1. **Problem Statement:** Given the limits \( \lim_{x \to 2} s(x) = 0 \) and \( \lim_{x \to 2} h(x) = -2 \), find the following limits: (i) \( \lim_{x \to 2} (s(x) + h(x)) \) (ii) \( \lim_{x \to 2} (5d(x) - 4h(x)) \) (Note: \(d(x)\) is not given, so we cannot evaluate this without more info.) (iii) \( \lim_{x \to 2} \frac{1}{2} h(x) \) (iv) \( \lim_{x \to 2} h(x)^4 \) (v) \( \lim_{x \to 2} \frac{s(x)}{3 h(x)} \) (vi) \( \lim_{x \to 2} (d(x) \cdot h(x)) \) (Again, \(d(x)\) unknown, cannot evaluate.) (vii) \( \lim_{x \to 2} (32 s(x) - h(x) + 3) \) --- 2. **Formulas and Rules Used:** - Sum Rule: \( \lim_{x \to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x) \) - Constant Multiple Rule: \( \lim_{x \to a} [c \cdot f(x)] = c \cdot \lim_{x \to a} f(x) \) - Product Rule: \( \lim_{x \to a} [f(x) \cdot g(x)] = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x) \) - Quotient Rule: \( \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)} \) if denominator limit \( \neq 0 \) - Power Rule: \( \lim_{x \to a} [f(x)]^n = [\lim_{x \to a} f(x)]^n \) --- 3. **Step-by-step Solutions:** (i) \( \lim_{x \to 2} (s(x) + h(x)) = \lim_{x \to 2} s(x) + \lim_{x \to 2} h(x) = 0 + (-2) = -2 \) (ii) Cannot evaluate \( \lim_{x \to 2} (5 d(x) - 4 h(x)) \) without \( \lim_{x \to 2} d(x) \). (iii) \( \lim_{x \to 2} \frac{1}{2} h(x) = \frac{1}{2} \lim_{x \to 2} h(x) = \frac{1}{2} \times (-2) = -1 \) (iv) \( \lim_{x \to 2} h(x)^4 = (\lim_{x \to 2} h(x))^4 = (-2)^4 = 16 \) (v) \( \lim_{x \to 2} \frac{s(x)}{3 h(x)} = \frac{\lim_{x \to 2} s(x)}{3 \times \lim_{x \to 2} h(x)} = \frac{0}{3 \times (-2)} = 0 \) (vi) Cannot evaluate \( \lim_{x \to 2} (d(x) \cdot h(x)) \) without \( \lim_{x \to 2} d(x) \). (vii) \( \lim_{x \to 2} (32 s(x) - h(x) + 3) = 32 \lim_{x \to 2} s(x) - \lim_{x \to 2} h(x) + 3 = 32 \times 0 - (-2) + 3 = 0 + 2 + 3 = 5 \) --- **Final answers:** (i) \( -2 \) (iii) \( -1 \) (iv) \( 16 \) (v) \( 0 \) (vii) \( 5 \)