Вопрос в картинках...

Решите задачу:

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