# Horizontal Asymptotes

• May 2nd 2009, 06:13 PM
ny_chow
Horizontal Asymptotes
Let f be the function given by f(x) =
$\displaystyle \frac {x}{sqrt(x^2 - 4)}$

write an equation for each horizontal asymptote.

How do I find these without graphing it on my calc?
• May 2nd 2009, 06:22 PM
TheEmptySet
Quote:

Originally Posted by ny_chow
Let f be the function given by f(x) =
$\displaystyle \frac {x}{sqrt(x^2 - 4)}$

write an equation for each horizontal asymptote.

How do I find these without graphing it on my calc?

Horizontal tangents occur when the derivative is eqal to zero.

So we need to take the derivative

$\displaystyle f'(x)=\frac{\sqrt{x^2-4}-x\frac{1}{2}\cdot (x^2-4)^{-\frac{1}{2}}2x}{x^2-4}$

Factoring out a $\displaystyle (x^2-4)^{-\frac{1}{2}}$ from the numerator we get

$\displaystyle (x^2-4)^{-\frac{1}{2}}\left( \frac{(x^2-4)-x^2}{x^2-4}\right)=\frac{-4}{(x^2-4)^{\frac{3}{2}}}$

Since this is never equal to zero the function has no horizontal tangents
• May 2nd 2009, 06:30 PM
TheEmptySet
sorry I answered the wrong question (Giggle)

I found the horiztonal tangents (or lack there of)
To find the horizontal asymptote take the limit as x goes to infinity and negative infinity

$\displaystyle \lim_{x \to \infty}\frac{x}{\sqrt{x^2}-4}=\frac{1}{\sqrt{1-\frac{4}{x^2}}}=1$

$\displaystyle \lim_{x \to -\infty}\frac{x}{\sqrt{x^2}-4}=\frac{1}{\sqrt{1-\frac{4}{x^2}}}=-1$

so they are $\displaystyle y= \pm 1$