Question: Evaluate lim x→∞ \frac{2 + 9x^3}{(1 + x)(3 + x^2)}.
1. **State the problem:** Evaluate the limit $$\lim_{x \to \infty} \frac{2 + 9x^3}{(1 + x)(3 + x^2)}.$$
2. **Rewrite the denominator:** Expand the denominator to understand its degree.
$$(1 + x)(3 + x^2) = 3 + x^2 + 3x + x^3 = x^3 + x^2 + 3x + 3.$$
3. **Identify the highest degree terms:** The highest degree term in numerator is $9x^3$, and in denominator is $x^3$.
4. **Divide numerator and denominator by $x^3$ (highest power in denominator):**
$$\lim_{x \to \infty} \frac{\frac{2}{x^3} + 9}{\frac{x^3}{x^3} + \frac{x^2}{x^3} + \frac{3x}{x^3} + \frac{3}{x^3}} = \lim_{x \to \infty} \frac{\frac{2}{x^3} + 9}{1 + \frac{1}{x} + \frac{3}{x^2} + \frac{3}{x^3}}.$$
5. **Evaluate the limit as $x \to \infty$:** Terms with $\frac{1}{x^n}$ go to zero.
$$= \frac{0 + 9}{1 + 0 + 0 + 0} = \frac{9}{1} = 9.$$
**Final answer:**
$$\boxed{9}.$$