1. **State the problem.** Find $\lim_{x\to 0}\dfrac{\sin(3x)-3x}{x^3}.$ 2. **Use a standard series rule (or small-angle approximation).** A key fact near $u=0$ is $$\sin(u)=u-\frac{u^3}{6}+\cdots$$ 3. **Substitute $u=3x$.** $$\sin(3x)=3x-\frac{(3x)^3}{6}+\cdots$$ 4. **Form the numerator and simplify.** $$\sin(3x)-3x=\left(3x-\frac{(3x)^3}{6}+\cdots\right)-3x=-\frac{(3x)^3}{6}+\cdots$$ 5. **Rewrite $(3x)^3$ and simplify the leading term.** $$-\frac{(3x)^3}{6}=-\frac{27x^3}{6}=-\frac{9}{2}x^3$$ 6. **Divide by $x^3$ and take the limit.** $$\frac{\sin(3x)-3x}{x^3}=\frac{-\frac{9}{2}x^3+\cdots}{x^3}=-\frac{9}{2}+\cdots$$ $$\lim_{x\to 0}\frac{\sin(3x)-3x}{x^3}=-\frac{9}{2}$$