1. The problem is to find the limit as $x$ approaches 3 for the expression $$\frac{9x^2 - 1}{27x^3 - 1}$$. 2. The formula for limits is to substitute the value of $x$ into the expression if it does not result in an indeterminate form. If it does, we simplify the expression. 3. Substitute $x = 3$: $$\frac{9(3)^2 - 1}{27(3)^3 - 1} = \frac{9 \times 9 - 1}{27 \times 27 - 1} = \frac{81 - 1}{729 - 1} = \frac{80}{728}$$ 4. Simplify the fraction: $$\frac{80}{728} = \frac{10}{91}$$ 5. Therefore, the limit is $$\frac{10}{91}$$. This means as $x$ gets closer to 3, the value of the expression approaches $\frac{10}{91}$.