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Evaluate the limit of y=(2x^2-x-3)/(x+1), x-->-1 using L'Hopital theorem.

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museinspires | High School Teacher | (Level 1) Honors

Posted February 11, 2011 at 12:40 AM via web

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Evaluate the limit of y=(2x^2-x-3)/(x+1), x-->-1 using L'Hopital theorem.

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted February 11, 2011 at 12:44 AM (Answer #1)

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We'll verify if the limit exists, for x = -1.

We'll substitute x by -1 in the expression of the function.

lim y = lim (2x^2-x-3)/(x+1)

lim (2x^2-x-3)/(x+1) =  (2+1-3)/(-1+1) = 0/0

We've get an indetermination case.

We could solve the problem in 2 ways, at least.

We'll apply L'Hospital rule:

lim f(x)/g(x) = lim f'(x)/g'(x)

f(x) = 2x^2-x-3 => f'(x) = 4x-1

g(x) = x+1 => g'(x) = 1

lim (2x^2-x-3)/(x+1) = lim (4x-1)

lim (2x^2-x-3)/(x+1) = 4*(-1) - 1 = -5

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted February 11, 2011 at 12:48 AM (Answer #2)

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We have to find the value of lim x--> -1 [(2x^2-x-3)/(x+1)]

If we substitute x = -1 in (2x^2-x-3)/(x+1) we get 0 / 0. So L'Hopital's theorem can be applied. We now have:

lim x--> -1 [(4x -1)/(1)]

substituing x = -1, we get (-4 - 1)/1

=> -5

Limit = -5

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