If \( f(x) = 3x - x^2 \), squeeze it between two functions to determine \( \lim_{x \to 1} f(x) \).

To find the limit \( \lim_{x \to 1} f(x) \) where \( f(x) = 3x - x^2 \), we can first evaluate \( f(1) \): \[ f(1) = 3(1) - (1)^2 = 3 - 1 = 2. \] Now, we will use the Squeeze Theorem to confirm this limit. To do this, we need to find two functions that bound \( f(x) \) as \( x \) approaches 1. First, we can rewrite \( f(x) \): \[ f(x) = 3x - x^2 = -x^2 + 3x. \] Next, we can factor \( f(x) \): \[ f(x) = - (x^2 - 3x) = - (x(x - 3)). \] To find bounds for \( f(x) \), we can analyze the behavior of \( f(x) \) around \( x = 1 \). We can evaluate \( f(x) \) at points close to 1: 1. For \( x = 0.5 \): \[ f(0.5) = 3(0.5) - (0.5)^2 = 1.5 - 0.25 = 1.25. \] 2. For \( x = 1.5 \): \[ f(1.5) = 3(1.5) - (1.5)^2 = 4.5 - 2.25 = 2.25. \] From these evaluations, we see that \( f(x) \) is increasing in the interval around \( x = 1 \). Now, we can establish bounds for \( f(x) \) as \( x \) approaches 1. We can observe that: \[ f(x) \text{ is continuous at } x = 1. \] Thus, we can use the fact that \( f(x) \) is continuous to apply the Squeeze Theorem. Since \( f(1) = 2 \) and \( f(x) \) is continuous, we can conclude: \[ \lim_{x \to 1} f(x) = f(1) = 2. \] Therefore, the limit is: \[ \boxed{2}. \]