1. We are asked to evaluate the limit: $$\lim_{x \to +\infty} -x \sqrt{\frac{x^4 - 3}{9x^3 + 2x}}$$ 2. First, simplify the expression inside the square root by factoring out the highest power of $x$ from numerator and denominator: $$\sqrt{\frac{x^4 - 3}{9x^3 + 2x}} = \sqrt{\frac{x^4(1 - \frac{3}{x^4})}{x^3(9 + \frac{2}{x^2})}} = \sqrt{\frac{x^4}{x^3} \cdot \frac{1 - \frac{3}{x^4}}{9 + \frac{2}{x^2}}} = \sqrt{x \cdot \frac{1 - \frac{3}{x^4}}{9 + \frac{2}{x^2}}}$$ 3. Substitute back into the limit: $$\lim_{x \to +\infty} -x \sqrt{x \cdot \frac{1 - \frac{3}{x^4}}{9 + \frac{2}{x^2}}} = \lim_{x \to +\infty} -x \cdot \sqrt{x} \cdot \sqrt{\frac{1 - \frac{3}{x^4}}{9 + \frac{2}{x^2}}} = \lim_{x \to +\infty} -x^{\frac{3}{2}} \cdot \sqrt{\frac{1 - \frac{3}{x^4}}{9 + \frac{2}{x^2}}}$$ 4. As $x \to +\infty$, terms with $\frac{1}{x^n}$ for $n>0$ go to zero: $$\sqrt{\frac{1 - 0}{9 + 0}} = \sqrt{\frac{1}{9}} = \frac{1}{3}$$ 5. So the limit simplifies to: $$\lim_{x \to +\infty} -x^{\frac{3}{2}} \cdot \frac{1}{3} = -\frac{1}{3} \lim_{x \to +\infty} x^{\frac{3}{2}}$$ 6. Since $x^{\frac{3}{2}} \to +\infty$ as $x \to +\infty$, this limit diverges to $-\infty$. **Final answer:** $$\lim_{x \to +\infty} -x \sqrt{\frac{x^4 - 3}{9x^3 + 2x}} = -\infty$$