1. **State the problem:** Find the limit as $x$ approaches infinity of $$\frac{-4x^2}{\sqrt{16x^4 + 3}}.$$\n\n2. **Recall the formula and rules:** When dealing with limits involving infinity and square roots, factor out the highest power of $x$ inside the root to simplify.\n\n3. **Simplify the denominator:**\n$$\sqrt{16x^4 + 3} = \sqrt{16x^4\left(1 + \frac{3}{16x^4}\right)} = \sqrt{16x^4} \sqrt{1 + \frac{3}{16x^4}}.$$\nSince $\sqrt{16x^4} = 4x^2$, we have\n$$\sqrt{16x^4 + 3} = 4x^2 \sqrt{1 + \frac{3}{16x^4}}.$$\n\n4. **Rewrite the original limit:**\n$$\lim_{x \to \infty} \frac{-4x^2}{4x^2 \sqrt{1 + \frac{3}{16x^4}}}.$$\n\n5. **Cancel common factors:**\n$$\lim_{x \to \infty} \frac{-\cancel{4x^2}}{\cancel{4x^2} \sqrt{1 + \frac{3}{16x^4}}} = \lim_{x \to \infty} \frac{-1}{\sqrt{1 + \frac{3}{16x^4}}}.$$\n\n6. **Evaluate the limit inside the square root:** As $x \to \infty$, $\frac{3}{16x^4} \to 0$, so\n$$\sqrt{1 + \frac{3}{16x^4}} \to \sqrt{1 + 0} = 1.$$\n\n7. **Final limit:**\n$$\lim_{x \to \infty} \frac{-1}{1} = -1.$$\n\n**Answer:** The limit is $-1$.