# Need a little more help

1. Dec 13, 2004

### ashleyk

Let f(x)= (2x-5)/(x^2-4)

I need help on finding the horizontal asymptote. I have already found the vertical. (I realize this doesn't really involve calculus but I already did other part of the problem that involved a derivative)

2. Dec 13, 2004

### Pyrrhus

the horizontal asymptote is

$$\lim_{x \rightarrow \infty} \frac{2x - 5}{x^2 -4} = 0$$

The horizontal asymptote is

$$y = 0$$

3. Dec 13, 2004

### Jameson

when the power in the denominator is larger than the numerator, you have a horizontal asymptote at y = 0.

4. Dec 13, 2004

### Sick0Fant

The general idea is to multiply the numerator and denominator by the inverse of x raised to the largest power of the denominator. Then, evaluate the limit.

5. Dec 14, 2004

### fourier jr

ya divide the numeraator & denominator by (1/x^2) & get that the numerator --> 0 as x --> infinity, and the denominator --> 1 or something. so the horizontal asymptote is y=0