Answer
$$-\frac{3}{4}$$
Work Step by Step
\begin{aligned}
\lim _{x \rightarrow-\infty}(2 x+\sqrt{4 x^{2}+3 x-2}) &=\lim _{x \rightarrow-\infty}[2 x+\sqrt{4 x^{2}+3 x-2}] \cdot\left[\frac{2 x-\sqrt{4 x^{2}+3 x-2}}{2 x-\sqrt{4 x^{2}+3 x-2}}\right]\\
&=\lim _{x \rightarrow-\infty} \frac{\left(4 x^{2}\right)-\left(4 x^{2}+3 x-2\right)}{2 x-\sqrt{4 x^{2}+3 x-2}} \\
&=\lim _{x \rightarrow-\infty} \frac{-3 x+2}{2 x-\sqrt{4 x^{2}+3 x-2}}\\
&=\lim _{x \rightarrow-\infty} \frac{\frac{-3 x+2}{\sqrt{4^{2}}}}{\frac{-3 x}{\sqrt{x^{2}}}-\sqrt{4+\frac{3}{x^{2}}} x^{2}}\\
&=\lim _{x \rightarrow-\infty} \frac{\frac{-3 x+2}{2 x} \sqrt{4+\frac{3}{x^{2}}-2}}{\frac{2 x}{-x} \sqrt{4+\frac{3}{x^{2}}-x^{2}}} \\ &=\lim _{x \rightarrow-\infty} \frac{3-\frac{2}{x}}{-2-\sqrt{4 \frac{3}{x}}}\\
&=\frac{3-0}{-2-2}\\
&=-\frac{3}{4}
\end{aligned}
