Cho hàm số \(f\left( x \right)\) xác định trên \(\mathbb{R}\) thỏa mãn \(\mathop {\lim }\limits_{x \to 2} \dfrac{{f\left( x \right) - 16}}{{x - 2}} = 12.\) Giới hạn \(\mathop {\lim...

* Hướng dẫn giải

Đặt \(g\left( x \right) = \dfrac{{f\left( x \right) - 16}}{{x - 2}}\) ta có: \(f\left( x \right) = \left( {x - 2} \right)g\left( x \right) + 16\).

\( \Rightarrow \mathop {\lim }\limits_{x \to 2} f\left( x \right) = \mathop {\lim }\limits_{x \to 2} \left[ {\left( {x - 2} \right)g\left( x \right) + 16} \right] = 16\).

Ta có:

\(\begin{array}{l}\,\,\,\,\mathop {\lim }\limits_{x \to 2} \dfrac{{\sqrt {2f\left( x \right) - 16}  - 4}}{{{x^2} + x - 6}}\\ = \mathop {\lim }\limits_{x \to 2} \dfrac{{2f\left( x \right) - 16 - 16}}{{\left( {{x^2} + x - 6} \right)\left( {\sqrt {2f\left( x \right) - 16}  + 4} \right)}}\\ = \mathop {\lim }\limits_{x \to 2} \dfrac{{2f\left( x \right) - 32}}{{\left( {x - 2} \right)\left( {x + 3} \right)\left( {\sqrt {2f\left( x \right) - 16}  + 4} \right)}}\\ = \mathop {\lim }\limits_{x \to 2} \dfrac{{f\left( x \right) - 16}}{{x - 2}}.\mathop {\lim }\limits_{x \to 2} \dfrac{2}{{\left( {x + 3} \right)\left( {\sqrt {2f\left( x \right) - 16}  + 4} \right)}}\\ = 12.\dfrac{2}{{5.\left( {\sqrt {2.16 - 16}  + 4} \right)}} = \dfrac{3}{5}\end{array}\)

Chọn B.