1. Stating the problem: We want to find the greatest limit of the function $$f(n) = \frac{\sqrt{n}}{9 + n}$$ as $$n$$ approaches infinity or other relevant points. 2. Consider the behavior as $$n \to \infty$$: For very large $$n$$, the term $$9$$ becomes negligible compared to $$n$$, so the function behaves like $$\frac{\sqrt{n}}{n} = \frac{n^{1/2}}{n^{1}} = n^{-1/2}.$$ 3. Evaluate the limit at infinity: $$\lim_{n \to \infty} f(n) = \lim_{n \to \infty} n^{-1/2} = 0.$$ This means as $$n$$ grows very large, the function approaches 0. 4. Check the limit as $$n \to 0^+$$: $$f(0) = \frac{\sqrt{0}}{9+0} = 0,$$ so at the start, the value is 0. 5. Analyze critical points for a maximum: Differentiate $$f(n)$$ to find any potential maximums. Let $$f(n) = \frac{\sqrt{n}}{9 + n} = \frac{n^{1/2}}{9 + n}.$$ Using the quotient rule: $$f'(n) = \frac{(9 + n) \cdot \frac{1}{2} n^{-1/2} - n^{1/2} \cdot 1}{(9 + n)^2} = \frac{\frac{9 + n}{2 \sqrt{n}} - \sqrt{n}}{(9 + n)^2}.$$ Simplify numerator: $$\frac{9 + n}{2 \sqrt{n}} - \sqrt{n} = \frac{9 + n - 2n}{2 \sqrt{n}} = \frac{9 - n}{2 \sqrt{n}}.$$ So, $$f'(n) = \frac{9 - n}{2 \sqrt{n} (9 + n)^2}.$$ 6. Set the derivative to zero to find critical points: $$9 - n = 0 \implies n = 9.$$ 7. Evaluate function at $$n = 9$$: $$f(9) = \frac{\sqrt{9}}{9 + 9} = \frac{3}{18} = \frac{1}{6}.$$ 8. Determine if this is a maximum: For $$n < 9$$, $$f'(n) > 0$$; for $$n > 9$$, $$f'(n) < 0$$, so $$n=9$$ is a local maximum. 9. Since limits at 0 and infinity are 0, the greatest limit (in fact, maximum value) of the function is $$\frac{1}{6}$$ at $$n = 9$$. Final answer: The greatest limit (maximum value) of $$\frac{\sqrt{n}}{9 + n}$$ is $$\frac{1}{6}$$ at $$n=9$$.