Giới hạn \(\underset{x\to -\infty }{\mathop{\lim }}\,\frac{\sqrt{{{x}^{2}}+x+1}}{2x+1}\) là :

* Hướng dẫn giải

Ta có: \(\underset{x\to -\infty }{\mathop{\lim }}\,\frac{\sqrt{{{x}^{2}}+x+1}}{2x+1}=\underset{x\to -\infty }{\mathop{\lim }}\,\frac{\sqrt{{{x}^{2}}\left( 1+\frac{1}{x}+\frac{1}{{{x}^{2}}} \right)}}{x\left( 2+\frac{1}{x} \right)}\)

                                      \(=\underset{x\to -\infty }{\mathop{\lim }}\,\frac{\left| x \right|\sqrt{1+\frac{1}{x}+\frac{1}{{{x}^{2}}}}}{x\left( 2+\frac{1}{x} \right)}\)

                                      \(=\underset{x\to -\infty }{\mathop{\lim }}\,\frac{-\sqrt{1+\frac{1}{x}+\frac{1}{{{x}^{2}}}}}{2+\frac{1}{x}}=-\frac{1}{2}\)