1. **Problem:** Find the limit $$\lim_{n \to \infty} \frac{(3 - n)^4 - (2 - n)^4}{(1 + n)^4 - (n - 1)^4}$$ 2. **Formula and rules:** When $n \to \infty$, dominant terms in polynomials determine the limit. Expand or factor expressions to identify leading terms. 3. **Step-by-step solution:** 1. Rewrite terms inside powers: $$(3 - n)^4 = (-n + 3)^4 = (-(n - 3))^4 = (n - 3)^4$$ $$(2 - n)^4 = (-(n - 2))^4 = (n - 2)^4$$ $$(1 + n)^4 = (n + 1)^4$$ $$(n - 1)^4$$ 2. Expand leading terms using binomial expansion for large $n$: $$(n - 3)^4 = n^4 - 4 \cdot 3 n^3 + \ldots = n^4 - 12 n^3 + \ldots$$ $$(n - 2)^4 = n^4 - 8 n^3 + \ldots$$ $$(n + 1)^4 = n^4 + 4 n^3 + \ldots$$ $$(n - 1)^4 = n^4 - 4 n^3 + \ldots$$ 3. Substitute expansions into numerator and denominator: $$\text{Numerator} = (n^4 - 12 n^3 + \ldots) - (n^4 - 8 n^3 + \ldots) = -12 n^3 + 8 n^3 + \ldots = -4 n^3 + \ldots$$ $$\text{Denominator} = (n^4 + 4 n^3 + \ldots) - (n^4 - 4 n^3 + \ldots) = 4 n^3 + 4 n^3 + \ldots = 8 n^3 + \ldots$$ 4. For large $n$, lower order terms vanish, so limit becomes: $$\lim_{n \to \infty} \frac{-4 n^3}{8 n^3} = \frac{-4}{8} = -\frac{1}{2}$$ **Final answer:** $$\boxed{-\frac{1}{2}}$$