1. Planteamos el problema: calcular los límites dados usando los valores $\lim_{x \to c} f(x) = -6$ y $\lim_{x \to c} g(x) = \frac{1}{2}$. 2. Recordamos las propiedades de límites importantes: - $\lim (f(x)g(x)) = \lim f(x) \cdot \lim g(x)$ - $\lim \frac{f(x)}{g(x)} = \frac{\lim f(x)}{\lim g(x)}$ si $\lim g(x) \neq 0$ - $\lim (f(x) + 2g(x)) = \lim f(x) + 2 \lim g(x)$ - $\lim (f(x))^2 = (\lim f(x))^2$ 3. Calculamos cada límite: **29.** $$\lim_{x \to c} [f(x)g(x)] = \lim_{x \to c} f(x) \cdot \lim_{x \to c} g(x) = (-6) \cdot \frac{1}{2} = -3$$ **30.** $$\lim_{x \to c} \frac{f(x)}{g(x)} = \frac{\lim_{x \to c} f(x)}{\lim_{x \to c} g(x)} = \frac{-6}{\frac{1}{2}}$$ Simplificamos usando cancelación: $$= \frac{-6}{\cancel{\frac{1}{2}}} \cdot \frac{\cancel{2}}{1} = -6 \cdot 2 = -12$$ **31.** $$\lim_{x \to c} [f(x) + 2g(x)] = \lim_{x \to c} f(x) + 2 \lim_{x \to c} g(x) = -6 + 2 \cdot \frac{1}{2} = -6 + 1 = -5$$ **32.** $$\lim_{x \to c} [f(x)]^2 = (\lim_{x \to c} f(x))^2 = (-6)^2 = 36$$ 4. Resumen de resultados: - $\lim_{x \to c} [f(x)g(x)] = -3$ - $\lim_{x \to c} \frac{f(x)}{g(x)} = -12$ - $\lim_{x \to c} [f(x) + 2g(x)] = -5$ - $\lim_{x \to c} [f(x)]^2 = 36$