1. The problem asks us to find the limit of the expression $$\frac{n+8}{n+4}$$ as $$n$$ approaches infinity.\n\n2. When $$n$$ becomes very large, the constants 8 and 4 become insignificant compared to the $$n$$ terms, because $$n$$ is growing without bound.\n\n3. To analyze the limit, divide the numerator and denominator by $$n$$ to simplify the expression:\n$$\frac{n+8}{n+4} = \frac{\frac{n}{n} + \frac{8}{n}}{\frac{n}{n} + \frac{4}{n}} = \frac{1 + \frac{8}{n}}{1 + \frac{4}{n}}$$\n\n4. As $$n \to \infty$$, the terms $$\frac{8}{n} \to 0$$ and $$\frac{4}{n} \to 0$$, so the fraction approaches:\n$$\frac{1 + 0}{1 + 0} = 1$$\n\n5. Therefore, the limit is $$1$$.