Answer
$f(x)$ has an irremovable jump discontinuity at $x=5.$
Work Step by Step
$f(x)=\dfrac{|x-5|}{x-5};x\ne5\to$
$\lim\limits_{x\to5^+}f(x)=\dfrac{|5^+-5|}{5^+-5}=\dfrac{0^+}{0^+}=1.$
$\lim\limits_{x\to5^-}f(x)=\dfrac{|5^--5|}{5^--5}=\dfrac{0^+}{0^-}=-1.$
Since $\lim\limits_{x\to5^+}f(x)\ne\lim\limits_{x\to5^-}f(x),$ then $\lim\limits_{x\to5}f(x)$ does not exist and the function has an irremovable (jump) discontinuity.
