You can put this solution on YOUR website! 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.
Edwin's solution.
Stanbon is right. You are finding the horizontal asymptote. Here's
a little more detailed way to look at it.
lim
x->OO
Finding the limit of a rational function as x approaches infinity depends
on this theorem that we accept on faith:
limx->oo = 0, if n>0 and k is any constant.
Method: Divide every term on top and bottom of
by the largest power of x that appears in there.
So,
limx->oo
becomes
limx->oo
and upon simplifying the fractions, you have:
limx->oo
Now every term on top approaches 0 as x approaches infinity.
So the whole numerator approaches 0.
And the two terms of the denominator other than the 8 both
approack 0, and so, the fraction approaches
So the limit is 0.
Note:
[Note here that the denominator did not approach
0 too, and we were saved from having to face that dilemma
of having both a numerator and a denominator approach 0.
But that dilemma occurs in other problems you'll be studying.]
Edwin
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