1. **State the problem:** We want to find the limit $$\lim_{x \to 0} \frac{e^x - \cos x - x}{x^2}$$ using series expansions. 2. **Recall the series expansions:** - The exponential function: $$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$$ - The cosine function: $$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots$$ 3. **Substitute the expansions into the expression:** $$e^x - \cos x - x = \left(1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots\right) - \left(1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots\right) - x$$ 4. **Simplify the numerator:** $$= 1 + x + \frac{x^2}{2} + \frac{x^3}{6} - 1 + \frac{x^2}{2} - \frac{x^4}{24} - x + \cdots$$ 5. **Cancel terms:** $$= \cancel{1} + \cancel{x} + \frac{x^2}{2} + \frac{x^3}{6} - \cancel{1} + \frac{x^2}{2} - \frac{x^4}{24} - \cancel{x} + \cdots = \frac{x^2}{2} + \frac{x^2}{2} + \frac{x^3}{6} - \frac{x^4}{24} + \cdots$$ 6. **Combine like terms:** $$= x^2 + \frac{x^3}{6} - \frac{x^4}{24} + \cdots$$ 7. **Divide numerator by $x^2$:** $$\frac{e^x - \cos x - x}{x^2} = \frac{x^2 + \frac{x^3}{6} - \frac{x^4}{24} + \cdots}{x^2} = 1 + \frac{x}{6} - \frac{x^2}{24} + \cdots$$ 8. **Take the limit as $x \to 0$:** $$\lim_{x \to 0} \left(1 + \frac{x}{6} - \frac{x^2}{24} + \cdots\right) = 1$$ **Final answer:** $$\boxed{1}$$