B

- Hướng dẫn giải

Chọn B

$$\begin{gathered} \underset{x\to +\infty }{\lim }x\left(\sqrt{x^2+16}-x\right)=\underset{x\to +\infty }{\lim }x\frac{x^2+16-x^2}{\sqrt{x^2+16}+x} \\ =\underset{x\to +\infty }{\lim }\frac{16x}{\sqrt{x^2+16}+x}=\underset{x\to +\infty }{\lim }\frac{16}{\sqrt{1+{\displaystyle \frac{16}{x^2}}}+1} \\ =\frac{16}{1+\hspace{0.17em}1}=8 \end{gathered}$$