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:

\[\begin{array}{*{20}{l}}{L = \mathop {\lim }\limits_{x \to {x_0}} \frac{{\sqrt {f\left( x \right) + 2} - f\left( x \right)}}{{f\left( x \right) - 2}}}\\{\,\,\,\,\, = \mathop {\lim }\limits_{x \to {x_0}} \frac{{f\left( x \right) + 2 - {f^2}\left( x \right)}}{{f\left( x \right) - 2}}.\frac{1}{{\sqrt {f\left( x \right) + 2} + f\left( x \right)}}}\\{\,\,\,\,\, = \mathop {\lim }\limits_{x \to {x_0}} \frac{{ - \left[ {f\left( x \right) + 1} \right]\left[ {f\left( x \right) - 2} \right]}}{{f\left( x \right) - 2}}.\frac{1}{{\sqrt {f\left( x \right) + 2} + f\left( x \right)}}}\\{\,\,\,\,\, = - \frac{3}{4}}\end{array}\]

\[\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {{x^2} + 1} + x - 1} \right)\]

\[ = \mathop {\lim }\limits_{x \to - \infty } \frac{{(\sqrt {{x^2} + 1} + x - 1)(\sqrt {{x^2} + 1} - x + 1)}}{{\sqrt {{x^2} + 1} - x + 1}}\]

\[\begin{array}{l} = \mathop {\lim }\limits_{x \to - \infty } \frac{{{x^2} + 1 - {{(x - 1)}^2}}}{{\sqrt {{x^2} + 1} - x + 1}}\\ = \mathop {\lim }\limits_{x \to - \infty } \frac{{{x^2} + 1 - {x^2} + 2x - 1}}{{\sqrt {{x^2} + 1} - x + 1}}\\ = \mathop {\lim }\limits_{x \to - \infty } \frac{{2x}}{{\sqrt {{x^2} + 1} - x + 1}}\\ = \mathop {\lim }\limits_{x \to - \infty } \frac{{\frac{{2x}}{x}}}{{\frac{{\sqrt {{x^2} + 1} }}{x} - \frac{x}{x} + \frac{1}{x}}}\\ = \mathop {\lim }\limits_{x \to - \infty } \frac{2}{{ - \sqrt {1 + \frac{1}{{{x^2}}}} - 1 + \frac{1}{x}}}\\ = \frac{2}{{ - 1 - 1 + 0}} = - 1\end{array}\]

Đáp án cần chọn là: A