Question 198498
lim 4x-1/8X^2+3X+1
->OO 
what is the trick to this
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Since x is getting larger without bound, 
it certainly is not zero.
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So divide numerator and denominator by x^2 to get:
{[4x-1]/x^2}/{[8x^2+3x+1]/x^2
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Expand that to get:
{(4/x - 1/x^2)] / [8 + 3/x + 1/x^2]
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Take the limit of that as x goes to infinity.
Notice that the fractions with x in the denominator
approach zero as x approaches infinity.
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The limit is (0 - 0)/(8 + 0 + 0) = 0/8 = 0
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Hope that helps.
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what you are really finding is the horizontal asymptote.
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Cheers,
Stan H.