A source claims that \[ f(x)=x^{-n},\qquad n\in\ZZ_{\geq0}, \] is uniformly continuous on \([0,\infty)\). Determine for which \(n\) this is correct, and explain the defect when it is false.
Solution
For \(n=0\), \(f\equiv1\), so \(f\) is uniformly continuous on \([0,\infty)\).
For every \(n\ge1\), the claim is already ill-posed at \(0\), since \(x^{-n}\) is not defined there. Even on its natural domain \((0,\infty)\), the function is not uniformly continuous. Take \[ x_k={1\over k}, \qquad y_k={1\over k+1}. \] Then \[ |x_k-y_k|={1\over k(k+1)}\to0, \] but \[ |f(x_k)-f(y_k)| =|k^n-(k+1)^n| \ge1 \] for every \(k\). Hence no uniform-continuity modulus can exist.