Question

\(\lim\limits_{x\rightarrow0}\frac{\sqrt{2x+1}-\sqrt[3]{3x+1}}{x^2}\)

Answer 1

- Sử dụng L'Hopital:

\(\lim\limits_{x\rightarrow0}\frac{\left(2x+1\right)^{\frac{1}{2}}-\left(3x+1\right)^{\frac{1}{3}}}{x^2}=\lim\limits_{x\rightarrow0}\frac{\frac{1}{2}\left(2x+1\right)^{\frac{-1}{2}}.2-\frac{1}{3}\left(3x+1\right)^{\frac{-2}{3}}.3}{2x}\)

\(=\lim\limits_{x\rightarrow0}\frac{-\frac{1}{2}\left(2x+1\right)^{\frac{-3}{2}}.2+\frac{2}{3}\left(3x+1\right)^{\frac{-5}{3}}.3}{2}=\frac{1}{2}\)