1. **State the problem:** Find the limit as $x \to \infty$ of the expression $$\ln(1 + x^2) - \ln(2 + x).$$ 2. **Use logarithm properties:** Recall that $$\ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right).$$ 3. **Rewrite the limit:** $$\lim_{x \to \infty} \left[\ln(1 + x^2) - \ln(2 + x)\right] = \lim_{x \to \infty} \ln\left(\frac{1 + x^2}{2 + x}\right).$$ 4. **Analyze the fraction inside the logarithm:** $$\frac{1 + x^2}{2 + x}.$$ 5. **Divide numerator and denominator by $x^2$ to understand behavior at infinity:** $$\frac{\frac{1}{x^2} + 1}{\frac{2}{x^2} + \frac{1}{x}}.$$ 6. **As $x \to \infty$, terms with $\frac{1}{x}$ or $\frac{1}{x^2}$ approach 0, so:** $$\frac{0 + 1}{0 + 0} = \frac{1}{0} = \infty.$$ 7. **Therefore, the fraction inside the logarithm grows without bound, so:** $$\lim_{x \to \infty} \ln\left(\frac{1 + x^2}{2 + x}\right) = \ln(\infty) = \infty.$$ **Final answer:** $$\boxed{\lim_{x \to \infty} \left[\ln(1 + x^2) - \ln(2 + x)\right] = \infty}.$$