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 $\mathop {\lim }\limits_{x \to 2} \dfrac{{\sqrt {2f\left( x \right) - 16} - 4}}{{{x^2} + x - 6}}$ bằng

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.