1. **Problem (a):** Given that $|f(x) - 3| \leq 4(x - 2)^2$, find $\lim_{x \to 2} f(x)$. 2. **Step 1:** Understand the inequality. The expression $|f(x) - 3| \leq 4(x - 2)^2$ means the difference between $f(x)$ and 3 is bounded by $4(x - 2)^2$. As $x$ approaches 2, $(x - 2)^2$ approaches 0, so the difference $|f(x) - 3|$ also approaches 0. 3. **Step 2:** Use the Squeeze Theorem. Since $|f(x) - 3|$ is squeezed between 0 and $4(x - 2)^2$, and $4(x - 2)^2 \to 0$ as $x \to 2$, it follows that $f(x) \to 3$. 4. **Answer (a):** $$\lim_{x \to 2} f(x) = 3.$$ --- 5. **Problem (b)(i):** Find $\lim_{x \to 3} \frac{|x - 3|}{x - 3}$. 6. **Step 1:** Consider the limit from the right ($x \to 3^+$). For $x > 3$, $|x - 3| = x - 3$, so $$\lim_{x \to 3^+} \frac{|x - 3|}{x - 3} = \lim_{x \to 3^+} \frac{x - 3}{x - 3} = 1.$$ 7. **Step 2:** Consider the limit from the left ($x \to 3^-$). For $x < 3$, $|x - 3| = -(x - 3)$, so $$\lim_{x \to 3^-} \frac{|x - 3|}{x - 3} = \lim_{x \to 3^-} \frac{-(x - 3)}{x - 3} = -1.$$ 8. **Step 3:** Since the left and right limits are not equal, the limit does not exist. 9. **Answer (b)(i):** The limit does not exist. --- 10. **Problem (b)(ii):** Find $\lim_{x \to \infty} \frac{\sqrt{x^4 - 1}}{(x + 1)^2}$. 11. **Step 1:** Simplify the expression by dividing numerator and denominator by $x^2$: $$\frac{\sqrt{x^4 - 1}}{(x + 1)^2} = \frac{\sqrt{x^4 - 1}}{x^2 (1 + \frac{1}{x})^2} = \frac{\sqrt{x^4(1 - \frac{1}{x^4})}}{x^2 (1 + \frac{1}{x})^2}.$$ 12. **Step 2:** Simplify inside the square root and cancel $x^2$: $$= \frac{x^2 \sqrt{1 - \frac{1}{x^4}}}{x^2 (1 + \frac{1}{x})^2} = \frac{\sqrt{1 - \frac{1}{x^4}}}{(1 + \frac{1}{x})^2}.$$ 13. **Step 3:** Take the limit as $x \to \infty$: $$\lim_{x \to \infty} \sqrt{1 - \frac{1}{x^4}} = 1,$$ $$\lim_{x \to \infty} (1 + \frac{1}{x})^2 = 1^2 = 1.$$ 14. **Answer (b)(ii):** $$\lim_{x \to \infty} \frac{\sqrt{x^4 - 1}}{(x + 1)^2} = \frac{1}{1} = 1.$$ --- 15. **Problem (b)(iii):** Find $\lim_{x \to \frac{\pi}{2}} \sin\left(\frac{x}{2} + \sin x\right)$. 16. **Step 1:** Evaluate the inner expression at $x = \frac{\pi}{2}$: $$\frac{\pi}{2} \div 2 = \frac{\pi}{4}, \quad \sin\left(\frac{\pi}{2}\right) = 1.$$ 17. **Step 2:** Sum inside the sine: $$\frac{\pi}{4} + 1.$$ 18. **Step 3:** Since sine is continuous, $$\lim_{x \to \frac{\pi}{2}} \sin\left(\frac{x}{2} + \sin x\right) = \sin\left(\frac{\pi}{4} + 1\right).$$ 19. **Answer (b)(iii):** $$\sin\left(\frac{\pi}{4} + 1\right).$$