Use l'Hopital theorem to find limit of (x^2-2x-8)/(x^3+8) x--> -2
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
The required value is lim x--> -2 [(x^2-2x-8)/(x^3+8)] = -1/2
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