1. **State the problem:** Find the limit $$\lim_{x \to 0} \frac{\ln(x+1)^2}{2x}$$. 2. **Rewrite the expression:** Note that $$\ln(x+1)^2 = (\ln(x+1))^2$$. 3. **Recall the limit property:** As $$x \to 0$$, $$\ln(1+x) \approx x$$ because $$\ln(1+x)$$ can be approximated by its first-order Taylor expansion. 4. **Substitute the approximation:** Replace $$\ln(1+x)$$ by $$x$$ in the numerator: $$\frac{(\ln(1+x))^2}{2x} \approx \frac{x^2}{2x}$$. 5. **Simplify the fraction:** $$\frac{x^2}{2x} = \frac{\cancel{x^2}}{2\cancel{x}} = \frac{x}{2}$$. 6. **Evaluate the limit:** As $$x \to 0$$, $$\frac{x}{2} \to 0$$. 7. **Conclusion:** Therefore, $$\lim_{x \to 0} \frac{(\ln(1+x))^2}{2x} = 0$$. This limit evaluates to zero because the numerator approaches zero faster than the denominator.