# Horizontal Asymptotes

• Nov 8th 2007, 05:04 PM
unluckykc
Horizontal Asymptotes
f(x)= 2x/(x^2 + 2)^(1/2).

How do I find the negative infinity limit part of the problem? I found one Horizontal asymptote at x =2, how do I find the other?
• Nov 8th 2007, 05:21 PM
swimmerxc

This procedure leads to the following general result:

* 1 * If the degree of the denominator exceeds that of the numerator, y = 0 is a horizontal asymptote;
* 2 * If the leading coefficient of the numerator is a and the leading coefficient of the denominator is b and their degrees are equal then http://www.math.csusb.edu/math110/im.../yeqaoverb.gif is a horizontal asymptote;

* 3 * If the degree of the numerator exceeds that of the denominator, there is no horizontal asymptote.

http://www.math.csusb.edu/math110/src/rationals/hasym.html
• Nov 8th 2007, 05:43 PM
unluckykc
How do I find the negative limit though?
• Nov 8th 2007, 06:01 PM
Jhevon
Quote:

Originally Posted by unluckykc
f(x)= 2x/(x^2 + 2)^(1/2).

How do I find the negative infinity limit part of the problem? I found one Horizontal asymptote at x =2, how do I find the other?

as x goes to infinity, the 2 in the denominator essentially does not matter. thus we have that:

$\displaystyle \lim_{x \to \pm \infty} \frac {2x}{\sqrt{x^2 + 2}} = \lim_{x \to \pm \infty} \frac {2x}{\sqrt{x^2}} =$ $\displaystyle \lim_{x \to \pm \infty} \frac {2x}{|x|} = \left \{ \begin{array}{lr} 2 & \mbox{ as } x \to \infty \\ -2 & \mbox { as } x \to - \infty \end{array} \right.$