Question

lim_x→0\(\dfrac{2\sqrt{1+x}-\sqrt[3]{8-x}}{x}\)

Answer 1

\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{4+2x}-\sqrt[3]{8-x}}{x}=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{4+2x}-2}{x}+\lim\limits_{x\rightarrow0}\dfrac{2-\sqrt[3]{8-x}}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{4+2x-4}{x\left(\sqrt{4+2x}+2\right)}+\lim\limits_{x\rightarrow0}\dfrac{8-8+x}{x\left(\sqrt[3]{\left(8-x\right)^2}+2.\sqrt[3]{8-x}+4\right)}\)

\(=\dfrac{2}{\sqrt{4}+2}+\dfrac{1}{\sqrt[3]{8^2}+2.\sqrt[3]{8}+4}=\dfrac{7}{12}\)