Question

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

Answer 1

\(\left(...\right)=\lim\limits_{x\rightarrow0}\dfrac{\left(\sqrt{1+2x}-x-1\right)+x+1-\sqrt[3]{1+3x}}{x^2}\)  

\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{1+2x-x^2-2x-1}{\sqrt{1+2x}+x+1}+\dfrac{x^3+3x^2+3x+1-1-3x}{...}}{x^2}\)

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

\(=\lim\limits_{x\rightarrow0}\dfrac{-1}{\sqrt{1+2x}+x+1}+\dfrac{x+3}{\left(x+1\right)^2+\left(\sqrt[3]{1+3x}\right)^2+\left(x+1\right)\sqrt[3]{1+3x}}\)

\(=\dfrac{-1}{2}+\dfrac{3}{1+1+1}=1-\dfrac{1}{2}=\dfrac{1}{2}\)