Horizontal Asymptotes

y=L

Such that

Translation: When x approaches +/-, there is a y value, L, which can be determined. This value is never a point on the graph, but rather L, the value from the limit, is where the horizontal asymptote is located.

In this example, there are both vertical and horizontal asymptotes. Let's focus on the horizontal one. Here, we see that as x approaches infinity on both sides of the graph (+ and -), there is a horizontal asymptote at y=3. In the equation, , x is said to approach infinity. To see how this is applied to the graph, plug in infinity for x into the equation. The result is an extremely small, negligible fraction, and 3. From this, the obvious horizontal asymptote, as x approaches infinity, must be 3, as it is the only numerical value left from the equation.

Vertical Asymptotes

x=C

Such that

Translation: As the x value approaches the asymptote, a vertical line at x=C, from either the left or right hand side of the equation, the y value approaches +/-. In the equation of the limit, this will be seen as x approaching a value that makes the denominator equal to zero, so the limit will approach infinity.

In this example, there is a vertical asymptote at x=2. As the x values approach 2 from both sides of the graph (+ and -), the y values veer off to + and - . To analyze this in terms of the equation, set the denominator of the equation, x-2 equal to 0; 0 = x-2. From this, you can find the x value of the vertical asymptote, as x=2.

SHORTCUTS for discovering Vertical and Horizontal asymptotes by looking at the equation

Answer: b, because 12/3=4, and both share the same exponential value, so their coefficients can be divided, as each is leading in the numerator and denominator.

Find the vertical asymptote and draw a graph.

Answer: x=2 is a vertical asymptote for f(x)

Explanation: Given that , we see that the function f is discontinuous at x=-2 and x=2. Since , x=-2 is a "removable" discontinuity. Since and , x=2 is the vertical asymptote of f.

Graph:

SOURCE: http://archives.math.utk.edu/visual.calculus/1/vertical.1/index.html

Slant (Oblique) Asymptotes

If the degree of the numerator is one more than the degree of the denominator, you have neither a horizontal or vertical asymptote, but a slant asymptote. To find the value of the asymptote, you must divide the numerator by the denominator using polynomial division, and you can ignore the remainder. The equation you get as a result will be the asymptote on the graph, which looks odd, but that's ok, it's supposed to be that way.

Example Problem: Draw a graph of and define the slant asymptote of .

Answer: The slant asymptote is at y=-3x-3, and the graph is drawn such:

SOURCE: http://www.purplemath.com/modules/asymtote3.htm

More Practice Problems

If f(x) and g(x) are such that

lim f(x) as x a =

and

lim g(x) as x a = 0

then

a. lim [ f(x) . g(x) ] as x a is always equal to 0

b. lim [ f(x) . g(x) ] as x a is never equal to 0

c. lim [ f(x) . g(x) ] as x a may be or

d. lim [ f(x) . g(x) ] as x a may be equal to a finite value.

Answer :

c and d. Try the following functions:

f(x) = 1 / x and g(x) = 2x as x approaches 0.

f(x) = 1 / x 2 and g(x) = x as x approaches 0.

Find the horizontal asymptote of the graph of

a. 0

b. +/-1

c. +/-4

d. nonexistent

Answer: b, there is a horizontal asymptote at y=+/-1.

Free Response

Determine the following limits, and graph the vertical asymptotes of the equation of f(x).

Answer: . There are asymptotes at x=4 and x=-3.

The limits are: and

and

LINKS

For a lot of good starting material on limits, http://www.mecca.org/~halfacre/MATH/limits.htm

For a good animated step-by-step tutorial on horizontal asymptotes, http://archives.math.utk.edu/visual.calculus/1/horizontal.5/3.html

For an animated tutorial on vertical asymptotes, http://archives.math.utk.edu/visual.calculus/1/vertical.4/3.html

For an applet to calculate limits for a variety of equations, http://math.dartmouth.edu/~klbooksite/appfolder/203unit/Limits.html

Bryce N.

i like cupcakes