- Hướng dẫn giải

chọn A

$\begin{array}{l}\underset{x\to -\infty }{\lim }\frac{1+3x}{\sqrt{2x^2+3}} = \underset{x\to -\infty }{\lim }\frac{1+3x}{\left|x\right|\sqrt{2+{\displaystyle \frac{3}{x^2}}}}\\ =\underset{x\to -\infty }{\lim }\frac{1+3x}{-x\sqrt{2+\frac{3}{x^2}}}\\ =\underset{x\to -\infty }{\lim }\frac{\frac{1}{x}+3}{-\sqrt{2+\frac{3}{x^2}}}=-\frac{3\sqrt{2}}{2}\end{array}$