1. **State the problem:** Evaluate the limit $$\lim_{n \to \infty} \ln \left( \sqrt{9n^2 + 18n} - 3n \right).$$ 2. **Rewrite the expression inside the logarithm:** We have $$\sqrt{9n^2 + 18n} - 3n.$$ To simplify, factor inside the square root: $$\sqrt{9n^2 + 18n} = \sqrt{9n^2(1 + \frac{2}{n})} = 3n \sqrt{1 + \frac{2}{n}}.$$ 3. **Rewrite the expression:** $$3n \sqrt{1 + \frac{2}{n}} - 3n = 3n \left( \sqrt{1 + \frac{2}{n}} - 1 \right).$$ 4. **Use the binomial expansion for large $n$:** For large $n$, $$\sqrt{1 + \frac{2}{n}} \approx 1 + \frac{1}{n} - \frac{1}{2n^2} + \cdots$$ So, $$\sqrt{1 + \frac{2}{n}} - 1 \approx \frac{1}{n} - \frac{1}{2n^2}.$$ 5. **Multiply by $3n$:** $$3n \left( \frac{1}{n} - \frac{1}{2n^2} \right) = 3 \left(1 - \frac{1}{2n} \right) = 3 - \frac{3}{2n}.$$ 6. **Take the limit inside the logarithm:** $$\lim_{n \to \infty} \left( \sqrt{9n^2 + 18n} - 3n \right) = \lim_{n \to \infty} \left(3 - \frac{3}{2n} \right) = 3.$$ 7. **Apply the logarithm limit:** $$\lim_{n \to \infty} \ln \left( \sqrt{9n^2 + 18n} - 3n \right) = \ln 3.$$ **Final answer:** $$\boxed{\ln 3}.$$