Answer by PierreCarre for Proving a function isn't bounded

If $f: \mathbb{R}\setminus \{0\} \to \mathbb{R}$ was bounded there would exist a constant $M >0$ such that$$|f(x)| < M, \quad \forall x\ne 0.$$

However, this generates a contradiction since $|f(1/M)| > M$.