Cho hàm số f(x) xác định trên \(\mathbb{R}\) thỏa mãn\[\mathop {\lim }\limits_{x \to 2} \frac{{f\left( x \right) - 16}}{{x - 2}} = 12\]. Giới hạn \[\mathop {lim}\limits_{x \to 2} \frac{{\sqrt {2f(x) - 16} - 4}}{{{x^2} + x - 6}}\] bằng \(\frac{a}{b}\)(phân số tối giản). Tổng \[{a^2} + {b^2}\;\]bằng:

Bước 1: Tính\[\mathop {\lim }\limits_{x \to 2} f\left( x \right)\]

Đặt\[g\left( x \right) = \frac{{f\left( x \right) - 16}}{{x - 2}}\]ta có:\[f\left( x \right) = \left( {x - 2} \right)g\left( x \right) + 16\]

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

Bước 2:

Ta có:

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

\[\begin{array}{l} = >{\rm{ }}a = 3;{\rm{ }}b = 5\\ \Rightarrow {a^2} + {b^2} = 34\end{array}\]