Use l'Hopital theorem to find limit of (x^2-2x-8)/(x^3+8) x--> -2
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We have to find lim x--> -2 [(x^2-2x-8)/(x^3+8)] using l'Hopital's rule.
First we find out if the l'Hopital's Rule can be used here. Substituting x = -2 we get the indeterminate form 0/0; therefore it can. We now use the derivatives of the numerator and the denominator.
lim x--> -2 [ (2x - 2)/(3x^2)]
substitute x = -2
=> (-4 - 2)/12
=> -6/12
=> -1/2
The required value is lim x--> -2 [(x^2-2x-8)/(x^3+8)] = -1/2
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We know that l'Hospital theorem could be applied if the limit gives an indetermination.
We'll verify if the limit exists, for x = -2.
We'll substitute x by -2 in the expression of the function.
lim y = lim (x^2-2x-8)/(x^3+8)
lim (x^2-2x-8)/(x^3+8) = (4+4-8)/(-8+8) = 0/0
We've get an indetermination case.
We'll apply L'Hospital rule:
lim f(x)/g(x) = lim f'(x)/g'(x)
f(x) = x^2-2x-8 => f'(x) = 2x-2
g(x) = x^3+8 => g'(x) = 3x^2
lim (x^2-2x-8)/(x^3+8) = lim (2x-2)/3x^2
We'll substitute x by -2:
lim (2x-2)/3x^2 = (-4-2)/12
lim (2x-2)/3x^2 = -6/12
lim (2x-2)/3x^2 = -1/2
The limit of the function, for x->-2, is: lim (x^2-2x-8)/(x^3+8) = -1/2.
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