Chọn phương án D.

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