1. **State the problem:** Find the limit as $x \to \infty$ of the expression $$\frac{1}{\sqrt{4x^2 - 3x} - 2x}.$$\n\n2. **Rewrite the expression:** The expression is $$\frac{1}{\sqrt{4x^2 - 3x} - 2x}.$$\n\n3. **Rationalize the denominator:** Multiply numerator and denominator by the conjugate $$\sqrt{4x^2 - 3x} + 2x$$ to simplify:\n$$\frac{1}{\sqrt{4x^2 - 3x} - 2x} \cdot \frac{\sqrt{4x^2 - 3x} + 2x}{\sqrt{4x^2 - 3x} + 2x} = \frac{\sqrt{4x^2 - 3x} + 2x}{(\sqrt{4x^2 - 3x})^2 - (2x)^2}.$$\n\n4. **Simplify the denominator:**\n$$ (\sqrt{4x^2 - 3x})^2 - (2x)^2 = (4x^2 - 3x) - 4x^2 = -3x.$$\n\nSo the expression becomes:\n$$\frac{\sqrt{4x^2 - 3x} + 2x}{-3x} = -\frac{\sqrt{4x^2 - 3x} + 2x}{3x}.$$\n\n5. **Divide numerator and denominator inside the fraction by $x$ to analyze the limit:**\n$$-\frac{\sqrt{4x^2 - 3x} + 2x}{3x} = -\frac{\frac{\sqrt{4x^2 - 3x}}{x} + 2}{3}.$$\n\n6. **Simplify $\frac{\sqrt{4x^2 - 3x}}{x}$:**\n$$\frac{\sqrt{4x^2 - 3x}}{x} = \sqrt{\frac{4x^2 - 3x}{x^2}} = \sqrt{4 - \frac{3}{x}}.$$\n\n7. **Take the limit as $x \to \infty$:**\nSince $\frac{3}{x} \to 0$,\n$$\lim_{x \to \infty} \sqrt{4 - \frac{3}{x}} = \sqrt{4} = 2.$$\n\n8. **Substitute back into the expression:**\n$$\lim_{x \to \infty} -\frac{2 + 2}{3} = -\frac{4}{3}.$$\n\n**Final answer:** $$\boxed{-\frac{4}{3}}.$$