What is limit (x^3+8)/(x+2) if x go to -2?

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You need to evaluate the given limit, hence, you to replace `-2` for `x` , such that:

`lim_(x->-2) (x^3+8)/(x+2) = ((-2)^3 + 8)/(-2+2) = (-8+8)/(-2+2) = 0/0`

The indetermination 0/0 requests for you either to use a special product `a^3 + b^3 = (a+b)(a^2-ab+b^2)` to numerator, or to use l'Hospital's theorem.

Converting the numerator `x^3+8` into the product `(x+2)(x^2 - 2x + 4)` yields:

`lim_(x->-2) ((x+2)(x^2 - 2x + 4))/(x+2)`

Reducing duplicate factors yields:

`lim_(x->-2) (x^2 - 2x + 4)`

Replacing -2 for x yields:

`lim_(x->-2) (x^2 - 2x + 4) = (-2)^2 - 2*(-2) + 4 = 12`

**Hence, evaluating the given limit, using the special product `a^3 + b^3 = (a+b)(a^2-ab+b^2)` , yields**` lim_(x->-2) (x^3+8)/(x+2) = 12.`

The limit `lim_(x->-2) (x^3+8)/(x+2)` has to be determined.

If we substitute x = -2 in `(x^3+8)/(x+2)` , the numerator as well as denominator are equal to 0. We get the indeterminate form `0/0` .

As this is the case, l'Hopital's rule can be used and the numerator and denominator substituted with their derivatives.

This gives: `lim_(x->-2)(3x^2)/1`

Substituting x = -2 gives 3*(-2)^2 = 3*4 = 12

The limit `lim_(x->-2) (x^3+8)/(x+2) = 12`

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