1. We are asked to find the limit: $$\lim_{x\to 0} \frac{\sin^3(4x)}{9x^2}$$. 2. Recall the standard limit: $$\lim_{t\to 0} \frac{\sin t}{t} = 1$$. 3. To use this, rewrite the expression by separating powers and terms: $$\frac{\sin^3(4x)}{9x^2} = \frac{(\sin(4x))^3}{9x^2} = \frac{(\sin(4x))^3}{x^2 \cdot 9}$$. 4. Write $(\sin(4x))^3$ as $(\sin(4x))^2 \cdot \sin(4x)$ and introduce $(4x)$ terms to use the standard limit: $$= \frac{(\sin(4x))^2 \cdot \sin(4x)}{9 x^2} = \frac{(\sin(4x))^2}{x^2} \cdot \frac{\sin(4x)}{9}$$. Alternatively, rewrite keeping the form for substitution: $$= \frac{(\sin(4x))^3}{(4x)^3} \cdot \frac{(4x)^3}{9 x^2}$$. 5. Since $$\lim_{x\to 0}\frac{\sin(4x)}{4x} = 1$$, then $$\lim_{x\to 0} \left( \frac{\sin(4x)}{4x} \right)^3 = 1^3 = 1$$. 6. Now, simplify the remaining factor: $$\frac{(4x)^3}{9x^2} = \frac{64 x^3}{9 x^2} = \frac{64}{9} x$$. 7. So the original limit is: $$\lim_{x\to 0} \left( \frac{\sin(4x)}{4x} \right)^3 \cdot \frac{64}{9} x = 1 \cdot 0 = 0$$. **Final answer:** $$\boxed{0}$$