Lời giải:
Có:
\(\frac{\sqrt{2x+1}.\sqrt[3]{3x+1}-1}{x}=\frac{\sqrt{2x+1}(\sqrt[3]{3x+1}-1)+\sqrt{2x+1}-1}{x}=\frac{\sqrt{2x+1}.3x}{x(\sqrt[3]{(3x+1)^2}+\sqrt[3]{3x+1}+1)}+\frac{2x}{x(\sqrt{2x+1}+1)}\)
\(=\frac{3\sqrt{2x+1}}{\sqrt[3]{(3x+1)^2}+\sqrt[3]{3x+1}+1}+\frac{2}{\sqrt{2x+1}+1}\)
Do đó:
\(\lim\limits_{x\to0}\frac{\sqrt{2x+1}.\sqrt[3]{3x+1}-1}{x}=\lim\limits_{x\to0}\frac{3\sqrt{2x+1}}{\sqrt[3]{\left(3x+1\right)^2}+\sqrt[3]{3x+1}+1}+\lim\limits_{x\to0}\frac{2}{\sqrt{2x+1}+1}\)
\(=\frac{3}{1+1+1}+\frac{2}{1+1}=2\)
