1. **Problem statement:** Evaluate the limits: (e) $$\lim_{x \to -1} \frac{f(x)}{g(x)}$$ (f) $$\lim_{x \to -1} (f(x))^{2}$$ (g) $$\lim_{x \to -1} (g(x))^{-3}$$ Given $$\lim_{x \to -1} f(x) = 2$$ and $$\lim_{x \to -1} g(x) = -2$$. 2. **Theorem used:** We use the **Limit Laws**, specifically: - Quotient Law: $$\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}$$ if $$\lim_{x \to a} g(x) \neq 0$$. - Power Law: $$\lim_{x \to a} (f(x))^{n} = (\lim_{x \to a} f(x))^{n}$$ for any integer $$n$$. 3. **Evaluate (e):** $$\lim_{x \to -1} \frac{f(x)}{g(x)} = \frac{\lim_{x \to -1} f(x)}{\lim_{x \to -1} g(x)} = \frac{2}{-2} = -1$$ 4. **Evaluate (f):** $$\lim_{x \to -1} (f(x))^{2} = (\lim_{x \to -1} f(x))^{2} = 2^{2} = 4$$ 5. **Evaluate (g):** $$\lim_{x \to -1} (g(x))^{-3} = (\lim_{x \to -1} g(x))^{-3} = (-2)^{-3} = \frac{1}{(-2)^{3}} = \frac{1}{-8} = -\frac{1}{8}$$ **Final answers:** (e) $$-1$$ (f) $$4$$ (g) $$-\frac{1}{8}$$