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

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greenbel | Honors

Posted July 16, 2013 at 4:51 PM via web

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

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted July 16, 2013 at 5:05 PM (Answer #1)

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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.`

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