Answer
$f(x)$ has an irremovable (vertical asymptote) discontinuity at $x=5.$
Work Step by Step
The only restriction on $f(x)$'s domain is that $x\ne5.$
When $x=5,$ the denominator is $0;$ hence, the function has a vertical asymptote at $x=5.$
This discontinuity is not removable.
