Factor the numerator using difference of cubes:
[tex]a^3 - b^3 = (a-b)(a^2 +ab+b^2)[/tex]
a = x
b = 2
[tex]\frac{x^3 - 8}{x-2} = \frac{(x-2)(x^2 +2x+4)}{x-2} = x^2 +2x+4[/tex]
Now you can evaluate limit as x=2.
[tex] \lim_{x \to 2} (x^2 +2x+4) = 2^2 +2(2)+4 = 12[/tex]
