(1.2) \( \lim _{x \rightarrow 1} \frac{\ln (-x+2)}{x-1} \)

To evaluate the limit \( \lim _{x \rightarrow 1} \frac{\ln (-x+2)}{x-1} \), we first substitute \( x = 1 \): \[ \frac{\ln(-1 + 2)}{1 - 1} = \frac{\ln(1)}{0} = \frac{0}{0} \] This indicates that we can apply L'Hôpital's Rule since both the numerator and the denominator approach 0. We need to differentiate the numerator and denominator separately. The derivative of the numerator, \( \ln(-x + 2) \), using the chain rule, is: \[ \frac{d}{dx} \ln(-x + 2) = \frac{1}{-x + 2} \cdot (-1) = \frac{-1}{-x + 2} = \frac{1}{x - 2} \] The derivative of the denominator \( x - 1 \) is: \[ \frac{d}{dx}(x - 1) = 1 \] Applying L'Hôpital's Rule gives us: \[ \lim_{x \rightarrow 1} \frac{\ln(-x + 2)}{x - 1} = \lim_{x \rightarrow 1} \frac{\frac{1}{x - 2}}{1} = \frac{1}{1 - 2} = \frac{1}{-1} = -1 \] Thus, the limit is \[ \boxed{-1} \]