Answer
$$\frac{3}{2}$$
Work Step by Step
Given $$ \lim _{x \rightarrow \infty}(\sqrt{4 x^{2}+3x}-2 x) $$
\begin{aligned}
\lim _{x \rightarrow \infty}(\sqrt{4 x^{2}+3x}-2 x) &=\lim _{x \rightarrow \infty} \frac{(\sqrt{4 x^{2}+3x}-2 x)(\sqrt{4 x^{2}+3x}+2 x)}{\sqrt{4 x^{2}+3x}+2 x}\\
&=\lim _{x \rightarrow \infty} \frac{ 4 x^{2}+3x -4 x^2 }{\sqrt{4 x^{2}+3x}+2 x}\\
&=\lim _{x \rightarrow \infty} \frac{ 3x/x }{\sqrt{4 x^{2}/x^2+3x/x^2}+2 x/x^2}\\
&=\lim _{x \rightarrow \infty} \frac{ 3 }{\sqrt{4 +3 /x }+2 /x }\\
&=\frac{3}{\sqrt{4}}\\
&=\frac{3}{2}
\end{aligned}
