Ta có $A=(\sqrt[3]{x^3+x^2} -x) - (\sqrt{x^2+1}-x)=\dfrac{x^2}{\sqrt[3]{(x^3+x^2)^2} +x\sqrt[3]{x^3+x^2} +x^2}-\dfrac{1}{\sqrt{x^2+1}+x}$
$=\dfrac{1}{\sqrt[3]{(1+\dfrac{1}{x})^2}+\sqrt[3]{1+\dfrac{1}{x}}+1} -\dfrac{\dfrac{1}{x}}{-\sqrt{1+\dfrac{1}{x^2}}+1}$
Vậy $\lim \limits_{x\to -\infty} A=\dfrac{1}{3}$