Lim x^3-3x-2/x^3-8 x->2

Решите задачу:

$\lim_{x \to 2} \dfrac{x^3 - 3x - 2}{x^3 - 8} = \lim_{x \to 2} \dfrac{x^3 + x^2 - x^2 - x - 2x - 2}{(x - 2)(x^2 + 2x + 4)} = \\ \\ = \lim_{x \to 2} \dfrac{x^2(x + 1) - x(x + 1) - 2(x + 1)}{(x^2 + 2x + 4)(x - 2)} = \\ \\ = \lim_{x \to 2} \dfrac{(x^2 - x - 2)(x + 1)}{(x - 2)(x^2 + 2x + 4)} = \\ \\ = \lim_{x \to 2} \dfrac{(x^2 + x - 2x - 2)(x + 1)}{(x - 2)(x^2 + 2x + 4)} = \\ \\ = \lim_{x \to 2} \dfrac{(x - 2)(x + 1)(x + 1)}{(x - 2)(x^2 + 2x + 4)} = \\ \\$
$= \lim_{x \to 2} \dfrac{(x + 1)^2 }{x^2 + 2x + 4} = \dfrac{(2 + 1)^2}{2^2 + 2^2 + 4} = \dfrac{9}{12} = \dfrac{3}{4}$