Question

Помогите решить предел. 6 вариант

Answer 1

$\lim\limits _{x \to 2}\frac{\sqrt[3]{6+x}-2}{x^2-4}=\lim\limits _{x \to 2}\frac{(\sqrt[3]{6+x}-2)(\sqrt[3]{(6+x)^2}+2\sqrt[3]{6+x}\, +\, 4)}{(x-2)(x+2)(\sqrt[3]{(6+x)^2}+2\sqrt[3]{6+x}\, +\, 4\, )}=\\\\\\=\lim\limits _{x \to 2}\, \frac{(6+x)-8}{(x-2)(x+2)(\sqrt[3]{(6+x)^2}+2\sqrt[3]{6+x}\, +\, 4\, )}=\lim\limits _{x \to 2}\, \frac{1}{(x+2)(\sqrt[3]{(6+x)^2}+2\sqrt[3]{6+x}\, +\, 4\, )}=\\\\\\=\frac{1}{4\cdot (\sqrt[3]{8^2}+2\sqrt[3]{8}\, +4\, )}=\frac{1}{4\cdot (4+4+4)}=\frac{1}{4\cdot 12}=\frac{1}{48}$