $$f(x) = \sin(x)/x$$ has even more interesting behavior than at just $$x=0\text{;}$$ as $$x$$ approaches $$\pm\infty\text{,}$$ $$f(x)$$ approaches $$0\text{,}$$ but oscillates as it does this.
We can analytically evaluate limits at infinity for rational functions once we understand $$\lim_{x\to\infty}\frac{1}{x}\text{.}$$ As $$x$$ gets larger and larger, $$1/x$$ gets smaller and smaller, approaching $$0\text{.}$$ We can, in fact, make $$1/x$$ as small as we want by choosing a large enough value of $$x\text{.}$$ Given $$\varepsilon\text{,}$$ we can make $$1/x\lt \varepsilon$$ by choosing $$x \gt 1/\varepsilon\text{.}$$ Thus we have $$\lim_{x\to\infty} 1/x=0\text{.}$$
It is now not much of a jump to conclude the following:
\begin{align*} \lim_{x\to\infty}\frac1{x^n}\amp=0\amp\lim_{x\to-\infty}\frac1{x^n}\amp=0\text{.} \end{align*}
Now suppose we need to compute the following limit:
\begin{equation*} \lim_{x\to\infty}\frac{x^3+2x+1}{4x^3-2x^2+9}\text{.} \end{equation*}
A good way of approaching this is to divide through the numerator and denominator by $$x^3$$ (hence multiplying by $$1$$), which is the largest power of $$x$$ to appear in the denominator. Doing this, we get
\begin{align*} \lim_{x\to\infty}\frac{x^3+2x+1}{4x^3-2x^2+9} \amp = \lim_{x\to\infty}\frac{1/x^3}{1/x^3}\cdot\frac{x^3+2x+1}{4x^3-2x^2+9}\\ \amp =\lim_{x\to\infty}\frac{x^3/x^3+2x/x^3+1/x^3}{4x^3/x^3-2x^2/x^3+9/x^3}\\ \amp = \lim_{x\to\infty}\frac{1+2/x^2+1/x^3}{4-2/x+9/x^3}\text{.} \end{align*}
Then using the rules for limits (which also hold for limits at infinity), as well as the fact about limits of $$1/x^n\text{,}$$ we see that the limit becomes
\begin{equation*} \frac{1+0+0}{4-0+0}=\frac14\text{.} \end{equation*}
This procedure works for any rational function. In fact, it gives us the following theorem.
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 $$b_mx^m/x^n\text{.}$$ Since $$n=m\text{,}$$ this will leave us with the limit $$a_n/b_m\text{.}$$ If $$n\lt m\text{,}$$ then after dividing through by $$x^m\text{,}$$ all the terms in the numerator will approach $$0$$ in the limit, leaving us with $$0/b_m$$ or $$0\text{.}$$ If $$n \gt m\text{,}$$ and we try dividing through by $$x^m\text{,}$$ we end up with the denominator tending to $$b_m$$ while the numerator tends to $$\infty\text{.}$$
Intuitively, as $$x$$ gets very large, all the terms in the numerator are small in comparison to $$a_nx^n\text{,}$$ and likewise all the terms in the denominator are small compared to $$b_nx^m\text{.}$$ If $$n=m\text{,}$$ looking only at these two important terms, we have $$(a_nx^n)/(b_nx^m)\text{.}$$ This reduces to $$a_n/b_m\text{.}$$ If $$n\lt m\text{,}$$ the function behaves like $$a_n/(b_mx^{m-n})\text{,}$$ which tends toward $$0\text{.}$$ If $$n \gt m\text{,}$$ the function behaves like $$a_nx^{n-m}/b_m\text{,}$$ which will tend to either $$\infty$$ or $$-\infty$$ depending on the values of $$n\text{,}$$ $$m\text{,}$$ $$a_n\text{,}$$ $$b_m$$ and whether you are looking for $$\lim_{x\to\infty} f(x)$$ or $$\lim_{x\to-\infty} f(x)\text{.}$$

Example1.6.22.Finding a limit of a rational function.

Confirm analytically that $$y=1$$ is the horizontal asymptote of $$f(x) = \frac{x^2}{x^2+4}\text{,}$$ as approximated in Example 1.6.16.
Solution 1. Video solution
Solution 2.
Before using Theorem 1.6.21, 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 $$2\text{,}$$ so divide the numerator and denominator of $$f$$ by $$x^2\text{,}$$ then take limits.
\begin{align*} \lim_{x\to\infty}\frac{x^2}{x^2+4} \amp = \lim_{x\to\infty}\frac{x^2/x^2}{x^2/x^2+4/x^2}\\ \amp =\lim_{x\to\infty}\frac{1}{1+4/x^2}\\ \amp =\frac{1}{1+0}\\ \amp = 1\text{.} \end{align*}
We can also use Theorem 1.6.21 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.

Example1.6.23.Finding limits of rational functions.

Use Theorem 1.6.21 to evaluate each of the following limits.
1. $$\displaystyle \lim_{x\to-\infty}\dfrac{x^2+2x-1}{x^3+1}$$
2. $$\displaystyle \lim_{x\to\infty}\dfrac{x^2+2x-1}{1-x-3x^2}$$
3. $$\displaystyle \lim_{x\to\infty}\dfrac{x^2-1}{3-x}$$
Solution.
1. The highest power of $$x$$ is in the denominator. Therefore, the limit is $$0\text{;}$$ see Figure 1.6.24.(a).
2. The highest power of $$x$$ is $$x^2\text{,}$$ which occurs in both the numerator and denominator. The limit is therefore the ratio of the coefficients of $$x^2\text{,}$$ which is $$-1/3\text{.}$$ See Figure 1.6.24.(b).
3. The highest power of $$x$$ is in the numerator so the limit will be $$\infty$$ or $$-\infty\text{.}$$ To see which, consider only the dominant terms from the numerator and denominator, which are $$x^2$$ and $$-x\text{.}$$ The expression in the limit will behave like $$x^2/(-x) = -x$$ for large values of $$x\text{.}$$ Therefore, the limit is $$-\infty\text{.}$$ See Figure 1.6.24.(c).
With care, we can quickly evaluate limits at infinity for a large number of functions by considering the long run behavior using “dominant terms” of $$f(x)\text{.}$$ For instance, consider again $$\lim_{x\to\pm\infty}\frac{x}{\sqrt{x^2+1}}\text{,}$$ graphed in Figure 1.6.19.(b). The dominant terms are $$x$$ in the numerator and $$\sqrt{x^2}$$ in the denominator. When $$x$$ is very large, $$x^2+1 \approx x^2\text{.}$$ Thus
\begin{align*} \sqrt{x^2+1}\amp\approx \sqrt{x^2} = \abs{x} \amp \frac{x}{\sqrt{x^2+1}} \amp\approx \frac{x}{\abs{x}}\text{.} \end{align*}
This expression is $$1$$ when $$x$$ is positive and $$-1$$ when $$x$$ is negative. Hence we get asymptotes of $$y=1$$ and $$y=-1\text{,}$$ respectively. We will show this more formally in the next example.

Example1.6.25.Finding a limit using dominant terms.

Confirm analytically that $$y=1$$ and $$y=-1$$ are the horizontal asymptote of $$\lim_{x\to\pm\infty}\frac{x}{\sqrt{x^2+1}}\text{,}$$ as graphed in Figure 1.6.19.(b).
Solution.
The dominating term of $$f$$ in the denominator is $$\sqrt{x^2}=\abs{x}$$ so divide the numerator and denominator of $$f$$ by $$\sqrt{x^2}\text{,}$$ then take limits.
\begin{align*} \lim_{x\to\infty}\frac{x}{\sqrt{x^2+1}} \amp = \lim_{x\to\infty}\frac{x}{\sqrt{x^2+1}}\cdot \frac{\frac{1}{\sqrt{x^2}}}{\frac{1}{\sqrt{x^2}}}\\ \amp = \lim_{x\to\infty} \frac{\frac{x}{\abs{x}}}{\sqrt{\frac{x^2+1}{x^2}}}\\ \amp = \lim_{x\to\infty} \frac{1}{\sqrt{1+\frac{1}{x^2}}} \text{ for } x\gt 0\\ \amp =\frac{1}{\sqrt{1+0}}\\ \amp = 1 \text{.} \end{align*}
As $$x \to -\infty\text{,}$$ the only thing that changes is the value of $$\frac{x}{\abs{x}}\text{.}$$ For $$x \lt 0\text{,}$$ we have $$\frac{x}{\abs{x}}=-1\text{,}$$ making $$\lim_{x\to-\infty} \frac{x}{\sqrt{x^2+1}}=-1\text{.}$$ Therefore, the horizontal asymptotes are $$y=1$$ and $$y=-1\text{.}$$
The video in Figure 1.6.26 provides another example similar to Example 1.6.25.

Exercises1.6.4Exercises

Terms and Concepts

1.
• True
• False
If $$\lim\limits_{x\to 5} f(x) = \infty\text{,}$$ then we are implicitly stating that the limit exists.
2.
• True
• False
If $$\lim\limits_{x\to 5} f(x) = 5\text{,}$$ then we are implicitly stating that the limit exists.
3.
• True
• False
If $$\lim\limits_{x\to 1^-} f(x) = -\infty\text{,}$$ then $$\lim\limits_{x\to 1^+} f(x) = \infty\text{.}$$
4.
• True
• False
If $$\lim\limits_{x\to 5} f(x) = \infty\text{,}$$ then $$f$$ has a vertical asymptote at $$x=5\text{.}$$
5.
• True
• False
$$\infty/0$$ is not an indeterminate form.
6.
List five indeterminate forms.
7.
Construct a function with a vertical asymptote at $$x=5$$ and a horizontal asymptote at $$y=5\text{.}$$
8.
Let $$\lim\limits_{x\to 7} f(x) = \infty\text{.}$$ Explain how we know that $$f$$ is or is not continuous at $$x=7\text{.}$$

Problems

Exercise Group.
Evaluate the given limits using the graph of the function.
9.
$$f(x) = {-\frac{1}{\left(x+2\right)^{2}}}$$ has the graph:
1. $$\displaystyle \lim\limits_{x\to -2^-} f(x)$$
2. $$\displaystyle \lim\limits_{x\to -2^+} f(x)$$
10.
$$f(x) = {\frac{1}{\left(x-3\right)\!\left(x-7\right)^{2}}}$$ has the graph:
1. $$\displaystyle \lim\limits_{x\to 3^-} f(x)$$
2. $$\displaystyle \lim\limits_{x\to 3^+} f(x)$$
3. $$\displaystyle \lim\limits_{x\to 3} f(x)$$
4. $$\displaystyle \lim\limits_{x\to 7^-} f(x)$$
5. $$\displaystyle \lim\limits_{x\to 7^+} f(x)$$
6. $$\displaystyle \lim\limits_{x\to 7} f(x)$$
11.
$$f(x) = {\frac{2}{e^{-x}+1}}$$ has the graph:
1. $$\displaystyle \lim\limits_{x\to-\infty} f(x)$$
2. $$\displaystyle \lim\limits_{x\to\infty} f(x)$$
3. $$\displaystyle \lim\limits_{x\to 0^{-}} f(x)$$
4. $$\displaystyle \lim\limits_{x\to 0^{+}} f(x)$$
12.
$$f(x) = {x\sin\!\left(3\pi x\right)}$$ has the graph:
1. $$\displaystyle \lim\limits_{x\to-\infty} f(x)$$
2. $$\displaystyle \lim\limits_{x\to\infty} f(x)$$
3. $$\displaystyle \lim\limits_{x\to 0^{-}} f(x)$$
4. $$\displaystyle \lim\limits_{x\to 0^{+}} f(x)$$
13.
$$f(x) = {\cos\!\left(4x\right)}$$ has the graph:
1. $$\displaystyle \lim\limits_{x\to-\infty} f(x)$$
2. $$\displaystyle \lim\limits_{x\to\infty} f(x)$$
14.
$$f(x) = {2.2^{-x}-4}$$ has the graph:
1. $$\displaystyle \lim\limits_{x\to-\infty} f(x)$$
2. $$\displaystyle \lim\limits_{x\to\infty} f(x)$$
Exercise Group.
Numerically approximate the limits.
15.
$$f(x)={\frac{x^{2}+7x-18}{x^{2}+x-42}}$$
1. $$\displaystyle \lim\limits_{x\to6^{-}}f(x)$$
2. $$\displaystyle \lim\limits_{x\to6^{+}}f(x)$$
3. $$\displaystyle \lim\limits_{x\to6}f(x)$$
16.
$$f(x)={\frac{x^{2}+7x-18}{x^{3}-21x^{2}+144x-320}}$$
1. $$\displaystyle \lim\limits_{x\to8^{-}}f(x)$$
2. $$\displaystyle \lim\limits_{x\to8^{+}}f(x)$$
3. $$\displaystyle \lim\limits_{x\to8}f(x)$$
17.
$$f(x)={\frac{x^{2}+3x}{x^{3}+16x^{2}+45x-162}}$$
1. $$\displaystyle \lim\limits_{x\to-9^{-}}f(x)$$
2. $$\displaystyle \lim\limits_{x\to-9^{+}}f(x)$$
3. $$\displaystyle \lim\limits_{x\to-9}f(x)$$
18.
$$f(x)={\frac{x^{2}-2x-63}{x^{2}-49}}$$
1. $$\displaystyle \lim\limits_{x\to-7^{-}}f(x)$$
2. $$\displaystyle \lim\limits_{x\to-7^{+}}f(x)$$
3. $$\displaystyle \lim\limits_{x\to-7}f(x)$$
Exercise Group.
Identify the horizontal and vertical asymptotes, if any, of the given function.
19.
$$f(x)={\frac{5x^{2}+22x-15}{x^{2}+x-12}}$$
20.
$$f(x)={\frac{3x^{2}+7x-6}{5x^{2}+40x+75}}$$
21.
$$f(x)={\frac{3x^{2}+x-2}{-7x^{3}+49x^{2}+56x}}$$
22.
$$f(x)={\frac{x^{2}-4x+4}{-x+1}}$$
23.
$$f(x)={\frac{5x^{2}-22x+8}{9x-36}}$$
24.
$$f(x)={\frac{3x^{2}-39x+126}{-x^{2}-2x-4}}$$
Exercise Group.
Evaluate the given limit.
25.
$$\lim\limits_{x\to{-\infty }}{\frac{x^{3}+x^{2}+6x-5}{2x-3}}$$
26.
$$\lim\limits_{x\to{\infty }}{\frac{x^{3}-6x^{2}-5x-9}{x+7}}$$
27.
$$\lim\limits_{x\to{\infty }}{\frac{x^{3}+7x^{2}+2x+4}{x^{2}-3}}$$
28.
$$\lim\limits_{x\to{\infty }}{\frac{x^{3}-x^{2}-9x+1}{x^{2}+7}}$$

Review

29.
Use an $$\varepsilon$$-$$\delta$$ proof to prove that $$\lim\limits_{x\to 1}(5x-2)=3\text{.}$$
30.
Let $$\lim\limits_{x\to0} f(x) = 6$$ and $$\lim\limits_{x\to0} g(x) = 3\text{.}$$ Evaluate the following limits.
1. $$\displaystyle \lim\limits_{x\to0}(f+g)(x)$$
2. $$\displaystyle \lim\limits_{x\to0}(fg)(x)$$
3. $$\displaystyle \lim\limits_{x\to0}(f/g)(x)$$
4. $$\displaystyle \lim\limits_{x\to0}f(x)^{g(x)}$$
31.
Let $$f(x) = \begin{cases}{-x^{2}+3x+10}\amp x\lt3\\{2x^{2}+x-1}\amp x\geq3\end{cases}\text{.}$$
Is $$f$$ continuous everywhere?
• yes
• no
32.
Find $$\lim\limits_{x\to e} \ln(x)\text{.}$$