1. **State the problem:** Find the limit $$\lim_{x \to \infty} \left(\frac{3x - 2}{3x + 9}\right)^{3 - 11x}$$. 2. **Rewrite the expression inside the limit:** Simplify the fraction inside the parentheses: $$\frac{3x - 2}{3x + 9} = \frac{3x(1 - \frac{2}{3x})}{3x(1 + \frac{3}{x})} = \frac{1 - \frac{2}{3x}}{1 + \frac{3}{x}}.$$ 3. **Analyze the base as $x \to \infty$:** As $x \to \infty$, $\frac{2}{3x} \to 0$ and $\frac{3}{x} \to 0$, so the base approaches: $$\frac{1 - 0}{1 + 0} = 1.$$ 4. **Rewrite the limit in exponential form:** Since the base approaches 1 and the exponent tends to $-\infty$ (because $3 - 11x \to -\infty$), this is an indeterminate form of type $1^{\infty}$. We use the standard limit formula: $$\lim_{x \to \infty} (1 + a_x)^{b_x} = e^{\lim_{x \to \infty} a_x b_x}$$ if the limit on the right exists. Set: $$a_x = \frac{3x - 2}{3x + 9} - 1 = \frac{3x - 2 - (3x + 9)}{3x + 9} = \frac{-11}{3x + 9}$$ Exponent: $$b_x = 3 - 11x.$$ 5. **Calculate the product $a_x b_x$:** $$a_x b_x = \frac{-11}{3x + 9} (3 - 11x) = \frac{-11(3 - 11x)}{3x + 9} = \frac{-33 + 121x}{3x + 9}.$$ 6. **Find the limit of $a_x b_x$ as $x \to \infty$:** Divide numerator and denominator by $x$: $$\lim_{x \to \infty} \frac{-33 + 121x}{3x + 9} = \lim_{x \to \infty} \frac{121 - \frac{33}{x}}{3 + \frac{9}{x}} = \frac{121}{3}.$$ 7. **Apply the exponential limit:** $$\lim_{x \to \infty} \left(\frac{3x - 2}{3x + 9}\right)^{3 - 11x} = e^{\frac{121}{3}}.$$ **Final answer:** $$\boxed{e^{\frac{121}{3}}}.$$