Помогите с решением предела

Решите задачу:

$\lim_{x \to 3} \bigg ( \dfrac{6}{x^2 - 9} - \dfrac{1}{x - 3} \bigg ) = \lim_{x \to 3} \bigg ( \dfrac{6}{(x - 3)(x + 3)} - \dfrac{1}{x - 3} \bigg) = \\ \\ \lim_{x \to 3} \dfrac{6 - x - 3 }{(x - 3)(x + 3)} = \lim_{x \to 3} \dfrac{-(x - 3)}{(x - 3)(x + 3)} = \\ \\ = \lim_{x \to 3} \dfrac{-1}{x + 3 } = -\dfrac{1}{3 + 3} = -\dfrac{1}{6}$