Tính các giới hạn sau:

$\mathop {\lim }\limits_{x \to 3} \frac{{ - {x^2} + 4x - 3}}{{{x^2} + 3x - 18}}$$= \mathop {\lim }\limits_{x \to 3} \frac{{\left( {x - 1} \right)\left( {3 - x} \right)}}{{\left( {x + 6} \right)\left( {x - 3} \right)}}$$= \mathop {\lim }\limits_{x \to 3} \frac{{ - \left( {x - 1} \right)}}{{x + 6}} = \frac{{ - 2}}{9}$.