1. **Problem Statement:** Evaluate the limit $$\lim_{x \to 0} \frac{\sin x - x}{x^3}$$ using L'Hôpital's rule. 2. **Recall the formula and rules:** When a limit results in an indeterminate form like $\frac{0}{0}$ or $\frac{\infty}{\infty}$, L'Hôpital's rule states that: $$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$$ provided the latter limit exists. 3. **Check the original limit form:** Evaluate numerator and denominator at $x=0$: $$\sin 0 - 0 = 0, \quad 0^3 = 0$$ So the limit is of the form $\frac{0}{0}$, which allows us to apply L'Hôpital's rule. 4. **First application of L'Hôpital's rule:** Differentiate numerator and denominator: $$f'(x) = \cos x - 1, \quad g'(x) = 3x^2$$ New limit: $$\lim_{x \to 0} \frac{\cos x - 1}{3x^2}$$ 5. **Check the new limit form:** At $x=0$: $$\cos 0 - 1 = 1 - 1 = 0, \quad 3 \cdot 0^2 = 0$$ Still $\frac{0}{0}$, so apply L'Hôpital's rule again. 6. **Second application of L'Hôpital's rule:** Differentiate numerator and denominator again: $$f''(x) = -\sin x, \quad g''(x) = 6x$$ New limit: $$\lim_{x \to 0} \frac{-\sin x}{6x}$$ 7. **Check the new limit form:** At $x=0$: $$-\sin 0 = 0, \quad 6 \cdot 0 = 0$$ Still $\frac{0}{0}$, so apply L'Hôpital's rule a third time. 8. **Third application of L'Hôpital's rule:** Differentiate numerator and denominator once more: $$f'''(x) = -\cos x, \quad g'''(x) = 6$$ New limit: $$\lim_{x \to 0} \frac{-\cos x}{6}$$ 9. **Evaluate the limit:** At $x=0$: $$\frac{-\cos 0}{6} = \frac{-1}{6}$$ 10. **Final answer:** $$\boxed{-\frac{1}{6}}$$ **Watch Out!** When applying L'Hôpital's rule multiple times, always check the form after each differentiation to ensure the rule applies. Also, be careful with signs and derivatives of trigonometric functions. This step-by-step approach helps you understand how to handle tricky limits patiently and correctly.