Dominant Terms

Date: 03/25/98 at 14:31:44
From: Addi Faerber
Subject: Dominant terms

Dear Dr. Math,

I am a senior in Calculus and we are working on the first and second 
derivative tests for increasing/decreasing values and concavity. We 
are now confronted with using these tests along with asymptotes and 
dominant terms to graph functions. I am confused on what dominant 
terms are and how to obtain their values, and once the values are 
obtained, what to do with them. Could you please help me?  

An example problem:

     f(x) = (x^4+1)\(x^2)

For the function, what are the dominant terms and the asymptotes?

Thank you for your time and effort. And thanks to the Math Forum for 
setting up this valuable tool.

Addi Faerber

Date: 03/25/98 at 15:39:41
From: Doctor Rob
Subject: Re: Dominant terms

A dominant term such as x -> a is a term which grows more rapidly than 
any other in the expression. In your example there are two terms:

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

As x -> infinity, the dominant term is x^2, because the ratio of x^2 
to 1/x^2 (which equals x^4) goes to infinity. The function behaves 
like x^2 for large values of x, because the other terms are of 
negligable size compared to the dominant one.

As x -> 0, the dominant term is 1/x^2, because the ratio of 1/x^2 to 
x^2 (which equals 1/x^4) goes to infinity. The function behaves like 
1/x^2 for small values of x, because the other terms are of negligible 
size compared to the dominant one.

There is a vertical asymptote at x = 0, as you might guess, because 
x^2 appears in the denominator. Vertical asymptotes occur where the 
numerator approaches infinity or the denominator approaches zero as
x -> a, a finite value. If this condition holds, the equation of the 
asymptote is x = a.

There are no other asymptotes. If there were a nonvertical asymptote, 
then f(x)/x would approach its slope, but f(x)/x approaches infinity 
as x -> infinity. If lim f(x)/x = m exists, and if b = lim [f(x)-m*x], 
then the equation of the nonvertical asymptote is y = m*x + b.

-Doctor Rob,  The Math Forum
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