1. **Problem:** Determine which inequalities can be used with the squeeze theorem to find the limit of the given functions as $x \to 0$. 2. **Recall the squeeze theorem:** If $g(x) \leq f(x) \leq h(x)$ for all $x$ near $a$ (except possibly at $a$), and if $\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L$, then $\lim_{x \to a} f(x) = L$. 3. **Analyze each inequality:** - I. $\frac{1}{3}(1 - x^2) \leq f(x) \leq \frac{1}{2}$ Calculate limits of bounds as $x \to 0$: $$\lim_{x \to 0} \frac{1}{3}(1 - x^2) = \frac{1}{3}(1 - 0) = \frac{1}{3}$$ $$\lim_{x \to 0} \frac{1}{2} = \frac{1}{2}$$ Since the lower and upper bounds approach different limits ($\frac{1}{3} \neq \frac{1}{2}$), the squeeze theorem cannot be applied here. - II. $-x^2 \leq g(x) \leq x^2$ Calculate limits of bounds as $x \to 0$: $$\lim_{x \to 0} -x^2 = 0$$ $$\lim_{x \to 0} x^2 = 0$$ Both bounds approach 0, so the squeeze theorem can be applied to find $\lim_{x \to 0} g(x) = 0$. - III. $-\frac{1}{|x|} \leq h(x) \leq \frac{1}{|x|}$ Calculate limits of bounds as $x \to 0$: $$\lim_{x \to 0} -\frac{1}{|x|} = -\infty$$ $$\lim_{x \to 0} \frac{1}{|x|} = +\infty$$ The bounds do not approach the same finite limit, so the squeeze theorem cannot be applied here. 4. **Conclusion:** Only inequality II can be used with the squeeze theorem. **Final answer:** B) II only