Đáp án: $B$
Giải thích các bước giải:
$\lim\limits_{x\to 0}\dfrac{2\sqrt{1+x}-\sqrt[3]{8-x}}{x}$
$=\lim\limits_{x\to 0}\dfrac{\sqrt{4+4x}-\sqrt[3]{8-x}}{x}$
$=\lim\limits_{x\to 0}\dfrac{\sqrt{4+4x}-2+2-\sqrt[3]{8-x}}{x}$
$=\lim\limits_{x\to 0}\Big( \dfrac{4+4x-4}{x(\sqrt{4+4x}+2)} +\dfrac{8-8+x}{x(4+2.\sqrt[3]{8-x}+\sqrt[3]{8-x}^2}\Big)$
$=\lim\limits_{x\to 0}\Big( \dfrac{4}{\sqrt{4+4x}+2} +\dfrac{1}{4+2.\sqrt[3]{8-x}+\sqrt[3]{8-x}^2}\Big)$
$=\dfrac{4}{\sqrt4+2}+\dfrac{1}{4+2.\sqrt[3]{8}+\sqrt[3]{8}^2}$
$=\dfrac{13}{12}$
