1. **Problem statement:** Given a continuous and strictly decreasing function $f$ on $(0,1)$ with $$\lim_{x \to 0} \frac{f(x)-5}{x} = 4$$ and $$2\sin(x-1) + 10(x-1)^3 \leq (x-1)f(x) \leq 8x^2 - 14x + 6, \quad \forall x \in (0,1),$$ find: (i) $\lim_{x \to 0^+} f(x)$ and $\lim_{x \to 1^-} f(x)$. 2. **Step for (i):** Use the limit definition: $$\lim_{x \to 0} \frac{f(x)-5}{x} = 4 \implies \lim_{x \to 0} (f(x)-5) = \lim_{x \to 0} 4x = 0,$$ so $$\lim_{x \to 0^+} f(x) = 5.$$ 3. **Step for $\lim_{x \to 1^-} f(x)$:** From the inequality, $$2\sin(x-1) + 10(x-1)^3 \leq (x-1)f(x) \leq 8x^2 - 14x + 6,$$ divide all sides by $(x-1)$ (noting $x \to 1^-$ so $x-1 < 0$ and inequalities reverse): $$\frac{2\sin(x-1) + 10(x-1)^3}{x-1} \geq f(x) \geq \frac{8x^2 - 14x + 6}{x-1}.$$ 4. Simplify limits: - For the left side numerator: $$\lim_{x \to 1} \frac{2\sin(x-1)}{x-1} = 2 \lim_{t \to 0} \frac{\sin t}{t} = 2,$$ - For the cubic term: $$\lim_{x \to 1} \frac{10(x-1)^3}{x-1} = \lim_{x \to 1} 10(x-1)^2 = 0,$$ so left side limit is $2 + 0 = 2$. 5. For the right side: $$\frac{8x^2 - 14x + 6}{x-1} = \frac{(8x^2 - 14x + 6)}{x-1}.$$ Factor numerator: $$8x^2 - 14x + 6 = 2(4x^2 - 7x + 3) = 2(4x - 3)(x - 1).$$ Cancel $(x-1)$: $$\frac{8x^2 - 14x + 6}{x-1} = 2(4x - 3).$$ 6. Taking limit as $x \to 1$: $$\lim_{x \to 1} 2(4x - 3) = 2(4 - 3) = 2.$$ 7. By squeeze theorem: $$\lim_{x \to 1^-} f(x) = 2.$$ 8. **Answer for (i):** $$\boxed{\lim_{x \to 0^+} f(x) = 5, \quad \lim_{x \to 1^-} f(x) = 2}.$$