1. **Problem Statement:** Given \(\lim_{x \to a} f(x) = 0\), \(\lim_{x \to a} h(x) = -7\), and \(\lim_{x \to a} g(x) = 3\), find the following limits if they exist. 2. **Recall Limit Properties:** - \(\lim (f(x) \cdot g(x)) = \lim f(x) \cdot \lim g(x)\) if both limits exist. - \(\lim (f(x) + h(x)) = \lim f(x) + \lim h(x)\). - \(\lim \frac{f(x)}{g(x)} = \frac{\lim f(x)}{\lim g(x)}\) if \(\lim g(x) \neq 0\). - \(\lim \sqrt{h(x)} = \sqrt{\lim h(x)}\) if the limit inside the root is non-negative. - \(\lim (h(x))^{-1} = \frac{1}{\lim h(x)}\) if \(\lim h(x) \neq 0\). 3. **Calculate each limit:** **a.** \(\lim_{x \to a} [f(x) \cdot g(x)] = \lim f(x) \cdot \lim g(x) = 0 \cdot 3 = 0\). **b.** \(\lim_{x \to a} [f(x) + h(x)] = \lim f(x) + \lim h(x) = 0 + (-7) = -7\). **c.** \(\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim f(x)}{\lim g(x)} = \frac{0}{3} = 0\). **d.** \(\lim_{x \to a} \frac{f(x)}{h(x)} = \frac{0}{-7} = 0\). **e.** \(\lim_{x \to a} \sqrt{h(x)} = \sqrt{\lim h(x)} = \sqrt{-7}\) which is not a real number, so the limit **does not exist (DNE)**. **f.** \(\lim_{x \to a} (h(x))^{-1} = \frac{1}{\lim h(x)} = \frac{1}{-7} = -\frac{1}{7}\). 4. **Final answers:** - a. 0 - b. -7 - c. 0 - d. 0 - e. DNE - f. -\frac{1}{7}