Đáp án:
$\begin{array}{l}
\mathop {\lim }\limits_{x \to 2} \frac{{\sqrt {4x + 1} + \sqrt[3]{{x - 1}} - 4}}{{{x^2} - x - 2}}\\
= \mathop {\lim }\limits_{x \to 2} \frac{{\sqrt {4x + 1} - 3 + \sqrt[3]{{x - 1}} - 1}}{{\left( {x - 2} \right)\left( {x + 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 2} \frac{{\frac{{4x + 1 - 9}}{{\sqrt {4x + 1} + 3}} + \frac{{\left( {x - 1} \right) - 1}}{{{{\left( {\sqrt[3]{{x - 1}}} \right)}^2} + \sqrt[3]{{x - 1}} + 1}}}}{{\left( {x - 2} \right)\left( {x + 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 2} \frac{{\frac{{4\left( {x - 2} \right)}}{{\sqrt {4x + 1} + 3}} + \frac{{x - 2}}{{{{\left( {\sqrt[3]{{x - 1}}} \right)}^2} + \sqrt[3]{{x - 1}} + 1}}}}{{\left( {x - 2} \right)\left( {x + 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 2} \frac{{\frac{4}{{\sqrt {4x + 1} + 3}} + \frac{1}{{{{\left( {\sqrt[3]{{x - 1}}} \right)}^2} + \sqrt[3]{{x - 1}} + 1}}}}{{x + 1}}\\
= \mathop {\lim }\limits_{x \to 2} \frac{{\frac{4}{{\sqrt {4.2 + 1} + 3}} + \frac{1}{{1 + 1 + 1}}}}{3}\\
= \frac{1}{3}
\end{array}$
