## Calculus 10th Edition

$f(x)$ has an irremovable (vertical asymptote) discontinuity at $x=5.$
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.