1. **Problem:** Find the limit $$\lim_{x \to 0} \frac{x - \sin x}{x^3}$$ without using L'Hospital's rule. 2. **Recall the Taylor series expansion:** For small $x$, $$\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots$$ 3. **Substitute the expansion into the expression:** $$\frac{x - \sin x}{x^3} = \frac{x - \left(x - \frac{x^3}{6} + \cdots \right)}{x^3} = \frac{x - x + \frac{x^3}{6} - \cdots}{x^3} = \frac{\frac{x^3}{6} - \cdots}{x^3}$$ 4. **Simplify the fraction:** $$= \frac{x^3}{6x^3} - \cdots = \frac{1}{6} - \cdots$$ 5. **As $x \to 0$, higher order terms vanish, so:** $$\lim_{x \to 0} \frac{x - \sin x}{x^3} = \frac{1}{6}$$ **Final answer:** $$\boxed{\frac{1}{6}}$$