1. **State the problem:** Find the limit as $x$ approaches $+\infty$ of the expression $$\lim_{x \to +\infty} \frac{3 - x}{\sqrt[3]{8x^3 - 3x^2 + x - 4}}.$$\n\n2. **Recall the formula and rules:** When evaluating limits involving polynomials and roots at infinity, the highest degree terms dominate. For cube roots, $$\sqrt[3]{a^3} = a.$$\n\n3. **Analyze numerator and denominator:**\n- Numerator: $3 - x$ behaves like $-x$ for large $x$.\n- Denominator: $\sqrt[3]{8x^3 - 3x^2 + x - 4}$. The dominant term inside the cube root is $8x^3$.\n\n4. **Simplify denominator:**\n$$\sqrt[3]{8x^3 - 3x^2 + x - 4} = \sqrt[3]{8x^3 \left(1 - \frac{3}{8x} + \frac{1}{8x^2} - \frac{4}{8x^3}\right)} = \sqrt[3]{8x^3} \cdot \sqrt[3]{1 - \frac{3}{8x} + \frac{1}{8x^2} - \frac{4}{8x^3}}.$$\n\n5. **Evaluate the cube root of the dominant term:**\n$$\sqrt[3]{8x^3} = 2x.$$\n\n6. **Evaluate the limit inside the cube root:**\nAs $x \to +\infty$, terms with $\frac{1}{x}$ and higher powers vanish, so\n$$\sqrt[3]{1 - \frac{3}{8x} + \frac{1}{8x^2} - \frac{4}{8x^3}} \to \sqrt[3]{1} = 1.$$\n\n7. **Rewrite the original limit:**\n$$\lim_{x \to +\infty} \frac{3 - x}{\sqrt[3]{8x^3 - 3x^2 + x - 4}} = \lim_{x \to +\infty} \frac{3 - x}{2x \cdot 1} = \lim_{x \to +\infty} \frac{3 - x}{2x}.$$\n\n8. **Simplify the fraction:**\n$$\frac{3 - x}{2x} = \frac{3}{2x} - \frac{x}{2x} = \frac{3}{2x} - \frac{1}{2}.$$\n\n9. **Evaluate the limit:**\nAs $x \to +\infty$, $\frac{3}{2x} \to 0$, so\n$$\lim_{x \to +\infty} \left(\frac{3}{2x} - \frac{1}{2}\right) = 0 - \frac{1}{2} = -\frac{1}{2}.$$\n\n**Final answer:** $$\boxed{-\frac{1}{2}}.$$