1. We need to find the limit as $x \to 0$ of the expression $\frac{1 - \cos x}{x^2}$. 2. Recall the Taylor series expansion of $\cos x$ around 0: $$\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots$$ 3. Substitute this approximation into the expression: $$\frac{1 - \cos x}{x^2} = \frac{1 - \left(1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots \right)}{x^2} = \frac{\frac{x^2}{2} - \frac{x^4}{24} + \cdots}{x^2}$$ 4. Simplify by dividing numerator terms by $x^2$: $$= \frac{x^2}{2x^2} - \frac{x^4}{24x^2} + \cdots = \frac{1}{2} - \frac{x^2}{24} + \cdots$$ 5. As $x \to 0$, the higher order terms vanish, so the limit is: $$\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$$ Therefore, the answer is $\boxed{\frac{1}{2}}$.