Đáp án: $\frac{2}{3}$
Giải thích các bước giải:
$\lim_{x \to 1} $ $\frac{\sqrt[3]{x}-1}{\sqrt{x}-1}$
= $\lim_{x \to 1} $ $\frac{(\sqrt[3]{x}-1)(\sqrt[3]{x^2} +\sqrt[3]{x}+1)(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt[3]{x^2} +\sqrt[3]{x}+1)(\sqrt{x}+1)}$
=$\lim_{x \to 1} $ $\frac{( (\sqrt[3]{x})^3 -1^3) (\sqrt{x}+1) }{(x-1)(\sqrt[3]{x^2} +\sqrt[3]{x}+1)}$
$\\$ =$\lim_{x \to 1} $ $\frac{(x-1)(\sqrt{x}+1)}{(x-1)(\sqrt[3]{x^2} +\sqrt[3]{x}+1)}$
=$\lim_{x \to 1} $ $\frac{(\sqrt{x}+1)}{(\sqrt[3]{x^2} +\sqrt[3]{x}+1)}$
= $\frac{(\sqrt{1}+1)}{(\sqrt[3]{1^2} +\sqrt[3]{1}+1)}$
= $\frac{2}{3}$
