\(\mathop {\lim }\limits_{x \to  + \infty } x\left...

B

- Hướng dẫn giải

\(\mathop {\lim }\limits_{x \to  + \infty } x\left( {\sqrt {{x^2} + 16}  - x} \right) = \mathop {\lim }\limits_{x \to  + \infty } \frac{{x\left( {{x^2} + 16 - {x^2}} \right)}}{{\sqrt {{x^2} + 16}  + x}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{16}}{{\sqrt {1 + \frac{{16}}{{{x^2}}}}  + 1}} = \frac{{16}}{2} = 8\)