Đáp án:
\[\lim\limits_{x\to 0}\dfrac{2\sqrt{x+1}-\sqrt[3]{8-x}}{x}=\dfrac{13}{12}\]
Giải thích các bước giải:
\(\lim\limits_{x\to 0}\dfrac{2\sqrt{x+1}-\sqrt[3]{8-x}}{x}\\=\lim_{x\to 0}\dfrac{\left(2\sqrt{x+1}-2\right)+\left(2-\sqrt[3]{8-x}\right)}{x}\)
\(=\lim\limits_{x\to 0}\left(\dfrac{2\left(\sqrt{x+1}-1\right)}{x}+\dfrac{2-\sqrt[3]{8-x}}{x}\right)\)
\(=\lim\limits_{x\to 0}\left(\dfrac{2x}{x\left(\sqrt{x+1}+1\right)}+\dfrac{8-(8-x)}{x\left(4+2\sqrt[3]{8-x}+\sqrt[3]{(8-x)^2}\right)}\right)\)
\(=\lim\limits_{x\to 0}\left(\dfrac 2{\sqrt{x+1}+1}+\dfrac{1}{4+2\sqrt[3]{8-x}+\sqrt[3]{\left(8-x\right)^2}}\right)\)
\(=\dfrac 2{\sqrt{0+1}+1}+\dfrac{1}{4+2\sqrt[3]{8-0}+\sqrt[3]{(8-0)^2}}\\ =1+\dfrac{1}{13}\)
\(=\dfrac{13}{12}\)
Vậy \(\lim\limits_{x\to 0}\dfrac{2\sqrt{x+1}-\sqrt[3]{8-x}}{x}=\dfrac{13}{12}\)
