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 $|f(x) - 3| \leq 4(x - 2)^2$. Recall that if $|f(x) - L| \leq g(x)$ and $\lim_{x \to a} g(x) = 0$, then $\lim_{x \to a} f(x) = L$. Here, $g(x) = 4(x - 2)^2$ and $\lim_{x \to 2} 4(x - 2)^2 = 0$. Therefore, by the Squeeze Theorem, $$\lim_{x \to 2} f(x) = 3.$$ 3. **Step (b)(i):** Evaluate $\lim_{x \to 3} \frac{|x - 3|}{x - 3}$. Consider the left-hand limit ($x \to 3^-$): For $x < 3$, $x - 3 < 0$, so $|x - 3| = -(x - 3)$. Thus, $$\lim_{x \to 3^-} \frac{|x - 3|}{x - 3} = \lim_{x \to 3^-} \frac{-(x - 3)}{x - 3} = \lim_{x \to 3^-} -1 = -1.$$ Consider the right-hand limit ($x \to 3^+$): For $x > 3$, $x - 3 > 0$, so $|x - 3| = x - 3$. Thus, $$\lim_{x \to 3^+} \frac{|x - 3|}{x - 3} = \lim_{x \to 3^+} \frac{x - 3}{x - 3} = \lim_{x \to 3^+} 1 = 1.$$ Since the left and right limits are not equal, 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: $$(x + 1)^2 = x^2 + 2x + 1 = x^2 \left(1 + \frac{2}{x} + \frac{1}{x^2}\right).$$ So the expression becomes: $$\frac{x^2 \sqrt{1 - \frac{1}{x^4}}}{x^2 \left(1 + \frac{2}{x} + \frac{1}{x^2}\right)} = \frac{\sqrt{1 - \frac{1}{x^4}}}{1 + \frac{2}{x} + \frac{1}{x^2}}.$$ Taking the limit as $x \to \infty$: $$\lim_{x \to \infty} \sqrt{1 - \frac{1}{x^4}} = \sqrt{1 - 0} = 1,$$ $$\lim_{x \to \infty} \left(1 + \frac{2}{x} + \frac{1}{x^2}\right) = 1 + 0 + 0 = 1.$$ Therefore, $$\lim_{x \to \infty} \frac{\sqrt{x^4 - 1}}{(x + 1)^2} = \frac{1}{1} = 1.$$ 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: $$\lim_{x \to \frac{\pi}{2}} \sin\left(\frac{x}{2} + \sin x\right) = \sin\left(\lim_{x \to \frac{\pi}{2}} \left(\frac{x}{2} + \sin x\right)\right).$$ Calculate the inner limit: $$\lim_{x \to \frac{\pi}{2}} \frac{x}{2} = \frac{\pi}{4},$$ $$\lim_{x \to \frac{\pi}{2}} \sin x = \sin \frac{\pi}{2} = 1.$$ So, $$\lim_{x \to \frac{\pi}{2}} \left(\frac{x}{2} + \sin x\right) = \frac{\pi}{4} + 1.$$ 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) $\sin\left(\frac{\pi}{4} + 1\right)$