# Thread: Finding horizontal asymptote

1. ## Finding horizontal asymptote

Equation: x^2 - 64 / ((3x+4)(x-5))

The horizontal asymptote is 1/3. I graphed this and for very large values of x, the graph equals appromixmately .354 and a little up. How is 1/3 an asymptote then? My books ays to make x equal a very large positive number, and then it will be (x^2)/(3x^(2)), which is 1/3. Am I looking at the graph wrongly?

2. ## Re: Finding horizontal asymptote

Originally Posted by benny92000
Equation: x^2 - 64 / ((3x+4)(x-5))

The horizontal asymptote is 1/3. I graphed this and for very large values of x, the graph equals appromixmately .354 and a little up. How is 1/3 an asymptote then? My books ays to make x equal a very large positive number, and then it will be (x^2)/(3x^(2)), which is 1/3. Am I looking at the graph wrongly?
$\displaystyle \frac{x^2 - 64}{3x^2-11x-20} = \frac{1 - \frac{64}{x^2}}{3 - \frac{11}{x} - \frac{20}{x^2}}$

what happens to each term in the rational expression as $\displaystyle x \to \infty$ ? what is left unchanged?

3. ## Re: Finding horizontal asymptote

Originally Posted by skeeter
$\displaystyle \frac{x^2 - 64}{3x^2-11x-20} ...}$
If I may borrow your markup...

Another way to think about this is to think of how "powerful" each term in a polynomial is. In the numerator, x^2 is more "powerful" than -64, since x^2 increases more rapidly (... well, -64 doesn't increase or decrease). So when x is large*, the x^2 term will dominate.

Likewise for the denominator. For x "sufficiently large", the 3x^2 term will dominate, so your fraction will be - roughly:

$\displaystyle \frac{x^2}{3x^2}$, and you can see that this is 3.

4. ## Re: Finding horizontal asymptote

That's how I would solve a limit. How do I know to make x approach positive infinity and say not negative infinity?

5. ## Re: Finding horizontal asymptote

Any terms involving x to an odd power go to 0 so it doen't matter if x goes to infinity or negative infinity.