Tính lim x→3 x^2−4x+3/x^2−9 bằng?

7/23

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

\[\frac{1}{5}.\]

\[\frac{2}{5}.\]

\(\frac{1}{2}\)

\[\frac{1}{3}.\]

Giải thích

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

Đáp án cần chọn là: D