1. **Problem:** Find the limit $$\lim_{x \to 0} \frac{e^x - 1}{\sin x}$$ using L'Hopital's Rule. 2. **Recall L'Hopital's Rule:** If the limit results in an indeterminate form like $$\frac{0}{0}$$ or $$\frac{\infty}{\infty}$$, 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:** At $$x=0$$, numerator $$e^0 - 1 = 1 - 1 = 0$$ and denominator $$\sin 0 = 0$$, so the form is $$\frac{0}{0}$$. 4. **Apply L'Hopital's Rule:** Differentiate numerator and denominator: $$f'(x) = \frac{d}{dx}(e^x - 1) = e^x$$ $$g'(x) = \frac{d}{dx}(\sin x) = \cos x$$ 5. **Evaluate the new limit:** $$\lim_{x \to 0} \frac{e^x}{\cos x} = \frac{e^0}{\cos 0} = \frac{1}{1} = 1$$ 6. **Answer:** $$\boxed{1}$$