Finding Horizontal Asymptote for f(x)=(2x-5)/(x^2-4): Calculus Help

[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)

 

[Pyrrhus]

the horizontal asymptote is

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

The horizontal asymptote is

y = 0

 

[Jameson]

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

 

[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.

 

[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