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ashleyk
Dec13-04, 08:52 PM
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)

Pyrrhus
Dec13-04, 09:00 PM
the horizontal asymptote is

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

The horizontal asymptote is

y = 0

Jameson
Dec13-04, 09:06 PM
when the power in the denominator is larger than the numerator, you have a horizontal asymptote at y = 0.

Sick0Fant
Dec13-04, 11:09 PM
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.

fourier jr
Dec14-04, 02:27 AM
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