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

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

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 $.

2025-01-06 19:31:44