As $x \to 0$, we get $\frac{0}{0}$ form. **Method 1: L'Hôpital's Rule** $$\lim_{x \to 0} \frac{e^x - 1 - x}{x^2} = \lim_{x \to 0} \frac{e^x - 1}{2x}$$ (still $\frac{0}{0}$) $$= \lim_{x \to 0} \frac{e^x}{2} = \frac{1}{2}$$ **Method 2: Taylor Series** $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots$ $e^x - 1 - x = \frac{x^2}{2} + \frac{x^3}{6} + \ldots$ $$\lim_{x \to 0} \frac{e^x - 1 - x}{x^2} = \lim_{x \to 0} \frac{\frac{x^2}{2} + \frac{x^3}{6} + \ldots}{x^2} = \frac{1}{2}$$