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Need a little more help

  1. Dec 13, 2004 #1
    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. jcsd
  3. Dec 13, 2004 #2


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    Homework Helper

    the horizontal asymptote is

    [tex] \lim_{x \rightarrow \infty} \frac{2x - 5}{x^2 -4} = 0[/tex]

    The horizontal asymptote is

    [tex] y = 0 [/tex]
  4. Dec 13, 2004 #3
    when the power in the denominator is larger than the numerator, you have a horizontal asymptote at y = 0.
  5. Dec 13, 2004 #4
    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.
  6. Dec 14, 2004 #5
    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
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