Tính \[\mathop {\lim }\limits_{x \to 3} \frac{{{x^2} - 4x + 3}}{{{x^2} - 9}}\]bằng?

\[\mathop {\lim }\limits_{x \to 3} \frac{{{x^2} - 4x + 3}}{{{x^2} - 9}} = \mathop {\lim }\limits_{x \to 3} \frac{{(x - 1)(x - 3)}}{{(x - 3)(x + 3)}} = \mathop {\lim }\limits_{x \to 3} \frac = \frac = \frac{1}{3}.\]

Đáp án cần chọn là: D