Đáp án: $1$
Giải thích các bước giải:
Ta có:
$\lim_{x\to0}\dfrac{\sqrt[3]{3x+1}-1}{x}$
$=\lim_{x\to0}\dfrac{\dfrac{3x+1-1}{(\sqrt[3]{3x+1})^2+\sqrt[3]{3x+1}+1}}{x}$
$=\lim_{x\to0}\dfrac{\dfrac{3x}{(\sqrt[3]{3x+1})^2+\sqrt[3]{3x+1}+1}}{x}$
$=\lim_{x\to0}\dfrac{3}{(\sqrt[3]{3x+1})^2+\sqrt[3]{3x+1}+1}$
$=\dfrac{3}{(\sqrt[3]{3\cdot 0+1})^2+\sqrt[3]{3\cdot 0+1}+1}$
$=1$
