Question:

We need to find $\lim_{x \to 0} \frac{e^{\sin x} - (1 + \sin x)}{x^2}$. Using L'Hôpital's rule or Taylor expansions, we can expand $e^{\sin x}$ around $x = 0$. Since $\sin x = x - \frac{x^3}{6} + O(x^5)$, we have $e^{\sin x} = 1 + \sin x + \frac{(\sin x)^2}{2} + \frac{(\sin x)^3}{6} + \ldots$. Therefore $e^{\sin x} - (1 + \sin x) = \frac{(\sin x)^2}{2} + \frac{(\sin x)^3}{6} + \ldots = \frac{x^2}{2} + O(x^4)$. Thus $\lim_{x \to 0} \frac{e^{\sin x} - (1 + \sin x)}{x^2} = \lim_{x \to 0} \frac{\frac{x^2}{2} + O(x^4)}{x^2} = \frac{1}{2}$. However, the given answer is 1000, indicating this may be part of a multi-part question or scaled problem.