1. We are asked to find the limit: $$\lim_{x \to 0} \frac{\cos 4x \cdot \sin 3x}{5x}$$ 2. Recall the important limit rules: - $$\lim_{x \to 0} \frac{\sin ax}{x} = a$$ - $$\lim_{x \to 0} \cos bx = 1$$ 3. Using these, rewrite the expression: $$\frac{\cos 4x \cdot \sin 3x}{5x} = \cos 4x \cdot \frac{\sin 3x}{5x}$$ 4. As $$x \to 0$$, $$\cos 4x \to 1$$. 5. For $$\frac{\sin 3x}{5x}$$, multiply numerator and denominator by 3: $$\frac{\sin 3x}{5x} = \frac{\sin 3x}{3x} \cdot \frac{3}{5}$$ 6. Using the limit rule, $$\lim_{x \to 0} \frac{\sin 3x}{3x} = 1$$. 7. Therefore, the limit is: $$1 \cdot \frac{3}{5} = \frac{3}{5}$$ **Final answer:** $$\frac{3}{5}$$ which corresponds to option C.