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Verified Answer :

wizard123

Factor the numerator using difference of cubes:
\(a^3 - b^3 = (a-b)(a^2 +ab+b^2)\)
a = x
b = 2

\(\frac{x^3 - 8}{x-2} = \frac{(x-2)(x^2 +2x+4)}{x-2} = x^2 +2x+4\)

Now you can evaluate limit as x=2.
\(\lim_{x \to 2} (x^2 +2x+4) = 2^2 +2(2)+4 = 12\)

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