1. **State the problem:** Evaluate the limit $$\lim_{x \to 0} \frac{\ln(1+x) - x}{x^2}$$ using L'Hôpital's Rule. 2. **Recall L'Hôpital's Rule:** If the limit results in an indeterminate form like $$\frac{0}{0}$$ or $$\frac{\infty}{\infty}$$, then $$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$$ provided the latter limit exists. 3. **Check the form:** Substitute $x=0$: $$\ln(1+0) - 0 = 0$$ and $$0^2 = 0$$, so the limit is $$\frac{0}{0}$$, an indeterminate form. 4. **Apply L'Hôpital's Rule first time:** Differentiate numerator and denominator: $$f'(x) = \frac{d}{dx}[\ln(1+x) - x] = \frac{1}{1+x} - 1$$ $$g'(x) = \frac{d}{dx}[x^2] = 2x$$ So the limit becomes: $$\lim_{x \to 0} \frac{\frac{1}{1+x} - 1}{2x}$$ 5. **Check the form again:** Substitute $x=0$: $$\frac{1}{1+0} - 1 = 1 - 1 = 0$$ and $$2 \times 0 = 0$$, again $$\frac{0}{0}$$. 6. **Apply L'Hôpital's Rule second time:** Differentiate numerator and denominator again: $$f''(x) = \frac{d}{dx}\left(\frac{1}{1+x} - 1\right) = -\frac{1}{(1+x)^2}$$ $$g''(x) = \frac{d}{dx}[2x] = 2$$ 7. **Evaluate the new limit:** $$\lim_{x \to 0} \frac{-\frac{1}{(1+x)^2}}{2} = \frac{-\frac{1}{(1+0)^2}}{2} = \frac{-1}{2}$$ **Final answer:** $$\boxed{-\frac{1}{2}}$$ which corresponds to option b).