1. **State the problem:** We are given the inequality $|f(x) - 3| \leq 4(x - 2)^2$ and asked to find $\lim_{x \to 2} f(x)$. 2. **Recall the squeeze theorem:** If $g(x) \leq f(x) \leq h(x)$ near $x = a$ and $\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L$, then $\lim_{x \to a} f(x) = L$. 3. **Apply the inequality:** From $|f(x) - 3| \leq 4(x - 2)^2$, we have $$-4(x - 2)^2 \leq f(x) - 3 \leq 4(x - 2)^2$$ which implies $$3 - 4(x - 2)^2 \leq f(x) \leq 3 + 4(x - 2)^2$$ 4. **Find the limits of the bounding functions:** $$\lim_{x \to 2} \left(3 - 4(x - 2)^2\right) = 3 - 4 \cdot 0 = 3$$ $$\lim_{x \to 2} \left(3 + 4(x - 2)^2\right) = 3 + 4 \cdot 0 = 3$$ 5. **Use the squeeze theorem:** Since both bounding functions approach 3 as $x \to 2$, by the squeeze theorem, $$\lim_{x \to 2} f(x) = 3$$ **Final answer:** $$\boxed{3}$$