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What is the limit of the fraction (x^3-2x^2)/(x-2) for x-->2 ?
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We notice that substituting x by 2, we'll get a indetermination case:
lim (x^3-2x^2)/(x-2) = (8 - 8)/(2-2) = 0/0
We'll factorize the numerator by x^2:
lim (x^3-2x^2)/(x-2) = lim x^2(x - 2)/(x-2)
We'll simplify and we'll get:
lim x^2(x - 2)/(x-2) = lim x^2
We'll substitute x by 2:
lim x^2 = 2^2
lim (x^3-2x^2)/(x-2) = 4
Posted by giorgiana1976 on February 26, 2011 at 6:59 PM (Answer #1)
We have to find the value of lim x-->2 [(x^3-2x^2)/(x-2)]
IF we substitute x=2, we get the indeterminate form 0/0. We can use the l'Hopital's Rule and replace the numerator and denominator with their derivatives.
We get :
lim x-->2 [(3x^2 - 4x)/1]
substitute x = 2
=> 3*2^2 - 4*2
=> 3*4 - 8
The required value of lim x-->2 [(x^3-2x^2)/(x-2)] = 4
Posted by justaguide on February 26, 2011 at 7:19 PM (Answer #2)
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