The function $f(x)=\frac{x-1}{x(x^2-1)},x≠1,f(1)=1$, is discontinuous at

Exactly three points

$f(x)=\frac{x-1}{x(x^2-1)}=\frac{x-1}{x(x-1)(x+1)},x≠1,f(1)=1$

so for $x=0,x=-1$

f(x) ⇒ not defined ⇒ it is discontinuous at $x = 0,-1$

$\underset{x→1}{\lim}f(x)=\underset{x→1}{\lim}\frac{x-1}{x(x-1)(x+1)}=\underset{x→1}{\lim}\frac{x-1}{x(x+1)}=\frac{1}{2}≠f(1)$

⇒ discontinuous at $x = 1$ also