Với \(\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right) - 10}}{{x - 1}} = 5\) và \(g\left( x \right) = \sqrt {f\left( x \right) + 6} - 2\sqrt[3]{{f\left( x \right) - 2}}\) thì \(\mathop {\lim }\limits_{x \to 1} \frac{1}{{\left( {\sqrt x - 1} \right)g\left( x \right)}}\) bằng:    

 \(\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right) - 10}}{{x - 1}} = 5{\rm{ n\^e n }}\mathop {\lim }\limits_{x \to 1} \left[ {f\left( x \right) - 10} \right] = 0 \Rightarrow \mathop {\lim }\limits_{x \to 1} f\left( x \right) = 10\).

Ta có \(g\left( x \right) = \sqrt {f\left( x \right) + 6}  - 2\sqrt[3]{{f\left( x \right) - 2}} = \left[ {\sqrt {f\left( x \right) + 6}  - 4} \right] - \left[ {2\sqrt[3]{{f\left( x \right) - 2}} - 4} \right]\)

            \( = \frac{{f\left( x \right) - 10}}{{\sqrt {f\left( x \right) + 6}  + 4}} - \frac{{2\left[ {f\left( x \right) - 10} \right]}}{{{{\left[ {\sqrt[3]{{f\left( x \right) - 2}}} \right]}^2} + 2\sqrt[3]{{f\left( x \right) - 2}} + 4}}\).

Suy ra \(\left( {\sqrt x  - 1} \right)g\left( x \right) = \left[ {\frac{{f\left( x \right) - 10}}{{\sqrt {f\left( x \right) + 6}  + 4}} - \frac{{2\left( {f\left( x \right) - 10} \right)}}{{{{\left( {\sqrt[3]{{f\left( x \right) - 2}}} \right)}^2} + 2\sqrt[3]{{f\left( x \right) - 2}} + 4}}} \right]\left( {\sqrt x  - 1} \right)\)

\[ = \frac{{f\left( x \right) - 10}}{{x - 1}}\left[ {\frac{1}{{\sqrt {f\left( x \right) + 6}  + 4}} - \frac{2}{{{{\left( {\sqrt[3]{{f\left( x \right) - 2}}} \right)}^2} + 2\sqrt[3]{{f\left( x \right) - 2}} + 4}}} \right]\left( {x - 1} \right)\left( {\sqrt x  - 1} \right)\]

\( = \frac{{f\left( x \right) - 10}}{{x - 1}}\left[ {\frac{1}{{\sqrt {f\left( x \right) + 6}  + 4}} - \frac{2}{{{{\left( {\sqrt[3]{{f\left( x \right) - 2}}} \right)}^2} + 2\sqrt[3]{{f\left( x \right) - 2}} + 4}}} \right]\left( {\sqrt x  + 1} \right){\left( {\sqrt x  - 1} \right)^2}\)

Khi đó \(\mathop {\lim }\limits_{x \to 1} \left( {\sqrt x  - 1} \right)g\left( x \right)\)

\( = \mathop {\lim }\limits_{x \to 1} \left[ {\frac{{f\left( x \right) - 10}}{{x - 1}}\left( {\frac{1}{{\sqrt {f\left( x \right) + 6}  + 4}} - \frac{2}{{{{\left( {\sqrt[3]{{f\left( x \right) - 2}}} \right)}^2} + 2\sqrt[3]{{f\left( x \right) - 2}} + 4}}} \right)\left( {\sqrt x  + 1} \right){{\left( {\sqrt x  - 1} \right)}^2}} \right]\)

\( = 5\left[ {\frac{1}{{\sqrt {10 + 6}  + 4}} - \frac{2}{{{{\left( {\sqrt[3]{{10 - 2}}} \right)}^2} + 2\sqrt[3]{{10 - 2}} + 4}}} \right]\left( {\sqrt 1  + 1} \right){\left( {\sqrt 1  - 1} \right)^2} = 0\).

Mặt khác \(\mathop {\lim }\limits_{x \to 1} \left[ {\frac{{f\left( x \right) - 10}}{{x - 1}}\left( {\frac{1}{{\sqrt {f\left( x \right) + 6}  + 4}} - \frac{2}{{{{\left( {\sqrt[3]{{f\left( x \right) - 2}}} \right)}^2} + 2\sqrt[3]{{f\left( x \right) - 2}} + 4}}} \right)\left( {\sqrt x  + 1} \right)} \right] =  - \frac{5}{{12}}{\rm{. }}\)

Và \({\left( {\sqrt x  - 1} \right)^2} > 0\) với \(\forall x \ne 1\) nên \(\mathop {\lim }\limits_{x \to 1} \frac{1}{{\left( {\sqrt x  - 1} \right)g\left( x \right)}} =  - \infty \). Chọn A.