Section 4.6 Limits at Infinity and Asymptotes
So far we have used the idea of limits to describe how a function behaves as its argument approaches a real value \(a\text{:}\) as \(x \to a\text{.}\) We also introduced the idea that if the value \(f(x)\) gets larger and larger without bound, we say that “\(f(x)\) is approaching infinity”, and use shorthand \(f(x) \to \infty\text{.}\)
We now consider what happens as the argument \(x\) gets larger and larger without bound, using the similar wording that “\(x\) is approaching infinity”, shorthand \(x \to \infty\text{.}\)
For \(x\) negative and increasing in magnitude, we talk of “\(x\) approaching negative infinty”, shorthand \(x \to -\infty\text{.}\)
The idea of limits at infinity is useful to describe what happens to the extreme right and left on the graph of a function like \(f(x) = \frac{x^2-1}{x^2+1}\text{,}\) whose value is very close to 1 for \(x\) values of large magnitude.
In the new shorthand, the value approaching 1 for ever larger positive \(x\) is “\(f(x) \to 1\) as \(x \to \infty\)”, or
\begin{equation*}
\lim_{x \to \infty}\frac{x^2-1}{x^2+1} = 1
\end{equation*}
The same happens going to the left, with \(x\) negative and of ever larger magnitude:
\begin{equation*}
\lim_{x \to -\infty}f(x) = 1.
\end{equation*}
Graphically, the curve gets very close to the horizontal line \(y = 1\) both to the right and left: this line is called a horizontal asymptote.
Limits at Infinity.
We measure \(x\) being close to \(\infty\) or \(-\infty\) by \(x > M\) for large \(M\) and \(x < M\) for large negative \(M\text{,}\) as we measured \(f(x)\) being close to infinity in the precise definition of infinite limits in Section 2.4. Thus, similar to that definition we have
Definition 4.6.1. Limits at Infinity.
For function \(f\) defined on infinite interval \((a,\infty)\) [i.e., for \(x > a\)], we say that the limit of \(f(x)\) as \(x\) goes to \(\infty\) is \(L\) if:
For any given positive number \(\epsilon\text{,}\) there is a number \(M\) so that
having \(x > M\) ensures that \(| f(x) - L | < \epsilon.\)
When this is true, we write
\begin{equation*}
\lim_{x \to \infty} f(x) = L.
\end{equation*}
Similarly for the limit at \(-\infty\text{,}\) using \(x < M\) instead.
Limit Laws for Limits at Infinity.
All the familiar limit laws apply for this new type of limits, so sums, products, compositions and such are all easily handled.
Example 4.6.2. The asymptotes of \(\mathbf \tan\) and \(\mathbf \tan^{-1}\).
The tangent function on interval \((-\pi/2,\pi/2)\) has one-sided limits
\begin{equation*}
\lim_{x \to -\pi/2^-}=-\infty, \quad \lim_{x \to -\pi/2^+}=\infty
\end{equation*}
and is one-to-one with range \((\infty,\infty)\text{.}\) Thus it has an inverse, \(\tan^{-1}\text{,}\) with domain \((\infty,\infty)\text{,}\) range \((-\pi/2,\pi/2)\text{,}\) and the vertical asymptotes flip over to give horizontal asymptotes \(y=-\pi/2\) to the left, and \(y=\pi/2\) to the right for this inverse.
These correspond to limits at infinity:
\begin{equation*}
\lim_{x \to -\infty}=-\pi/2, \quad \lim_{x \to \infty}=\pi/2.
\end{equation*}
\(\lim_{x \to \infty} \frac{1}{x}\) and \(\lim_{x \to -\infty} \frac{1}{x}\) are both zero; these facts that "As \(x \to \pm \infty\text{,}\) \(1/x \to 0\)" are useful building blocks in computing the horizontal asymptotes of other rational functions.
For example, these zero limits also occur for any negative power of \(x\text{:}\)
\begin{equation*}
\text {For any } r > 0, \lim_{x \to \infty} \frac{1}{x^r} = \lim_{x \to -\infty} \frac{1}{x^r} = 0.
\end{equation*}
which follows from the above by using the power rule for limits.
Checkpoint 4.6.3. Calculate \(\dsp \lim_{x \to \infty} \frac{3x^2-x-2}{5x^2+4x+1}\).
Solution.
As so often, the key idea is finding a useful simplification, so that limit laws and results above then give the answer easily. Here this means getting rid of division by infinity, by dividing top and bottom by \(x^{2}\text{.}\) Thus the limit is
\begin{equation*}
\lim_{x \to \infty} \frac{3-1/x-2/x^2}{5+4/x+1/x^2} = \lim_{x \to \infty}\frac{3-1/x-2(1/x)^2}{5+4(1/x)+(1/x)^2}
= \frac{3-0-2(0)^2}{5+4(0)+(0)^2} = \frac{3}{5}.
\end{equation*}
Exponential functions like \(e^x\) have a horizontal asymptote \(y=0\) to the left, so
\begin{equation*}
\lim_{x \to -\infty} e^x = 0 \text {, and in fact } \lim_{x \to -\infty} a^x = 0 \text { for any } a > 1.
\end{equation*}
Checkpoint 4.6.4. Calculate \(\lim_{x \to 0^{-}} e^{1/x}\).
Hint.Convert this to a limit at infinity with the change of variable \(t = 1/x\text{.}\)
Infinite limits at infinity.
Many familiar functions have values that grow without bound (“\(f(x) \to \infty\)”) as their argument grows without bound (“\(x \to \infty\)”). The simplest example is \(f(x)=x\text{,}\) and \(g(x)=e^x\) is another. This situation combines function values going to infinity (as seen with infinite limits in Section 2.2) with the argument \(x\) going to infinity (as just seen with limits at infinity).
Combining these ideas and the notation for them, we say that \(\lim_{x \to \infty} x = \infty, \; \lim_{x \to \infty} e^{x} = \infty.\) In general in this situation, we write that
\begin{equation*}
\lim_{x \to \infty} f(x) = \infty.
\end{equation*}
and likewise with the various \(-\infty\) options.
Example 4.6.5.
Calculate \(\lim_{x \to \infty} x^3\text{,}\) \(\lim_{x \to -\infty} x^3\text{,}\) \(\lim_{x \to \infty} x^2\text{,}\) and \(\lim_{x \to -\infty} x^2\text{.}\)
Example 4.6.6. Calculate \(\lim_{x \to \infty} x^2-x\).
Checkpoint 4.6.7. Sketching \(\dsp y = f(x) = \frac{x^2-1}{x^2+1}\).
When a function has horizontal asymptotes or infinite limits at infinity, we can enhance the sketching procedure from
Section 4.5 with the help of fictitious "infinite endpoints" at
\(x = \pm\infty\) added to the table described there.
First, we need the derivatives, to seek critical and inflection points:
\begin{equation*}
y' = \frac{d}{dx}\left(\frac{x^2-1}{x^2+1}\right)
= \frac{2x(x^2+1) - (x^2-1)2x}{(x^2+1)^2}
= \frac{4x}{(x^2+1)^2}
\end{equation*}
so the only critical point is at \(x=0\text{:}\) the point \((0, -1)\text{.}\) Next
\begin{equation*}
y'' = \frac{d}{dx}\left(\frac{4x}{(x^2+1)^2}\right)
= \frac{4(x^2+1)^2 - 4x 2(x^2+1) 2x}{(x^2+1)^4}
= \frac{4x^2 + 4 - 16x^2}{(x^2+1)^3}
= \frac{4(1 - 3x^2)}{(x^2+1)^3}
\end{equation*}
so the second derivative is zero at \(x = \pm 1/\sqrt{3}\text{,}\) giving the points \((\pm 1/\sqrt{3}, -1/2)\text{.}\)
In this case, we can also easily get the x-axis intercepts, where \(y=0\text{:}\) they are at \(x = \pm 1\) Thus there are five "interesting" \(x\)-values, or seven counting the infinities; arranged in order left-to-right, \(-\infty, -1, -1/\sqrt{3}, 0, 1/\sqrt{3}, 1, \infty\text{.}\)
| \(x\) |
\(-\infty\) |
\(\qquad\qquad\) |
\(-1\) |
\(\qquad\qquad\) |
\(-1/\sqrt{3}\) |
\(\qquad\qquad\) |
\(0\) |
\(\qquad\qquad\) |
\(1/\sqrt{3}\) |
\(\qquad\qquad\) |
\(1\) |
\(\qquad\qquad\) |
\(\infty\) |
| \(y\) |
\(1\) |
|
\(0\) |
|
\(-1/2\) |
|
\(-1\) |
|
\(-1/2\) |
|
\(0\) |
|
\(1\) |
| \(y'\) |
|
|
|
|
|
|
\(0\) |
|
|
|
|
|
|
| \(y''\) |
|
|
|
|
\(0\) |
|
|
|
\(0\) |
|
|
|
|
It is easy to check that \(y' < 0\) for \(x < 0\) and \(y'>0\) for \(x>0\)
\(y''\)
| \(x\) |
\(-\infty\) |
\(\qquad\qquad\) |
\(-1\) |
\(\qquad\qquad\) |
\(-1/\sqrt{3}\) |
\(\qquad\qquad\) |
\(0\) |
\(\qquad\qquad\) |
\(1/\sqrt{3}\) |
\(\qquad\qquad\) |
\(1\) |
\(\qquad\qquad\) |
\(\infty\) |
| \(y\) |
\(1\) |
|
\(0\) |
|
\(-1/2\) |
|
\(-1\) |
|
\(-1/2\) |
|
\(0\) |
|
\(1\) |
| \(y'\) |
|
\(-\) |
\(-\) |
\(-\) |
\(-\) |
\(-\) |
\(0\) |
\(+\) |
\(+\) |
\(+\) |
\(+\) |
\(+\) |
|
| \(y''\) |
|
\(-\) |
\(-\) |
\(-\) |
\(0\) |
\(+\) |
\(+\) |
\(+\) |
\(0\) |
\(-\) |
\(-\) |
\(-\) |
|
| Sketches |
With this information we can sketch the function; do this in three ways:
Using just the axis intercepts and the increasing/decreasing information given by the signs of the first derivative, \(y'\text{.}\)
Using just the axis intercepts and the concavity information given by the signs of the second derivative, \(y''\text{.}\)
Using all the above information.
Exercises Exercises
Study
Calculus Volume 1, Section 4.6 2 ; particularly Examples 21 to 26, 28, 29 and 31 and Checkpoints 20, 23–25, 27, 28 and 30 (We omit
oblique asymptotes, so skip Example 30 and Checkpoint 29), and a selection from Exercises 251–255, 256–260, 261–270, 271–281, 285–288 and 294–298.
Here the exercises are grouped in ranges by "question type", so start by trying one or two from each of the seven ranges; some suggested selections are Exercises 251, 256, 257, 259, 261, 263, 265, 267, 271, 279, 281, 285, 306 and 307.
openstax.org/books/calculus-volume-1/pages/4-6-limits-at-infinity-and-asymptotes
openstax.org/books/calculus-volume-1/pages/4-6-limits-at-infinity-and-asymptotes
