Question

первый пример,правилом Лопиталя

Answer 1

$\lim\limits _{x \to 0}\frac{\sqrt[3]{1-6x}-1+2x}{x^2}=[\frac{0}{0}]=\lim\limits _{x \to 0}\frac{\frac{1}{3}\cdot (1-6x)^{-\frac{2}{3}}\cdot (-6)+2}{2x}=\lim\limits _{x \to 0}\frac{-2((1-6x)^{-\frac{2}{3}}-1)}{2x}=\\\\=\lim\limits _{x \to 0}\frac{(1-6x)^{-\frac{2}{3}}-1}{x}=[\frac{0}{0} ]=lim\limits _{x \to 0}\frac{-\frac{2}{3}(1-6x)^{-\frac{5}{3}}\cdot (-6)}{1}=4\cdot \lim\limits _{x \to 0}\frac{1}{\sqrt[3]{(1-6x)^5}}=4$