Chọn phương án B.

\(\begin{aligned}
\lim\limits_{x\to+\infty}\dfrac{1+3x-2x^2}{x^2+5}&=\lim\limits_{x\to+\infty}\dfrac{\dfrac{1}{x^2}+\dfrac{3}{x}-2}{1+\dfrac{5}{x^2}}\\
&=\dfrac{0+0-2}{1+0}=-2.
\end{aligned}\)