1. **Problem 43:** Given the inequalities $$\sqrt{5 - 2x^2} \leq f(x) \leq \sqrt{5 - x^2}$$ for $$-1 \leq x \leq 1$$, find $$\lim_{x \to 0} f(x)$$. 2. **Problem 44:** Given $$2 - x^2 \leq g(x) \leq 2 \cos x$$ for all $$x$$, find $$\lim_{x \to 0} g(x)$$. 3. **Problem 45a:** Given the inequalities $$1 - \frac{x^2}{6} < \frac{x \sin x}{2 - 2 \cos x} < 1$$ hold for all $$x$$ close to zero, find $$\lim_{x \to 0} \frac{x \sin x}{2 - 2 \cos x}$$ and explain. --- ### Step-by-step solutions: **1. Problem 43:** 1. The Sandwich Theorem states if $$h(x) \leq f(x) \leq k(x)$$ and $$\lim_{x \to a} h(x) = \lim_{x \to a} k(x) = L$$, then $$\lim_{x \to a} f(x) = L$$. 2. Here, $$h(x) = \sqrt{5 - 2x^2}$$ and $$k(x) = \sqrt{5 - x^2}$$. 3. Evaluate the limits of $$h(x)$$ and $$k(x)$$ as $$x \to 0$$: $$\lim_{x \to 0} \sqrt{5 - 2x^2} = \sqrt{5 - 0} = \sqrt{5}$$ $$\lim_{x \to 0} \sqrt{5 - x^2} = \sqrt{5 - 0} = \sqrt{5}$$ 4. Since both limits are equal, by the Sandwich Theorem: $$\lim_{x \to 0} f(x) = \sqrt{5}$$. **2. Problem 44:** 1. Given $$2 - x^2 \leq g(x) \leq 2 \cos x$$. 2. Evaluate the limits of the bounding functions as $$x \to 0$$: $$\lim_{x \to 0} (2 - x^2) = 2 - 0 = 2$$ $$\lim_{x \to 0} 2 \cos x = 2 \cdot 1 = 2$$ 3. Both limits are equal to 2, so by the Sandwich Theorem: $$\lim_{x \to 0} g(x) = 2$$. **3. Problem 45a:** 1. Given inequalities: $$1 - \frac{x^2}{6} < \frac{x \sin x}{2 - 2 \cos x} < 1$$ for $$x$$ close to zero. 2. Evaluate the limits of the bounding functions as $$x \to 0$$: $$\lim_{x \to 0} \left(1 - \frac{x^2}{6}\right) = 1 - 0 = 1$$ $$\lim_{x \to 0} 1 = 1$$ 3. Since the function is squeezed between two expressions both tending to 1, by the Sandwich Theorem: $$\lim_{x \to 0} \frac{x \sin x}{2 - 2 \cos x} = 1$$. 4. This tells us the limit exists and equals 1. --- **Final answers:** $$\lim_{x \to 0} f(x) = \sqrt{5}$$ $$\lim_{x \to 0} g(x) = 2$$ $$\lim_{x \to 0} \frac{x \sin x}{2 - 2 \cos x} = 1$$