1. Problem: Calculate the limit \(\lim_{n \to \infty} \frac{2n^2 + 4n + 3}{(n + 3)(\sqrt{16n^2 + 2n - 1} + n - 4)}\). 2. Formula and rules: When calculating limits involving polynomials and square roots as \(n \to \infty\), divide numerator and denominator by the highest power of \(n\) present to simplify. 3. Work: - Numerator: \(2n^2 + 4n + 3\) - Denominator: \((n + 3)(\sqrt{16n^2 + 2n - 1} + n - 4)\) 4. Simplify inside the square root: $$\sqrt{16n^2 + 2n - 1} = n\sqrt{16 + \frac{2}{n} - \frac{1}{n^2}} = n(4 + \frac{1}{4n} + o(\frac{1}{n}))$$ 5. So denominator becomes: $$(n + 3)(n(4 + \frac{1}{4n}) + n - 4) = (n + 3)(n(5 + \frac{1}{4n}) - 4)$$ 6. Approximate denominator: $$(n + 3)(5n + \frac{1}{4} - 4) = (n + 3)(5n - 3.75) = 5n^2 + 15n - 3.75n - 11.25 = 5n^2 + 11.25n - 11.25$$ 7. Divide numerator and denominator by \(n^2\): $$\frac{2 + \frac{4}{n} + \frac{3}{n^2}}{5 + \frac{11.25}{n} - \frac{11.25}{n^2}}$$ 8. Taking limit as \(n \to \infty\), terms with \(\frac{1}{n}\) and \(\frac{1}{n^2}\) vanish: $$\lim_{n \to \infty} \frac{2}{5} = \frac{2}{5}$$ Final answer: \(\boxed{\frac{2}{5}}\)