Помогите решить Lim x=3 (6/(x^2-9)-1/(x-3))

Решите задачу:

$\lim_{x \to \ 3} ( \frac{6}{ x^{2} -9}- \frac{1}{x-3} ) = \frac{6}{ 3^{2}-9 } - \frac{1}{3-3} = \frac{6}{0} - \frac{1}{0}$
$\lim_{x \to \ 3}( \frac{6}{(x-3)*(x+3)}- \frac{1 ^{(x+3} }{x-3} )= \lim_{x \to \ 3} \frac{6-x-3}{(x-3)*(x+3)} =$
$= \lim_{x \to \ 3} \frac{-x+3}{(x-3)*(x+3)} = \lim_{x \to \ 3} \frac{-(x-3)}{(x-3)*(x+3)} = \lim_{x \to \ 3} \frac{-1}{x+3} = \frac{-1}{3+3} =$
$=- \frac{1}{6}$