Answer
False.
Work Step by Step
$\lim\limits_{x\to1^-}f(x)=\lim\limits_{x\to1^-}\dfrac{|x-1|}{x-1}=\dfrac{|1^--1|}{1^--1}=\dfrac{0^+}{0^-}=-1.$
$\lim\limits_{x\to1^+}f(x)=\lim\limits_{x\to1^+}\dfrac{|x-1|}{x-1}=\dfrac{|1^+-1|}{1^+-1}=\dfrac{0^+}{0^+}=1.$
Since $\lim\limits_{x\to1^+}f(x)\ne\lim\limits_{x\to1^-}f(c),$ then $\lim\limits_{x\to1}f(x)$ does not exist and the function has a discontinuity at $x=1.$
The function is continuous over the interval $(-\infty, 1)$ U $(1, \infty).$
