The horizontal asymptote is an imaginary line that a function will go to as the domain approaches infinity. However, unlike the vertical asymptote, the function may cross the asymptote at some time and then proceed to approach it. To find the horizontal asymptote, you take the lim (x→∞) f(x) = L (the limit as x goes to infinity of f(x)). y = L is the horizontal asymptote.

An example exercise of this could be:
Find the horizontal asymptote(s) of

2x^2 - 5
---------------------.
3x^2 + x + 2

First divide the the numerator and denominator by x^2 to get

(2 - 5/x^2)
------------------.
(3 + 1/x + 2/x^2)

Take the limit of the function by taking the limit of each term of the numerator and denominator

lim (x→∞)2 - lim (x→∞)5/x^2
---------------------------------------------------------
lim (x→∞)3 + (x→∞)1/x + (x→∞)2/x^2

thus getting

2 - 0
-----------
3 + 0 + 0

2/3
so the horizontal asymptote is y = 2/3.

Now this is a long and tedious task. There is a shortcut, the Angelini Method, that my Calculus teacher taught us.

First you look at the highest exponent of x in the numerator and denominator. If the highest exponent is the same in the numerator and denominator, (such as (ax^c + dx^(c-1))/(bx^c - e)) the coefficient of x with the highest exponent in the numerator divided by the coefficient of x with the highest exponent in the denominator is the asymptote. (y = a/b would be the asymptote) As we can see from the previous example the highest exponent of the x in the numerator and denominator is both 2. The coefficient of the x^2 in the numerator is 2 and in the denominator is 3, thus, making the horizontal asymptote y = 2/3.

If you have not figured it out yet, if the highest exponents of the numerator and denominator are different, you "make" them have the same exponents by adding 0x^c (c being the highest exponent of the numerator and denominator) to the side that has the lower exponent. Then you proceed with the instructions above. It may sound confusing, but it really is not.

Suppose you want to find the horizontal asymptote of

2x^3 - 5
------------.
3x^2 + x + 2

You see that the numerator has the highest exponent, 3, and the denominator only has 2. So you add 0x^3 in the bottom and and you see that the horizontal asymptote is 3/0 or it does not exist. If you were to find the asymptote using the "long way," you would find that the asymptote would be ∞/3 or, again, it does not exist. If the function was

3x^2 + x + 2
------------,
2x^3 - 5

you will see that the asymptote would turn out to be y = 0/3, or y = 0.