1. **State the problem:** Find the limit $$\lim_{x\to0} \frac{\sin 5x}{3x}$$. 2. **Recall the standard limit formula:** $$\lim_{x\to0} \frac{\sin x}{x} = 1$$. 3. **Rewrite the expression to use the standard limit:** $$\lim_{x\to0} \frac{\sin 5x}{3x} = \lim_{x\to0} \frac{\sin 5x}{5x} \cdot \frac{5x}{3x}$$. 4. **Simplify the fraction:** $$\frac{5x}{3x} = \cancel{\frac{5\cancel{x}}{3\cancel{x}}} = \frac{5}{3}$$. 5. **Apply the limit:** Since $$\lim_{x\to0} \frac{\sin 5x}{5x} = 1$$, we have $$\lim_{x\to0} \frac{\sin 5x}{3x} = 1 \cdot \frac{5}{3} = \frac{5}{3}$$. **Final answer:** $$\boxed{\frac{5}{3}}$$