Limits Help?

There are two ways to approach this, as we get an indeterminate form (0/0 or ∞/∞) when we sub in the limit we can apply L'Hopitals rule by differentiating the numerator and denominator:

(-1/x^2) / (3x^2)

= -1 / (3x^4)

Now sub in the limit:

lim x -> -2 = -1 / [3(-2)^4]

= -1/48

The other way is to multiply the fraction through by 2x, factor the sum of two cubes in the denominator and simplify.

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EDIT (in response to askers comment):

Sure thing, multiply the fraction through by 2x:

(2 + x) / 2x(x^3 + 8)

Factor the sum of two cubes in the denominator:

(2 + x) / 2x(x + 2)(x^2 - 2x + 4)

Simplify:

1 / 2x(x^2 - 2x + 4)

Now sub in the limit:

lim x -> -2 = 1 / [2(-2) * ((-2)^2 - 2(-2) + 4))]

= 1 / [-4 * (4 + 4 + 4)]

= -1/48

((1/x) + (1/2)) / (x^3 + 8) =>

((2 + x) / (2x)) / (x^3 + 8) =>

(x + 2) / (2x * (x^3 + 8)) =>

(x + 2) / (2x * (x + 2) * (x^2 - 2x + 4)) =>

1 / (2x * (x^2 - 2x + 4))

x goes to -2

1 / (2 * (-2) * (4 + 4 + 4)) =>

1 / (-4 * 12) =>

-1/48

Whoa!!! Definitely looks like Freaky Greeky to me!!