Giới hạn lim(căn(x^2+x+1)/(2x+1)) là :

* Hướng dẫn giải

Chọn D.

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

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

$=\underset{x\to -\infty }{\lim }\frac{-\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}}{2+\frac{1}{x}}=-\frac{1}{2}$