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

Chọn D.

Ta có: limx→−∞x2+x+12x+1=limx→−∞x2(1+1x+1x2)x(2+1x)

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

=limx→−∞−1+1x+1x22+1x=−12