Câu hỏi 8

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

Chọn D.