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

\[ \begin{array}{c|c} x & g(x)=\frac {1}{x} \\\hline 10 & \answer [given]{0.1} \\ 100 & \answer [given]{0.01} \\ 1000 & \answer [given]{0.001} \\ 10000 & \answer [given]{0.0001} \\ 100000 & \answer [given]{0.00001}\end{array} \]

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*}

Notice in the first step of this computation, we divided the numerator and denominator by $x$ (by multiplying them by $1/x$). This is a “trick” that frequently helps in the calculation of limits as $x\to \pm \infty $, when the function is a fraction. The important point of using $x$ here, is that the denominator has degree $1$. When we divide the denominator by $x$, the denominator approaches a nonzero constant. (We’re dividing the numerator and denominator by the term type with highest growth order from the denominator.)

Sometimes one must be careful when attempting to apply this trick. Consider this example of a non-rational function.

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

The trick we applied above to the rational function will work here, but notice that the denominator is inside a radical. In order to determine the correct factor to divide by, we will need to rewrite the radical.

\begin{align*} \sqrt {x^6 + 5} &= \sqrt { x^6\left (\dfrac {x^6}{x^6} + \dfrac {5}{x^6}\right )}\\ &= \sqrt { x^6\left ( 1 + \dfrac {5}{x^6}\right ) }\\ &= \sqrt {x^6} \sqrt {1 + \dfrac {5}{x^6} }\\ &= \sqrt {(x^3)^2} \sqrt {1 + \dfrac {5}{x^6} }\\ &= |x^3|\sqrt {1 + \dfrac {5}{x^6} }\\ \end{align*}

For this last step, remember that $\sqrt {a^2} = |a|$ when $a$ could have a negative value. Since we’re looking at $x\to -\infty $, we know that eventually $x$ will be negative and $x^3$ will be, too.

This simplification means the denominator will, in the limit as $x\to -\infty $, act like $-x^3\sqrt {1+\dfrac {5}{x^6}}$. Since the square root factor here has limit $\displaystyle \lim _{x\to -\infty } \sqrt {1 + \dfrac {5}{x^6} } = \sqrt {1 + \lim _{x\to \infty } \dfrac {5}{x^6}} = \sqrt {1+0}=1$, the growth of the denominator is controlled by the $-x^3$ factor. In this case we divide the numerator and denominator by $x^3$,

\begin{align*} \lim _{x\to -\infty } \dfrac {x^3+1}{\sqrt {x^6+5}} &= \lim _{x\to -\infty } \frac {x^3+1}{-x^3 \sqrt {1 + \dfrac {5}{x^6} }} \cdot \frac {1/x^3}{1/x^3}\\ &= \lim _{x\to -\infty } \dfrac {1+\dfrac {1}{x^3}}{\dfrac {-x^3}{x^3}\cdot \sqrt {1+\dfrac {5}{x^6}}}\\ &= \lim _{x\to -\infty } \dfrac {1+\dfrac {1}{x^3}}{-1\cdot \sqrt {1+\dfrac {5}{x^6}}} \\ &= \dfrac { \lim _{x\to -\infty }\left (1+\dfrac {1}{x^3}\right )}{\lim _{x\to -\infty }-1\cdot \sqrt {1+\dfrac {5}{x^6}}}\\ &= \dfrac {1}{-1\cdot 1} = -1. \end{align*}

In the previous example, we used the fact that $|x^3|= -x^3$ when $x$ is negative. If the limit were as $x\to \infty $, then $x$ and $x^3$ would each eventually be positive valued. This would have made $|x^3|=x^3$. Repeat the calculation from the previous example, finding the limit $\displaystyle \lim _{x\to \infty } \frac {x^3+1}{\sqrt {x^6+5}}$ taking note to use $|x^3|=x^3$ in your simplification. You’ll find that the two limits (as $x\to \infty $ and as $x \to -\infty $) have different values.

Notice that 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, $\lim _{x\to \infty }\frac {\sin (7x)}{x}+4 = 4$.

\[ \lim _{x\to \infty } f(x) = L \qquad \text {or}\qquad \lim _{x\to -\infty } f(x) = L, \]

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 $\lim _{x\to \infty } f(x) = \answer [given]{4}$. Hence $y=\answer [given]{4}$ is a horizontal asymptote of $f(x)$.

We conclude with an infinite limit at infinity.

Compute

\[ \lim _{x\to \infty } \ln (x) \]

The function $\ln (x)$ grows very slowly, and seems like it may have a horizontal asymptote, see the graph above. However, if we consider the definition of the natural log as the inverse of the exponential function

$\ln (x) = y$ means that $e^y =x$ and that $x$ is positive.

We see that we may raise $e$ to higher and higher values to obtain larger numbers. This means that $\ln (x)$ is unbounded, and hence $\lim _{x\to \infty }\ln (x)=\infty $.