1. **State the problem:** Evaluate the limit $$\lim_{x\to 0}\frac{\sin(5x)-5x+\frac{(5x)^3}{6}}{x^5}$$. 2. **Recall the Taylor series expansion for sine:** $$\sin(5x) = 5x - \frac{(5x)^3}{3!} + \frac{(5x)^5}{5!} - \cdots$$ where $3! = 6$ and $5! = 120$. 3. **Substitute the expansion into the numerator:** $$\sin(5x) - 5x + \frac{(5x)^3}{6} = \left(5x - \frac{(5x)^3}{6} + \frac{(5x)^5}{120} - \cdots\right) - 5x + \frac{(5x)^3}{6}$$ 4. **Simplify the numerator by canceling terms:** $$5x - 5x - \frac{(5x)^3}{6} + \frac{(5x)^3}{6} + \frac{(5x)^5}{120} - \cdots = \frac{(5x)^5}{120} - \cdots$$ 5. **Focus on the leading term in the numerator:** $$\frac{(5x)^5}{120} = \frac{5^5 x^5}{120}$$ 6. **Rewrite the limit using this leading term:** $$\lim_{x\to 0} \frac{\frac{5^5 x^5}{120} + \text{higher order terms}}{x^5} = \lim_{x\to 0} \frac{5^5 x^5}{120 x^5} + \lim_{x\to 0} \frac{\text{higher order terms}}{x^5}$$ 7. **Cancel $x^5$ in the main term:** $$\lim_{x\to 0} \frac{\cancel{5^5} \cancel{x^5}}{120 \cancel{x^5}} = \frac{5^5}{120}$$ 8. **Since higher order terms vanish as $x \to 0$, the limit is:** $$\frac{5^5}{120} = \frac{3125}{120} = \frac{625}{24}$$ **Final answer:** $$\boxed{\frac{625}{24}}$$