Tìm giới hạn \(D=\lim\limits _{x \rightarrow-\infty}\left(\sqrt[3]{x^{3}+x^{2}+1}+\sqrt{x^{2}+x+1}\right)\)

* Hướng dẫn giải

\(\begin{array}{l} D=\lim\limits _{x \rightarrow-\infty}\left(\sqrt[3]{x^{3}+x^{2}+1}-x\right)+\lim\limits _{x \rightarrow-\infty}\left(\sqrt{x^{2}+x+1}+x\right)=M+N \\ M=\lim\limits _{x \rightarrow-\infty} \frac{x^{2}+1}{\sqrt[3]{\left(x^{3}+x^{2}+1\right)^{2}}+x \cdot \sqrt[3]{x^{3}+x^{2}+1}+x^{2}}=\frac{1}{3} \\ N=\lim \limits_{x \rightarrow-\infty} \frac{x+1}{\sqrt{x^{2}+x+1}-x}=\lim \limits_{x \rightarrow-\infty} \frac{1+\frac{1}{x}}{-\sqrt{1+\frac{1}{x}+\frac{1}{x^{2}}}-1}=-\frac{1}{2} \\ \text { Do đó: } B=\frac{1}{3}-\frac{1}{2}=-\frac{1}{6} \end{array}\)