Oblique Asymptotes

Date: 04/03/2002 at 23:47:45
From: Jo
Subject: Oblique limits

Hi, 

I'm really stuck on how to calculate oblique limits. Vertical and 
horizontal limits are fine, but I don't even know where to start with 
oblique ones. For example this question.

   x^2 / (x-3)

It seems simple but there is an oblique limit somewhere that I can't 
find. How do I find it? What method do I use?

Thanks for your time.
Jo

Date: 04/04/2002 at 12:56:03
From: Doctor Peterson
Subject: Re: Oblique limits

Hi, Jo.

I'm assuming that by "limit" you mean what I call an "asymptote."

Here is how I would do it:

First, I notice that the denominator has a lower degree than the 
numerator, and I divide both by x so that the highest power of x in 
the former is 0:

            x^2       x
    f(x) = ----- = -------
           x - 3   1 - 3/x

When x gets large, 3/x gets very small, so this will get close to 
x/1 = x. The asymptote therefore is parallel to y = x.

To find the actual asymptote, look at the limit of f(x) - x:

           x^2              x^2 - x(x - 3)          3x            3
    lim  (----- - x) = lim  -------------- = lim  ----- = lim  -------
   x->oo  x - 3       x->oo      x - 3      x->oo x - 3  x->oo 1 - 3/x

This limit is 3; so for large x, f(x) is close to x + 3, which is the 
asymptote.

Rather than this two-step method (finding the slope and then the 
offset, or y-intercept), you can just use polynomial division:

    f(x) = x^2/(x-3) = x + 3 + 9/(x-3)

The last term becomes negligible for large x, so the asymptote is 
x + 3 again.

Here are some other versions of these two methods:

   Asymptotes
   http://mathforum.org/dr.math/problems/moshfeghian4.13.96.html   

   Nonvertical Asymptotes
   http://mathforum.org/dr.math/problems/amrita.10.16.01.html   

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/   

Date: 04/04/2002 at 22:00:16
From: jo
Subject: oblique limits

Question submitted via WWW:
thanks very much for your help. i haven't found another place more 
useful for help with my questions. 
its very much appreciated.
thanks.