Provide a rigorous definition of infinite limits at points \(a\in \R\) and at infinity.
Investigate infinite limits graphically.
Define vertical asymptotes using language of infinite limits.
Develop computation techniques for infinite limits.
Definition9.1.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.
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.
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.
Remark9.2.Infinite limits.
It is important to observe that the various notions of infinite limit defined in Definition 9.1 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
its failure to exist is due to values of the function getting arbitrarily large (in positive or negative direction) as \(x\) approaches \(a\text{.}\)
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.
Remark9.3.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 \(a\in \R\text{:}\)
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:
for all \(n\text{,}\) and hence that each line \(x=a_n\) is a vertical asymptote of the graph of \(f\text{.}\)
As illustrated by Example 9.5, infinite limit formulas can be easily deduced from graphs of familiar functions. Theorem 9.7 can be thought of as directly translating properties of the graphs of power functions \(f(x)=x^n\) and their reciprocals \(g(x)=\frac{1}{x^n}\)into limit statements.
Theorem9.7.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*}
The next theorem helps us to compute the limit of functions built from other functions that may have infinite limits. Instead of trying to commit all the details of this theorem to memory, it is easier to understand the simple arithmetical arguments that go into establishing these results. For example, the fact that if \(\lim\limits_{x\to a}f(x)=\pm \infty\) and \(\lim\limits_{x\to a}p(x)=c\text{,}\) then \(\lim\limits_{x\to a}\frac{p(x)}{f(x)}=0\text{,}\) essentially follows from the fact that if the numerator \(p(x)\) is approaching some finite number \(c\) as \(x\to a\text{,}\) while the denominator \(f(x)\) gets arbitrarily large, then the quotient \(p(x)/f(x)\) is roughly described as \(c\) divided by a very large number, which is very small. Thus the limit is equal to 0.
All of the arguments behind the results of Theorem 9.8 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. Note that the type descriptions alone don’t indicate the sign (\(\pm\)) of your result: e.g., a limit computation of type \(\infty\cdot \infty\text{,}\) can yield \(\infty\) or \(-\infty\text{,}\) depending on the behavior of the functions near the limit point \(a\text{.}\)
Theorem9.8.Infinite limit formulas.
In what follows \(a\) denotes either a real number or \(\pm \infty\text{.}\) Assume that \(\lim\limits_{x\to a}f(x)\) and \(\lim\limits_{x\to a}g(x)\) are both infinite.
Type \(\infty+c\).
If \(\lim\limits_{x\to a}p(x)=c\) for some \(c\in \R\text{,}\) then
\begin{equation*}
\lim\limits_{x\to a}f(x)g(x)=\begin{cases}
\infty \amp \text{if } f(x)g(x)>0 \text{ for all } x \text{ near } a\\
-\infty \amp \text{if } f(x)g(x)<0 \text{ for all } x \text{ near } a.\\
\end{cases}
\end{equation*}
Type \(\infty\cdot c\).
If \(\lim\limits_{x\to a}p(x)=c\) for some \(c\in \R\text{,}\) then
\begin{equation*}
\lim\limits_{x\to a}f(x)p(x)=\begin{cases}
\infty \amp \text{if } f(x)p(x)> 0 \text{ for all } x \text{ near } a\\
-\infty \amp \text{if } f(x)p(x)< 0 \text{ for all } x \text{ near } a.
\end{cases}
\end{equation*}
Type \(\frac{c}{\infty}\).
If \(\lim\limits_{x\to a}p(x)=c\) for some \(c\in \R\text{,}\) then
If \(\lim\limits_{x\to a}p(x)=c\) and \(\lim\limits_{x\to a}q(x)=0\text{,}\) then
\begin{equation*}
\lim\limits_{x\to a}\frac{p(x)}{q(x)}=\begin{cases}
\infty\amp \text{if }\frac{p(x)}{q(x)}> 0 \text{ for all } x \text{ near } a \\
-\infty\amp \text{if } \frac{p(x)}{q(x)}< 0 \text{ for all } x \text{ near } a.
\end{cases}
\end{equation*}
Type \(\infty^n\).
If \(n\) is a positive integer, then
\begin{equation*}
\lim\limits_{x\to a}\left(f(x)\right)^n=\begin{cases}
\infty \amp \text{if } \lim\limits_{x\to a}f(x)=\infty \text{ or } n \text{ is even} \\
-\infty \amp \text{if } \lim\limits_{x\to a}f(x)=-\infty \text{ and } n \text{ is odd}.
\end{cases}
\end{equation*}
Type \(\sqrt[n]{\infty}\).
If \(n\) is a positive integer, then
\begin{equation*}
\lim\limits_{x\to a}\sqrt[n]{f(x)}=\begin{cases}
\infty \amp \text{if } \lim\limits_{x\to a}f(x)=\infty \\
-\infty \amp \text{if } \lim\limits_{x\to a}f(x)=-\infty \text{ and } n \text{ is odd}.
\end{cases}
\end{equation*}
Let’s see how to write up our work when making use of the results of Theorem 9.8. 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!
Example9.9.Infinite limit: elementary examples.
Compute the following limits. Your answer should be either a real number, \(\infty\text{,}\) or \(-\infty\text{.}\)
Our thinking is as follows: the function in question if of the form \(p(x)/f(x)\text{,}\) where \(p(x)\to 1\) as \(x\to 0^+\) and \(f(x)\to 0\) as \(x\to 0^+\text{.}\) This is thus a limit of type \(c/0\text{,}\) according to Theorem 9.8, and so should be equal to \(\pm \infty\text{.}\) Let’s write this up more concisely, and determine the correct sign:
since \(\lim\limits_{x\to 0^+}\cos x=1\text{,}\)\(\lim\limits_{x\to 0^+}x^3=0\text{,}\) and \(\cos x/x^3 > 0\) for positive \(x\) close to \(0\text{.}\)
In this case a quick appraisal tells us that the limit in question will fall under type \(\sqrt[n]{\infty}\text{,}\) and thus should be equal to \(\pm \infty\) (sign to be determined). Here is how we can formally write this up:
Furthermore, this last limit holds since \(\lim_{x\to 1^-}x-1=0\text{,}\) and \(3/(x-1)\) is negative for all \(x\) near to and less than \(1\text{.}\)
Example9.10.Infinite limit: polynomial.
Compute \(\lim_{x\to -\infty}x^3+7x^2\text{.}\) Your answer should be a real number or \(\pm \infty\text{.}\)
Solution.
Note first that \(\lim\limits_{x\to -\infty}x^3=-\infty\) (9.7), 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 9.8 directly. Instead, we first do some algebra, using our intuition that the \(x^3\) term “dominates” the \(7x^2\) term:
since \(\lim\limits_{x\to -\infty}x^3=-\infty\) and \(\lim\limits_{x\to -\infty}1+\frac{7}{x}=1+0=1\text{.}\)
Theorem9.11.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.
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.
Since both limits at infinity do not exist, \(f\) does not have any horizontal asymptotes.
Example9.12.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.
Solution.
Horizontal asymptotes.
For our horizontal asymptote investigation we compute the limits at infinity of \(f\text{.}\) At \(\infty\) we have
Since \(f\) is continuous everywhere on its domain, the only candidates for vertical asymptotes are the lines \(x=2\) and \(x=1\text{.}\) We investigate the limits at these points:
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 \(f\text{.}\)
We are asked to compute the other one-sided limit. The computation is similar:
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 \(1\text{.}\)
