Tính giới hạn sau \(\mathop {\lim }\limits_{x \to \,\,3} \dfrac{{\sqrt {5x - 6} .\sqrt[3]{{3x - 1}} - 2x}}{{{x^2} - x - 6}}\).

* Hướng dẫn giải

\(\mathop {\lim }\limits_{x \to \,\,3} \dfrac{{\sqrt {5x - 6} .\left( {\sqrt[3]{{3x - 1}} - 2} \right) + 2\sqrt {5x - 6}  - 2x}}{{{x^2} - x - 6}}\) \( = \mathop {\lim }\limits_{x \to \,\,3} \left[ {\dfrac{{\sqrt {5x - 6} .\left( {\sqrt[3]{{3x - 1}} - 2} \right)}}{{{x^2} - x - 6}} + \dfrac{{2\sqrt {5x - 6}  - 2x}}{{{x^2} - x - 6}}} \right]\)

\( = \mathop {\lim }\limits_{x \to \,\,3} \left[ {\dfrac{{3\sqrt {5x - 6} \left( {x - 3} \right)}}{{\left( {x - 3} \right)\left( {x + 2} \right)\left[ {{{\left( {\sqrt[3]{{3x - 1}}} \right)}^2} + 2\sqrt[3]{{3x - 1}} + 4} \right]}} + \dfrac{{2\left( {x - 3} \right)\left( { - x + 2} \right)}}{{\left( {x - 3} \right)\left( {x + 2} \right)\left( {\sqrt {5x - 6}  + x} \right)}}} \right]\)

\( = \mathop {\lim }\limits_{x \to \,\,3} \left[ {\dfrac{{3\sqrt {5x - 6} }}{{\left( {x + 2} \right)\left[ {{{\left( {\sqrt[3]{{3x - 1}}} \right)}^2} + 2\sqrt[3]{{3x - 1}} + 4} \right]}} + \dfrac{{2\left( { - x + 2} \right)}}{{\left( {x + 2} \right)\left( {\sqrt {5x - 6}  + x} \right)}}} \right]\)\( = \dfrac{1}{{12}}\)

Chọn C.