A function f satisfies |f(x) - 3| ≤ |x - 1| for all x. What can be concluded about lim_(x→1) f(x)?
Correct: B
Given |f(x) - 3| ≤ |x - 1| for all x. As x → 1, the right side |x - 1| → 0. By the Squeeze Theorem, |f(x) - 3| is squeezed between 0 and something that approaches 0, so |f(x) - 3| → 0. This implies f(x) → 3 as x → 1. Therefore lim_(x→1) f(x) = 3. The correct answer is B.