Câu 18
Ta có
$\underset{x \to 0}{\lim} \dfrac{\sqrt{4 + 4x} - \sqrt[3]{8 + 12x}}{x^2} = \underset{x \to 0}{\lim} \dfrac{\sqrt{4 + 4x} - 2 - (\sqrt[3]{8 + 12x} - 2)}{x^2}$
$= \underset{x \to 0}{\lim} \dfrac{(4 + 4x - 4) - (8 + 12x - 8)}{x^2(\sqrt{4 + 4x} + 2)(\sqrt[3]{8 + 12x} + 2)}$
$= \underset{x \to 0}{\lim} \dfrac{-8x}{x^2(\sqrt{4 + 4x} + 2)(\sqrt[3]{8 + 12x} + 2)}$
$= \underset{x \to 0}{\lim} \dfrac{-8}{x(\sqrt{4 + 4x} + 2)(\sqrt[3]{8 + 12x} + 2)}$
$= -\infty$
