Chọn phương án C.

\(\begin{aligned}
\lim\limits_{x\to-\infty}\dfrac{2x^3+5x^2-3}{x^2+6x+3}&=\lim\limits_{x\to-\infty}\dfrac{x^3\left(2+\dfrac{5}{x}-\dfrac{3}{x^3}\right)}{x^2\left(1+\dfrac{6}{x}+\dfrac{3}{x^2}\right)}\\
&=\lim\limits_{x\to-\infty}x\cdot\dfrac{2+\dfrac{5}{x}-\dfrac{3}{x^3}}{1+\dfrac{6}{x}+\dfrac{3}{x^2}}\\
&=-\infty.
\end{aligned}\)

Vì \(\begin{cases}
\lim\limits_{x\to-\infty}x&=-\infty\\
\lim\limits_{x\to-\infty}\dfrac{2+\dfrac{5}{x}-\dfrac{3}{x^3}}{1+\dfrac{6}{x}+\dfrac{3}{x^2}}&=2>0.
\end{cases}\)