Although this was posted a long time ago I have given the definitions
and some examples as they could be useful for exam revision
A) Horizontal asymptotes
A function f(x) has a horizontal asymptote of value y = a
if the limit of f(x) as x tends to plus infinity is a.
A function f(x) has a horizontal asymptote of value y = b
if the limit of f(x) as x tends to minus infinity is b.
B) Vertical asymptotes
A function f(x) has a vertical asymptote at x = a
if the limit of f(x) as x tends to a, is plus or minus infinity
(That also applies to a+ or a-)
Vertical asymptotes are easily identified if the denominator factors.
C) Oblique asymptotes
If ax +b is an oblique asymptote to f(x) then
i) The limit of [f(x)/x] as x tends to plus infinity is a
ii) The limit of [f(x) - ax] as x tends to plus infinity is b
Oblique asymptotes occur if the degree is greater in the numerator
Images and more explanations/examples at
http://www.ping.be/~ping1339/asym.htm#Oblique-asymptotes
Your questions: Neither function has oblique asymptotes as need numerator to be of greater degree. You are OK with vertical asymptotes. (If in doubt factor the denominator).
1) Rewrite as f(x) = (2 + 1/3x + 12/3x^2)/(1 - 5/3x - 2 /3x^2)
This function has horizontal asymptotes at x equals plus and minus infinity of value y = 2
2) Rewrite as f(x) = (-1 + 1/2x^2)/(x + 2)
This function has horizontal asymptotes at x equals plus and minus infinity of value y = 0
More help with Oblique asymptotes
The functions with oblique asymptotes that you will be given will have the numerator degree exactly one more than the degree of the denominator. To apply C) ii) easily, it can be useful to know how to calculate the equivalent improper polynomial P.
This is how, (without using long division).
Example 1: f(x) = (3x^2)/(x - 1) = P
Use the difference of squares formula to write
3x^2 = 3(x - 1)(x + 1) +3 then it is easy to see
P = 3x +3 + 3/(x - 1)
It follows from C) ii) that the oblique asymptote is y = 3x +3
The next one is a little harder but the idea is similar.
Example 2: f(x) = (x^3 - 5x^2 + 7)/(x^2 + 2) = P
Using (x^2 + 2)(x - 5) = x^3 - 5x^2 + 2x - 10
x^3 - 5x^2 + 7 = (x^2 + 2)(x - 5) - 2x +17 then dividing by (x^2 + 2)
P = x - 5 - (2x - 17)/(x^2 + 2)
It follows from C) ii) that the oblique asymptote is y = x - 5
I hope this lot is of some use to you now or later.
Regards - Ian
