Cho hàm số f(x) có \(f(1) = 3\) và \(f'(1) = 2\)

Chọn A

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

\( = \mathop {\lim }\limits_{x \to 1} \frac{{\left[ {f(x) - f(1)} \right]}}{{x - 1}}.\left[ {f(1) + 3} \right] = f'\left( 1 \right).\left( {3 + 3} \right) = 2.6 = 12\).