Nonvertical Asymptotes

Date: 10/16/2001 at 10:23:27
From: Amrita
Subject: Slant asymptotes.

I love the idea of your Web site - the archives are really helpful. 

I need to know how to find slant asymptotes. For example, does this 
problem have slant asymptotes?

   f(x)=(x^(4/3)+x^(1/3)-2)/(x^(4/3)-16).

I'm also having trouble taking the derivative of this problem:

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

I've started it, but it's so long I get confused. 

Thanks a million!
-Amrita

Date: 10/16/2001 at 14:48:51
From: Doctor Rob
Subject: Re: Slant asymptotes.

Thanks for writing to Ask Dr. Math, Amrita.

One way to find nonvertical asymptotes is to take the derivative
of f(x) and see what

       lim    f(x)/x,
   x->infinity

is. If it exists and equals m, that is the slope of an asymptote.
The y-intercept b of the asymptote is then found as

       lim    [f(x)-m*x] = b.
   x->infinity

Then the asymptote is y = m*x + b.

For example, the function f(x) = (3*x^2+1)/x has

   m =     lim    (3*x^2+1)/x^2,
       x->infinity

           lim    3 + 1/x^2 = 3.
       x->infinity

   b =     lim    [(3*x^2+1)/x-3*x],
       x->infinity

     =     lim    1/x = 0.
       x->infinity

Thus the line y = 3*x is a nonvertical asymptote.

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