Lim căn 3]{{27{x^3} - 4{x^2} + 5} / {x - 6} bằng

Chọn D

\[\mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt[3]{{27{x^3} - 4{x^2} + 5}}}}{{x - 6}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt[3]{{{x^3}\left( {27 - \frac{4}{x} + \frac{5}{{{x^3}}}} \right)}}}}{{x - 6}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{x\sqrt[3]{{\left( {27 - \frac{4}{x} + \frac{5}{{{x^3}}}} \right)}}}}{{x - 6}} = 3\].