\(\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + x} - x} \right)\)bằng:

* Hướng dẫn giải

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