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

* Hướng dẫn giải

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