Section 1.17 Infinite limits
Objectives
-
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.
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).
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.
-
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.
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.
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:
\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*}
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.)
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*}
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 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. 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
\(a\text{.}\)
Theorem 1.17.6. 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)+p(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.
\end{cases}
\end{equation*}
-
Type \(\infty+\infty\).
\begin{equation*}
\lim\limits_{x\to a}f(x)+g(x)=\begin{cases}
\infty \amp \text{if } \lim\limits_{x\to a}f(x)=\lim\limits_{x\to a}g(x)=\infty\\
-\infty \amp \text{if } \lim\limits_{x\to a}f(x)=\lim\limits_{x\to a}g(x)=-\infty.
\end{cases}
\end{equation*}
-
Type \(\infty\cdot \infty\).
\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
\begin{equation*}
\lim\limits_{x\to a}\frac{p(x)}{f(x)}=0\text{.}
\end{equation*}
-
Type \(\frac{c}{0}\).
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 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!
Example 1.17.7. Infinite limit.
Compute the following limits. Your answer should be either a real number, \(\infty\text{,}\) or \(-\infty\text{.}\)
-
\(\displaystyle \lim\limits_{x\to 1^-}\sqrt[5]{\frac{3}{x-1}}\)
-
\(\displaystyle \lim\limits_{x\to -\infty}x^3+7x^2\)
Solution.
-
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:
\begin{align*}
\lim\limits_{x\to 1^+}\sqrt[5]{\frac{3}{x-1}} \amp = -\infty \amp (\text{type } \sqrt[n]{\infty}) \text{,}
\end{align*}
since
\begin{align*}
\lim\limits_{x\to 1^-}\frac{3}{x-1} \amp = -\infty \amp (\text{type } c/0) \text{.}
\end{align*}
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 \(1\text{.}\)
-
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:
\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 c)\text{,}
\end{align*}
since \(\lim\limits_{x\to -\infty}x^3=-\infty\) and \(\lim\limits_{x\to -\infty}1+\frac{7}{x}=1+0=1\text{.}\)
The argument above can be extended even further to compute limits at infinity of an arbitrary
rational function.
Theorem 1.17.9. 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*}
-
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*}
-
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.
Since both limits at infinity do not exist,
\(f\) does not have any horizontal asymptotes.
The next example will have us put
Theorem 1.17.9 to good use in the context of computing horizontal and vertical asymptotes of a rational function.
Example 1.17.10. 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
\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 1.17.9}}) \\
\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 \(x\to \infty\text{.}\)
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 1.17.9}}) \\
\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 \(x\to -\infty\text{.}\)
Vertical asymptotes.
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*}
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:
\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 here, the line \(x=2\) is not a vertical asymptote.
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 \(f\text{.}\)
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 \(1\text{.}\)
