1. **Problem:** Find the limit $$\lim_{n \to \infty} \frac{(n+7)^2 + (3n+1)^2}{(n+4)^3 - (1+n)^3}$$. 2. **Formula and rules:** For limits at infinity involving polynomials, divide numerator and denominator by the highest power of $n$ in the denominator to simplify. 3. **Step-by-step solution:** 1. Expand numerator: $$(n+7)^2 = n^2 + 14n + 49$$ $$(3n+1)^2 = 9n^2 + 6n + 1$$ Sum: $$n^2 + 14n + 49 + 9n^2 + 6n + 1 = 10n^2 + 20n + 50$$ 2. Expand denominator: $$(n+4)^3 = n^3 + 12n^2 + 48n + 64$$ $$(1+n)^3 = n^3 + 3n^2 + 3n + 1$$ Difference: $$(n+4)^3 - (1+n)^3 = (n^3 + 12n^2 + 48n + 64) - (n^3 + 3n^2 + 3n + 1) = 9n^2 + 45n + 63$$ 3. Rewrite limit: $$\lim_{n \to \infty} \frac{10n^2 + 20n + 50}{9n^2 + 45n + 63}$$ 4. Divide numerator and denominator by $n^2$: $$\lim_{n \to \infty} \frac{10 + \frac{20}{n} + \frac{50}{n^2}}{9 + \frac{45}{n} + \frac{63}{n^2}}$$ 5. As $n \to \infty$, terms with $\frac{1}{n}$ and $\frac{1}{n^2}$ go to zero: $$\frac{10}{9}$$ 4. **Answer:** $$\boxed{\frac{10}{9}}$$