1. We are given the sequence defined by $$a_n = \frac{7n - 3}{4n + 1}$$ and asked to find its limit as $$n$$ approaches infinity. 2. The formula for the limit of a rational sequence where numerator and denominator are linear functions of $$n$$ is: $$\lim_{n \to \infty} \frac{an + b}{cn + d} = \frac{a}{c}$$ provided $$c \neq 0$$. 3. Here, the numerator is $$7n - 3$$ and the denominator is $$4n + 1$$. 4. To find the limit, divide numerator and denominator by $$n$$: $$a_n = \frac{7n - 3}{4n + 1} = \frac{\cancel{n}(7 - \frac{3}{n})}{\cancel{n}(4 + \frac{1}{n})}$$ 5. Simplifying by canceling $$n$$: $$a_n = \frac{7 - \frac{3}{n}}{4 + \frac{1}{n}}$$ 6. As $$n \to \infty$$, $$\frac{3}{n} \to 0$$ and $$\frac{1}{n} \to 0$$, so: $$\lim_{n \to \infty} a_n = \frac{7 - 0}{4 + 0} = \frac{7}{4}$$ 7. The limit of the sequence is $$\frac{7}{4}$$, which is already in simplest form. Final answer: $$\boxed{\frac{7}{4}}$$