Use l'Hopital theorem to find limit of (x^2-2x-8)/(x^3+8) x--> -2

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.