1. **State the problem:** Find the limit $$\lim_{x \to \infty} \frac{\ln(x^2 + 3x + 2) - \ln(x^2 - 4x + 2)}{x^2}$$ 2. **Use logarithm properties:** Recall that $\ln a - \ln b = \ln \frac{a}{b}$. So rewrite the expression as $$\lim_{x \to \infty} \frac{\ln \left( \frac{x^2 + 3x + 2}{x^2 - 4x + 2} \right)}{x^2}$$ 3. **Analyze the fraction inside the logarithm:** $$\frac{x^2 + 3x + 2}{x^2 - 4x + 2} = \frac{x^2(1 + \frac{3}{x} + \frac{2}{x^2})}{x^2(1 - \frac{4}{x} + \frac{2}{x^2})} = \frac{1 + \frac{3}{x} + \frac{2}{x^2}}{1 - \frac{4}{x} + \frac{2}{x^2}}$$ 4. **Take the limit inside the logarithm as $x \to \infty$:** $$\lim_{x \to \infty} \frac{1 + \frac{3}{x} + \frac{2}{x^2}}{1 - \frac{4}{x} + \frac{2}{x^2}} = \frac{1 + 0 + 0}{1 - 0 + 0} = 1$$ 5. **Use the fact that $\ln(1) = 0$, so the numerator tends to $\ln(1) = 0$.** 6. **Since the numerator tends to 0 and the denominator tends to infinity $(x^2 \to \infty)$, the whole fraction tends to 0.** 7. **To be rigorous, use the expansion for logarithm near 1:** Set $$y = \frac{1 + \frac{3}{x} + \frac{2}{x^2}}{1 - \frac{4}{x} + \frac{2}{x^2}} = 1 + \frac{7}{x} + O\left(\frac{1}{x^2}\right)$$ Then $$\ln(y) \approx \frac{7}{x} + O\left(\frac{1}{x^2}\right)$$ 8. **Substitute back into the limit:** $$\lim_{x \to \infty} \frac{\ln(y)}{x^2} \approx \lim_{x \to \infty} \frac{\frac{7}{x} + O(\frac{1}{x^2})}{x^2} = \lim_{x \to \infty} \left( \frac{7}{x^3} + O\left(\frac{1}{x^4}\right) \right) = 0$$ **Final answer:** $$\boxed{0}$$