\( \lim _{x \rightarrow 1} \frac{|x-1|}{x-1}=\square \) Question Help: Submit Question Jump to Answer

To evaluate \( \lim _{x \rightarrow 1} \frac{|x-1|}{x-1} \), we need to consider the behavior of the function as \( x \) approaches 1 from both sides. When \( x \) approaches 1 from the left (i.e., \( x < 1 \)), \( |x-1| = -(x-1) \), so: \[ \lim _{x \rightarrow 1^-} \frac{|x-1|}{x-1} = \lim _{x \rightarrow 1^-} \frac{-(x-1)}{x-1} = \lim _{x \rightarrow 1^-} -1 = -1. \] When \( x \) approaches 1 from the right (i.e., \( x > 1 \)), \( |x-1| = x-1 \), therefore: \[ \lim _{x \rightarrow 1^+} \frac{|x-1|}{x-1} = \lim _{x \rightarrow 1^+} \frac{x-1}{x-1} = \lim _{x \rightarrow 1^+} 1 = 1. \] Since the left-hand limit and right-hand limit do not match, the overall limit does not exist. Thus, the limit is: \[ \lim _{x \rightarrow 1} \frac{|x-1|}{x-1} = \text{undefined or does not exist.} \]