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} $.

Thu, 16 Apr 2026 19:03:19 +0000

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