Solution: Divide numerator and denominator by $ n^2 $: $ g(n) = \frac{3 + \frac{5}{n}}{1 + \frac{1}{n^2}} $. As $ n \to \infty $, $ \frac{5}{n} \to 0 $ and $ \frac{1}{n^2} \to 0 $. Thus, the limit is $ \frac{3}{1} = \boxed{3} $.

["Title: Evaluating the Limit of a Rational Function: Simplifying $ g(n) = \frac{3 + \frac{5}{n}}{1 + \frac{1}{n^2}} $ as $ n \ o \infty $", "Understanding limits in sequences and functions is fundamental in calculus, especially when analyzing behavior as variables grow arbitrarily large. A common technique involves simplifying complex expressions by dividing numerator and denominator by a term that shows dominant growth—often $ n^2 $ in rational functions involving $ n $. This article explores the behavior of the function\n$$\ng(n) = \frac{3 + \frac{5}{n}}{1 + \frac{1}{n^2}}\n$$\nas $ n \ o \infty $, demonstrating how dividing by $ n^2 $ clarifies the limit.", "## Simplifying the Expression by Dividing by $ n^2 $", "As $ n $ becomes very large (i.e., approaches infinity), the terms $ \frac{5}{n} $ and $ \frac{1}{n^2} $ approach zero:\n- $ \frac{5}{n} \ o 0 $\n- $ \frac{1}{n^2} \ o 0 $", "To clarify this limit analytically, we divide both the numerator and the denominator by $ n^2 $:", "$$\ng(n) = \frac{3 + \frac{5}{n}}{1 + \frac{1}{n^2}} = \frac{\frac{3}{n^2} + \frac{5}{n^3}}{\frac{1}{n^2} + \frac{1}{n^4}}\n$$", "This transformation reveals the leading behavior of each term:\n- In the numerator: $ \frac{3}{n^2} \ o 0 $ and $ \frac{5}{n^3} \ o 0 $\n- In the denominator: $ \frac{1}{n^2} \ o 0 $ and $ \frac{1}{n^4} \ o 0 $", "Thus, as $ n \ o \infty $, both numerator and denominator tend toward 0 — but crucially, the structure shows that $ g(n) $ approaches a finite limit rather than $ \frac{0}{0} $ indeterminacy.", "## Computing the Limit", "Using the known fact that:\n- $ \frac{5}{n} \ o 0 $ and $ \frac{1}{n^2} \ o 0 $ as $ n \ o \infty $,", "we directly substitute:\n$$\n\lim_{n \ o \infty} g(n) = \frac{3 + 0}{1 + 0} = \frac{3}{1} = \boxed{3}\n$$", "This result means that as $ n $ increases without bound, the function $ g(n) $ settles steadily at 3.", "## Why Dividing by $ n^2 $ Helps", "Dividing by $ n^2 $ (or the dominant term $ n^2 $ in the denominator) standardizes the scaling of each component, making asymptotic behavior immediately visible. It converts expressions involving inverse powers of $ n $ into forms like $ \frac{\ ext{small}}{0} $, clarifying convergence without relying on L’Hôpital’s Rule or direct substitution.", "This technique is especially useful in sequences and power series, where such simplifications enable clean, intuitive limit evaluation.", "## Conclusion", "Evaluating limits by strategically dividing numerator and denominator by $ n^2 $ reveals clean, asymptotic behavior. In this case, dividing $ g(n) = \frac{3 + \frac{5}{n}}{1 + \frac{1}{n^2}} $ by $ n^2 $ shows that both numerator and denominator approach zero through negligible terms, yielding a finite, well-defined limit:", "$$\n\lim_{n \ o \infty} g(n) = \boxed{3}\n$$", "Mastering such algebraic manipulations empowers faster, deeper insight into function behavior at infinity — a cornerstone of calculus and mathematical analysis.", "---", "Keywords:\nlimit as $ n \ o \infty $, $ g(n) = \frac{3 + \frac{5}{n}}{1 + \frac{1}{n^2}} $, divide numerator and denominator by $ n^2 $, asymptotic analysis, calculus limit, infinite limits, rational functions."]