Solution: Divide numerator and denominator by $ n^2 $: $ g(n) = \frac3 + \frac5n1 + \frac1n^2 $. As $ n \to \infty $, $ \frac5n \to 0 $ and $ \frac1n^2 \to 0 $. Thus, the limit is $ \frac31 = \boxed3 $. - Appfinity Technologies