$\lim_{x \to+\infty} \bigg [(\sqrt[3]{x^3 +x^2} -x) - (\sqrt{x^2+ 1 } -x )\bigg ] =\lim A$
Mà $A=\dfrac{x^2}{\sqrt[3]{(x^3+x^2)^2} +x\sqrt[3]{x^3+x^2} +x^2} -\dfrac{1}{\sqrt{x^2+1}+x}$
$=\dfrac{1}{\sqrt[3]{(1+\dfrac{1}{x})^2} +\sqrt[3]{1+\dfrac{1}{x}} +1} -\dfrac{1}{\sqrt{x^2+1}+x}$
Vậy $\lim_{x\to + \infty} A =\dfrac{1}{3}$