Tính x^2 - 4x + 3 / x^2 - 9

24/33

Tính \(\mathop {\lim }\limits_{x \to 3} \frac{{{x^2} - 4x + 3}}{{{x^2} - 9}}\).

\[\frac{1}{2}\].

\(\frac{2}{5}\).

\(\frac{1}{3}\).

\(\frac{1}{5}\).

Giải thích

Chọn C

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