1. **State the problem:** We want to find the limit $$\lim_{x \to 0} \frac{\sinh x - \sin x}{x^3}$$ using L'Hopital's rule. 2. **Recall definitions and formulas:** - Hyperbolic sine: $$\sinh x = \frac{e^x - e^{-x}}{2}$$ - Sine function: $$\sin x$$ - L'Hopital's rule states that if $$\lim_{x \to a} f(x) = 0$$ and $$\lim_{x \to a} g(x) = 0$$ or both limits are $$\pm \infty$$, then $$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$$ provided the latter limit exists. 3. **Check the form of the limit:** - As $$x \to 0$$, $$\sinh 0 = 0$$ and $$\sin 0 = 0$$, so numerator $$\sinh x - \sin x \to 0$$. - Denominator $$x^3 \to 0$$. - So the limit is of the form $$\frac{0}{0}$$, suitable for L'Hopital's rule. 4. **Apply L'Hopital's rule first time:** - Derivative numerator: $$\frac{d}{dx}(\sinh x - \sin x) = \cosh x - \cos x$$ - Derivative denominator: $$\frac{d}{dx}(x^3) = 3x^2$$ - New limit: $$\lim_{x \to 0} \frac{\cosh x - \cos x}{3x^2}$$ 5. **Check the form again:** - At $$x=0$$, $$\cosh 0 = 1$$ and $$\cos 0 = 1$$, so numerator $$\cosh x - \cos x \to 0$$. - Denominator $$3x^2 \to 0$$. - Again, form $$\frac{0}{0}$$, apply L'Hopital's rule again. 6. **Apply L'Hopital's rule second time:** - Derivative numerator: $$\frac{d}{dx}(\cosh x - \cos x) = \sinh x + \sin x$$ - Derivative denominator: $$\frac{d}{dx}(3x^2) = 6x$$ - New limit: $$\lim_{x \to 0} \frac{\sinh x + \sin x}{6x}$$ 7. **Check the form again:** - At $$x=0$$, $$\sinh 0 = 0$$ and $$\sin 0 = 0$$, numerator $$0$$. - Denominator $$6 \times 0 = 0$$. - Form $$\frac{0}{0}$$ again, apply L'Hopital's rule third time. 8. **Apply L'Hopital's rule third time:** - Derivative numerator: $$\frac{d}{dx}(\sinh x + \sin x) = \cosh x + \cos x$$ - Derivative denominator: $$\frac{d}{dx}(6x) = 6$$ - New limit: $$\lim_{x \to 0} \frac{\cosh x + \cos x}{6}$$ 9. **Evaluate the limit:** - At $$x=0$$, $$\cosh 0 = 1$$ and $$\cos 0 = 1$$ - So numerator $$1 + 1 = 2$$ - Denominator is 6 - Therefore, limit $$= \frac{2}{6} = \frac{1}{3}$$ **Final answer:** $$\boxed{\frac{1}{3}}$$