calculuspowerup.com

Asymptotes

April 8th, 2009 |

Author: Merrin

An asymptote is the limiting value of a function where x or f(x) approaches infinity. There are three varieties of asymptotes that we will consider: horizontal, vertical, and oblique.

Definition 1. A horizontal asymptote occurs when at least one of these conditions is met.

$\mathop{\lim }\limits_{x\to \infty } f(x)=a\quad {\rm or}\quad \quad \mathop{\lim }\limits_{x\to -\infty } f(x)=b$

Each limit corresponds to a horizontal line at plus or minus infinity.

Definition 2. A vertical asymptote is the vertical line x = a when one of these conditions is met.

$\mathop{\lim }\limits_{x\to a^{+} } f(x)=\pm \infty \quad \quad \mathop{\lim }\limits_{x\to a^{-} } f(x)=\pm \infty$

These limits indicate that a vertical line is the limiting behavior of the function at the location x = a.

Definition 3. An oblique asymptote is the slanted line y = mx + b when either of these conditions is met.

$\mathop{\lim }\limits_{x\to \infty } (f(x)-mx-b)=0\quad \quad \mathop{\lim }\limits_{x\to -\infty } (f(x)-mx-b)=0\,$

With an oblique asymptote the limiting behavior of the function is a sloped line. The value m cannot be zero because then the asymptote would be horizontal.

Example 1. Show that the function f(x)=1/x has both a horizontal and vertical asymptote.

Solution 1. Often it is easier to see asymptotes by graphing functions limiting behaviors.

Make a sketch of f(x)=1/x. Near x = 0, f(x) approaches infinity from the right and negative infinity from the left.

$\mathop{\lim }\limits_{x\to 0^{+} } \cfrac{1}{x} =\infty \quad \, \mathop{\lim }\limits_{x\to 0^{-} } \cfrac{1}{x} =-\infty$

Both are conditions for a vertical asymptote at the line x = 0. Now we also see that the graph levels out horizontally as the function approaches positive or negative infinity.

$\mathop{\lim }\limits_{x\to \infty } \cfrac{1}{x} =0\quad \mathop{\lim }\limits_{x\to -\infty }\cfrac{1}{x}=0$

According to the definition, either of these conditions is sufficient to identify $y=0$ as a horizontal asymptote.

Example 2. What is the maximum number of horizontal asymptotes a function can have?

Solution 2. Since there are only two possible conditions for the limits at positive and negative infinity, then a function can have at most two horizontal asymptotes. Two asymptotes occur when

$\mathop{\lim }\limits_{x\to \infty } f(x)=a\quad {\rm or}\quad \, \mathop{\lim }\limits_{x\to -\infty } f(x)=b\quad \quad a\ne b$

An example of a function with two horizontal asymptotes is arctangent. Check back on the plot if you forgot how it looks.

$\mathop{\lim }\limits_{x\to \infty } \arctan x=\cfrac{\pi }{2} \quad \mathop{\lim }\limits_{x\to -\infty } \arctan x=-\cfrac{\pi }{2} .$

Example 3. What is the maximum number of vertical asymptotes a function can have?
Solution 3. A function can have infinitely many vertical asymptotes. Take for example the functions tangent or secant either of which has vertical asymptotes at the vertical lines

$x=\pi /2+n\pi , n\in {\mathbb Z}$

Example 4. Find the asymptotes of the function

$f(x)=\cfrac{(x-1)(x+1)}{(x-1)(x+2)}$

Solution 4. This function appears to have a potential singularity at x = -2 and x = 1. If we examine the limit at x = 1 we find

$\mathop{\lim }\limits_{x\to 1^{-} } \cfrac{(x-1)(x+1)}{(x-1)(x+2)} =\mathop{\lim }\limits_{x\to 1^{+} } \cfrac{(x-1)(x+1)}{(x-1)(x+2)} =\cfrac{2}{3}$

The left hand and right hand limits are equal although the function is undefined at x = 1. Neither of these limits meets the criteria for a vertical asymptote at the vertical line x = 1. Now we examine the behavior at x = -2

$\mathop{\lim }\limits_{x\to -2^{+} } \cfrac{(x-1)(x+1)}{(x-1)(x+2)} =\mathop{\lim }\limits_{\varepsilon \to 0^{+} } \cfrac{-1}{-2+\varepsilon +2} =-\infty$

One condition is only necessary to identify a vertical asymptote, so x = -2 is a vertical asymptote from the behavior of the limit from the right. Now we should also check for horizontal asymptotes. Since the limit of f(x) as x goes to plus or minus infinity is one, then y = 1 is the horizontal asymptote in both cases.

Example 5. Find the oblique asymptotes of the following function

$f(x)=\cfrac{(3x^{3} +7x^{2} +9x+3)}{x^{2} +2x+2}$

Solution 5. The degree of the numerator is greater than the denominator. We can perform polynomial division to find a linear term and the remainder.

$f(x)=3x+1+\cfrac{x+1}{x^{2} +2x+2}$