Lời giải:
\(\lim\limits_{x\to+\infty}\left(\sqrt[3]{x^3+4x^2}-\sqrt{x^2+3}\right)=\lim\limits_{x\to+\infty}x\left(\sqrt[3]{1+\frac{4}{x}}-\sqrt{1+\frac{3}{x^2}}\right)\)
\(=\lim\limits_{x\to+\infty}x\left[\left(\sqrt[3]{1+\frac{4}{x}}-1\right)-\left(\sqrt{1+\frac{3}{x^2}}-1\right)\right]=\lim\limits_{x\to+\infty}x\left[\frac{\frac{4}{x}}{\sqrt[3]{\left(1+\frac{4}{x}\right)^2}+\sqrt[3]{1+\frac{4}{x}}+1}-\frac{\frac{3}{x^2}}{\sqrt{1+\frac{3}{x^2}}+1}\right]\)
\(=\lim\limits_{x\to+\infty}\left[\frac{4}{\sqrt[3]{\left(1+\frac{4}{x}\right)^2}+\sqrt[3]{1+\frac{4}{x}}+1}-\frac{\frac{3}{x}}{\sqrt{1+\frac{3}{x}}+1}\right]\) \(=\frac{4}{3}-\frac{0}{2}=\frac{4}{3}\)
