\(f\left(x\right)=\dfrac{\sqrt{1+2x}-\sqrt[3]{1+3x}}{\sqrt{1-x}-x-1}\)
\(=\dfrac{\sqrt{1+2x}-\left(x+1\right)}{\sqrt{1-x}-x-1}+\dfrac{\left(x+1\right)-\sqrt[3]{1+3x}}{\sqrt{1-x}-x-1}\)
\(=\dfrac{\left(1+2x-x^2-2x-1\right)\left(\sqrt{1-x}+x+1\right)}{\left(1-x-x^2-2x-1\right)\left(\sqrt{1+2x}+x+1\right)}+\dfrac{\left(x^3+3x^2+3x+1-1-3x\right)\left(\sqrt{1-x}+x+1\right)}{\left(1-x-x^2-2x-1\right)\left[\left(x+1\right)^2+\sqrt[3]{\left(1+3x\right)^2}+\left(x+1\right)\sqrt[3]{1+3x}\right]}\)
\(=\dfrac{x^2\left(\sqrt{1-x}+x+1\right)}{\left(x^2+3x\right)\left(\sqrt{1+2x}+x+1\right)}-\dfrac{\left(x^3+3x^2\right)\left(\sqrt{1-x}+x+1\right)}{\left(x^2+3x\right)\left[\left(x+1\right)^2+\sqrt[3]{\left(1+3x\right)^2}+\left(x+1\right)\sqrt[3]{1+3x}\right]}\)
\(=\dfrac{x\left(\sqrt{1-x}+x+1\right)}{\left(x+3\right)\left(\sqrt{1+2x}+x+1\right)}-\dfrac{\left(x+3\right)\left(\sqrt{1-x}+x+1\right)}{\left(x+1\right)^2+\sqrt[3]{\left(1+3x\right)^2}+\left(x+1\right)\sqrt[3]{1+3x}}\)
Khi đó:
\(A=\lim\limits_{x\rightarrow0}f\left(x\right)=-\dfrac{3.2}{1+1+1}=-2\)
