Horizontal asymptotes

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We explore functions that behave like horizontal lines as the input grows without bound.

Let’s start with an example:

Consider the function: $g(x) = \frac {1}{x}$ with the table below $\begin {array}{c|c} x & g(x)=\frac {1}{x} \\\hline 10 & \answer [given]{0.1} \\ 100 & \answer [given]{0.01} \\ 1000 & \answer [given]{0.001} \\ 10000 & \answer [given]{0.0001} \\ 100000 & \answer [given]{0.00001}\\ 1000000 & \answer [given]{0.000001} \end {array}$

What does the table tell us about $g(x)$ as $x$ grows bigger and bigger?

As $x$ grows bigger and bigger, it seems that $g(x)$ approaches $0$ . It seems that we should write $\lim _{x\to \infty }\frac {1}{x}=0$

Now consider a different function: $f(x) = \frac {6x-9}{x-1}$

Fill in the table below, rounding to $5$ decimal places. $\begin {array}{c|c} x & f(x)=\frac {6x-9}{x-1} \\ \hline 10 & \answer [given]{5.66667}\\ 100 & \answer [given]{5.96970}\\ 1000 & \answer [given]{5.99700}\\ 10000 & \answer [given]{5.99970}\\ 100000 & \answer [given]{5.99997}\\ \end {array}$ What does the table tell us about $f(x)$ as $x$ grows bigger and bigger?

As $x$ grows bigger and bigger, it seems like $f(x)$ approaches $6$ . It seems that we should write $\lim _{x\to \infty }\frac {6x-9}{x-1}=6$

If $f(x)$ becomes arbitrarily close to a specific value $L$ by making $x$ sufficiently large, we write $\lim _{x\to \infty } f(x) = L$ and we say, the limit at infinity of $f(x)$ is $L$ .

If $f(x)$ becomes arbitrarily close to a specific value $L$ by making $x$ sufficiently large and negative, we write $\lim _{x\to -\infty } f(x) = L$ and we say, the limit at negative infinity of $f(x)$ is $L$ .

Compute the limit. $\lim _{x\to \infty } \frac {6x-9}{x-1}.$ We will confirm what we guessed earlier by computing the limit at infinity.
We can assume that all the Limit Laws also apply to limits at infinity.

Write $\begin{align*} \lim_{x\to\infty}\frac{6x-9}{x-1} &= \lim_{x\to\infty}\frac{6x-9}{x-1}\cdot \frac{1/x}{1/x}\\ &=\lim_{x\to\infty}\frac{\frac{6x}{x} - \frac{9}{x}}{\frac{x}{x} - \frac{1}{x}}\\ &=\frac{\lim_{x\to\infty}\Bigl(6- \frac{9}{x}\Bigr)}{\lim_{x\to\infty}\Bigl(1- \frac{1}{x}\Bigr)}\\ &=\frac{6-9\lim_{x\to\infty} \frac{1}{x}}{1-\lim_{x\to\infty} \frac{1}{x}}\\ &=\frac{6-9(0)}{1-0}\\ &= \frac{6}{1}\\ &= 6. \end{align*}$

Sometimes one must be careful, consider this example.

$\lim _{x\to -\infty } \frac {x^3+1}{\sqrt {x^6+5}}$

In this case we multiply the numerator and denominator by $-1/x^3$ , which is a positive number as since $x\to -\infty$ , $x^3$ is a negative number. $\begin{align*} \lim_{x\to -\infty} \frac{x^3+1}{\sqrt{x^6+5}} &= \lim_{x\to -\infty} \frac{x^3+1}{\sqrt{x^6+5}} \cdot \frac{-1/x^3}{-1/x^3}\\ &= \lim_{x\to -\infty} \frac{-1-1/x^3}{\sqrt{x^6/x^6+5/x^6}}\\ &= \lim_{x\to -\infty} \frac{-1-1/x^3}{\sqrt{1+5/x^6}}\\ &= \frac{ \lim_{x\to -\infty}\Bigl(-1-1/x^3\Bigr)}{\lim_{x\to -\infty}\sqrt{1+5/x^6}}\\ &= \frac{ -1-\lim_{x\to -\infty}(1/x)^3}{\sqrt{\lim_{x\to -\infty}\Bigl(1+5/x^6\Bigr)}}\\ &= \frac{ -1-(\lim_{x\to -\infty}1/x)^3}{\sqrt{1+5(\lim_{x\to -\infty}1/x)^6}}\\ &= \frac{ -1-(0)^3}{\sqrt{1+5(0)^6}}\\ &= \frac{ -1-0}{\sqrt{1+0}}\\ &= -1. \end{align*}$

Note, since $\lim _{x\to \infty } f(x) = \lim _{x\to 0^+} f\left (\frac {1}{x}\right )$ and $\lim _{x\to -\infty } f(x) = \lim _{x\to 0^-} f\left (\frac {1}{x}\right )$ we can also apply the Squeeze Theorem when taking limits at infinity. Here is an example of a limit at infinity that uses the Squeeze Theorem, and shows that functions can, in fact, cross their horizontal asymptotes.

Compute: $\lim _{x\to \infty } \frac {\sin (7x)+4x}{x}$

We can bound our function $\frac {-1+4x}{x} \le \frac {\sin (7x)+4x}{x} \le \frac {1+4x}{x}.$ Now write with me $\begin{align*} \lim_{x\to\infty}\frac{-1+4x}{x} \cdot \frac{1/x}{1/x} &= \lim_{x\to\infty}\frac{-1/x+4}{1}\\ &=4. \end{align*}$

And we also have

$\begin{align*} \lim_{x\to\infty}\frac{1+4x}{x} \cdot \frac{1/x}{1/x} &= \lim_{x\to\infty}\frac{1/x+4}{1}\\ &=4. \end{align*}$

Since $\lim _{x\to \infty } \frac {-1+4x}{x} = 4 = \lim _{x\to \infty }\frac {1+4x}{x}$ we conclude by the Squeeze Theorem, $\lim _{x\to \infty }\frac {\sin (7x)}{x}+4 = 4$ .

If $\lim _{x\to \infty } f(x) = L \qquad \text {or}\qquad \lim _{x\to -\infty } f(x) = L,$ then the line $y=L$ is a horizontal asymptote of $f(x)$ .

Give the horizontal asymptotes of $f(x) = \frac {6x-9}{x-1}$

From our previous work, we see that $\lim _{x\to \infty } f(x) = 6$ , and upon further inspection, we see that $\lim _{x\to -\infty } f(x) = 6$ . Hence the horizontal asymptote of $f(x)$ is the line $y=6$ .

It is a common misconception that a function cannot cross an asymptote. As the next example shows, a function can cross a horizontal asymptote, and in the example this occurs an infinite number of times!

Give a horizontal asymptote of $f(x) = \frac {\sin (7x)+4x}{x}.$

Again from previous work, we see that $\lim _{x\to \infty } f(x) = \answer {4}$ . Hence $y=\answer {4}$ is a horizontal asymptote of $f(x)$ .

We conclude with an infinite limit at infinity.

Compute $\lim _{x\to \infty }\sqrt {x}.$

The function $\sqrt {x}$ grows slowly, and seems like it may have a horizontal asymptote, see the graph above. However, if we consider the square root as the inverse of the function $f(x) = x^2 , \quad x \geq 0$ .

$\sqrt {x} = y$ means that $y^2 = x$ and that $y \geq 0$ .

We see that we may square higher and higher values to obtain larger outputs. This means that $\sqrt {x}$ is unbounded, and hence $\displaystyle \lim _{x\to \infty }\sqrt {x}=\infty$ .