1. **State the problem:** Find the limit $$\lim_{x \to \infty} \frac{\sqrt{1 + 4x^6}}{2 - x^3}$$ as $x$ approaches infinity. 2. **Recall the rule for limits at infinity involving radicals and polynomials:** When $x$ becomes very large, the highest power terms dominate the behavior of the function. 3. **Analyze the numerator:** $$\sqrt{1 + 4x^6}$$ For large $x$, $4x^6$ dominates $1$, so $$\sqrt{1 + 4x^6} \approx \sqrt{4x^6} = 2x^3$$ 4. **Analyze the denominator:** $$2 - x^3$$ For large $x$, $-x^3$ dominates $2$, so $$2 - x^3 \approx -x^3$$ 5. **Rewrite the limit using dominant terms:** $$\lim_{x \to \infty} \frac{\sqrt{1 + 4x^6}}{2 - x^3} \approx \lim_{x \to \infty} \frac{2x^3}{-x^3}$$ 6. **Simplify the fraction:** $$\frac{2x^3}{-x^3} = \frac{\cancel{2} \cancel{x^3}}{-\cancel{x^3}} = -2$$ 7. **Conclusion:** The limit is $$\boxed{-2}$$. This means as $x$ grows very large, the expression approaches $-2$.