find the horizontal asymptote of f(x)= 8x/2x^2+1

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asked by azmeena
  1. as x gets huge, all but the highest powers don't matter, so you have

    f(x) -> 8x/2x^2 = 4/x -> 0

    whenever the denominator is a higher power than the numerator, the asymptote is y=0.

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    posted by Steve

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