1. **Problem statement:** (a) Given that $|f(x) - 3| \leq 4(x - 2)^2$, find $\lim_{x \to 2} f(x)$. (b) Find the following limits if they exist: i. $\lim_{x \to 3} \frac{|x - 3|}{x - 3}$ ii. $\lim_{x \to \infty} \frac{\sqrt{x^4 - 1}}{(x + 1)^2}$ iii. $\lim_{x \to \frac{\pi}{2}} \sin\left(\frac{x}{2} + \sin x\right)$ 2. **Step (a):** We have the inequality $|f(x) - 3| \leq 4(x - 2)^2$. As $x \to 2$, note that $(x - 2)^2 \to 0$. By the Squeeze Theorem, since $|f(x) - 3|$ is squeezed by $4(x - 2)^2$ which tends to 0, it follows that $$\lim_{x \to 2} |f(x) - 3| = 0,$$ which implies $$\lim_{x \to 2} f(x) = 3.$$ 3. **Step (b)(i):** Consider $\lim_{x \to 3} \frac{|x - 3|}{x - 3}$. For $x > 3$, $|x - 3| = x - 3$, so the expression becomes $$\frac{x - 3}{x - 3} = 1.$$ For $x < 3$, $|x - 3| = -(x - 3)$, so the expression becomes $$\frac{-(x - 3)}{x - 3} = -1.$$ Since the left-hand limit is $-1$ and the right-hand limit is $1$, the limit does not exist. 4. **Step (b)(ii):** Evaluate $$\lim_{x \to \infty} \frac{\sqrt{x^4 - 1}}{(x + 1)^2}.$$ Rewrite numerator: $$\sqrt{x^4 - 1} = \sqrt{x^4(1 - \frac{1}{x^4})} = x^2 \sqrt{1 - \frac{1}{x^4}}.$$ Rewrite denominator: $$(x + 1)^2 = x^2 + 2x + 1.$$ Divide numerator and denominator by $x^2$: $$\frac{x^2 \sqrt{1 - \frac{1}{x^4}}}{x^2 + 2x + 1} = \frac{\sqrt{1 - \frac{1}{x^4}}}{1 + \frac{2}{x} + \frac{1}{x^2}}.$$ As $x \to \infty$, terms with $\frac{1}{x}$ and higher powers go to 0, so $$\lim_{x \to \infty} \frac{\sqrt{1 - \frac{1}{x^4}}}{1 + \frac{2}{x} + \frac{1}{x^2}} = \frac{\sqrt{1 - 0}}{1 + 0 + 0} = 1.$$ 5. **Step (b)(iii):** Evaluate $$\lim_{x \to \frac{\pi}{2}} \sin\left(\frac{x}{2} + \sin x\right).$$ Since sine is continuous, we can substitute the limit inside: $$\sin\left(\lim_{x \to \frac{\pi}{2}} \left(\frac{x}{2} + \sin x\right)\right) = \sin\left(\frac{\pi}{4} + \sin \frac{\pi}{2}\right).$$ Calculate inside: $$\sin \frac{\pi}{2} = 1,$$ so $$\frac{\pi}{4} + 1.$$ Therefore, $$\lim_{x \to \frac{\pi}{2}} \sin\left(\frac{x}{2} + \sin x\right) = \sin\left(\frac{\pi}{4} + 1\right).$$ **Final answers:** (a) $\lim_{x \to 2} f(x) = 3$ (b)(i) Limit does not exist (b)(ii) $1$ (b)(iii) $\sin\left(\frac{\pi}{4} + 1\right)$