Tính \[\mathop {\lim }\limits_{x \to 2} \frac{{\sqrt[3]{{ - 2x + 12}} - \sqrt { - x + 6} }}{{{x^2} - 2x}}\]

Ta có: \[\mathop {\lim }\limits_{x \to 2} \frac{{\sqrt[3]{{ - 2x + 12}} - \sqrt { - x + 6} }}{{{x^2} - 2x}} = \mathop {\lim }\limits_{x \to 2} \left( {\frac{{\sqrt[3]{{ - 2x + 12}} - 2}}{{{x^2} - 2x}} + \frac{{2 - \sqrt { - x + 6} }}{{{x^2} - 2x}}} \right)\]

\[\begin{array}{l}\mathop {\lim }\limits_{x \to 2} \frac{{\sqrt[3]{{ - 2x + 12}} - 2}}{{{x^2} - 2x}} = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {\sqrt[3]{{ - 2x + 12}} - 2} \right)\left( {\sqrt[3]{{{{\left( { - 2x + 12} \right)}^2}}} + 2\sqrt[3]{{ - 2x + 12}} + 4} \right)}}{{x\left( {x - 2} \right)\left( {\sqrt[3]{{{{\left( { - 2x + 12} \right)}^2}}} + 2\sqrt[3]{{ - 2x + 12}} + 4} \right)}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to 2} \frac{{ - 2(x - 2)}}{{x\left( {x - 2} \right)\left( {\sqrt[3]{{{{\left( { - 2x + 12} \right)}^2}}} + 2\sqrt[3]{{ - 2x + 12}} + 4} \right)}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{x \to 2} \frac{{ - 2}}{{x\left( {\sqrt[3]{{{{\left( { - 2x + 12} \right)}^2}}} + 2\sqrt[3]{{ - 2x + 12}} + 4} \right)}} = - \frac{1}{{12}}\end{array}\]

\[\mathop {\lim }\limits_{x \to 2} \frac{{2 - \sqrt { - x + 6} }}{{{x^2} - 2x}} = \mathop {\lim }\limits_{x \to 2} \frac{{4 - ( - x + 6)}}{{x(x - 2)(2 + \sqrt { - x + 6} )}} = \mathop {\lim }\limits_{x \to 2} \frac{1}{{x(2 + \sqrt { - x + 6} )}} = \frac{1}{8}\]

\[\mathop { \Rightarrow \lim }\limits_{x \to 2} \frac{{\sqrt[3]{{ - 2x + 12}} - \sqrt { - x + 6} }}{{{x^2} - 2x}} = - \frac{1}{{12}} + \frac{1}{8} = \frac{1}{{24}}\]