1. **Problem:** Find $\lim_{x \to 0} \frac{\sin(5x)}{2x}$.\n\n2. **Formula and rules:** We use the standard limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$.\n\n3. **Rewrite the limit:**\n$$\lim_{x \to 0} \frac{\sin(5x)}{2x} = \lim_{x \to 0} \frac{\sin(5x)}{5x} \cdot \frac{5x}{2x}$$\n\n4. **Simplify the fraction:**\n$$\frac{5x}{2x} = \cancel{\frac{5\cancel{x}}{2\cancel{x}}} = \frac{5}{2}$$\n\n5. **Apply the limit:**\n$$\lim_{x \to 0} \frac{\sin(5x)}{5x} = 1$$ by the standard limit.\n\n6. **Multiply results:**\n$$1 \cdot \frac{5}{2} = \frac{5}{2}$$\n\n**Final answer:** $\boxed{\frac{5}{2}}$