(January 2009 #2a, obfuscated) Is \(f(x)=\ln x\) uniformly continuous on \((0,1]\)? Prove your answer.
Solution
1. \(f(x) = \ln x\) is NOT uniformly continuous on \((0,1]\).Proof: take \(x_n = e^{-n}\) and \(y_n = e^{-n-1}\).Then \(x_n, y_n \in (0,1]\) and \[|x_n - y_n| = e^{-n-1}(e - 1) \to 0,\] but \(|\ln x_n - \ln y_n| = |-n - (-n-1)| = 1\) for all \(n\).Hence for \(\epsilon = 1\) (say \(\epsilon = 1/2\)) there is no \(\delta > 0\) working for all pairs; the \(\epsilon\)-\(\delta\) definition of uniform continuity fails.
2. \(g(x) = \sin(1/x)\) is also NOT uniformly continuous on \((0,\pi]\).Proof: take \(x_n = \frac{1}{\pi/2 + 2\pi n}\) and \(y_n = \frac{1}{3\pi/2 + 2\pi n}\).Both lie in \((0,\pi]\), and \(|x_n - y_n| \to 0\), while \(g(x_n) = \sin(\pi/2 + 2\pi n) = 1\) and \(g(y_n) = \sin(3\pi/2 + 2\pi n) = -1\), so \(|g(x_n) - g(y_n)| = 2\) for all \(n\).Not uniformly continuous.
3. Q.E.D.