1. **Problem:** Find $$\lim_{x \to 0} \frac{\sin(5x)}{2x}$$. 2. **Formula and rules:** We use the standard limit $$\lim_{x \to 0} \frac{\sin x}{x} = 1$$. 3. **Rewrite the limit:** $$\lim_{x \to 0} \frac{\sin(5x)}{2x} = \lim_{x \to 0} \frac{\sin(5x)}{5x} \cdot \frac{5x}{2x}$$ 4. **Simplify the fraction:** $$\frac{5x}{2x} = \cancel{\frac{5\cancel{x}}{2\cancel{x}}} = \frac{5}{2}$$ 5. **Apply the limit:** $$\lim_{x \to 0} \frac{\sin(5x)}{5x} = 1$$ by the standard limit. 6. **Multiply results:** $$1 \cdot \frac{5}{2} = \frac{5}{2}$$ **Final answer:** $$\boxed{\frac{5}{2}}$$