Question

Помогите найти предел:

Answer 1

$\lim_{x \to 8} \frac{\sqrt[3]{x} - 2}{x - 8}=\lim_{x \to 8} \frac{\sqrt[3]{x} - 2}{(\sqrt[3]{x}-2)(\sqrt[3]{x^2}+2\sqrt[3]{x}+4)} =\\\\=\lim_{x \to 8} \frac{1}{\sqrt[3]{x^2}+2\sqrt[3]{x}+4}=\frac{1}{\sqrt[3]{64}+2\sqrt[3]{8}+4} =\frac{1}{4+4+4} =\frac{1}{12}$