Let \(f\) be continuous on \([0,\infty)\) and suppose the finite limit \[ L=\lim_{x\to\infty}f(x) \] exists. Prove that \(f\) is uniformly continuous on \([0,\infty)\).
Solution
Fix \(\varepsilon>0\). Since \(f(x)\to L\), choose \(R>0\) such that \[ |f(x)-L|<\frac\varepsilon2 \qquad(x\ge R). \] Then whenever \(x,y\ge R\), \[ |f(x)-f(y)| \le |f(x)-L|+|f(y)-L| <\varepsilon. \]
On the compact interval \([0,R+1]\), the function \(f\) is uniformly continuous by the Heine–Cantor theorem. Hence there exists \(\delta_0>0\) such that \[ |x-y|<\delta_0, \qquad x,y\in[0,R+1], \] implies \[ |f(x)-f(y)|<\varepsilon. \] Set \[ \delta=\min\{1,\delta_0\}. \] Suppose \(x,y\ge0\) and \(|x-y|<\delta\). If both \(x,y\ge R\), the tail estimate above applies. Otherwise, one of the two points is \(<R\); because \(|x-y|<1\), both points then belong to \([0,R+1]\), so the compact-interval estimate applies. Thus in all cases \[ |x-y|<\delta\implies |f(x)-f(y)|<\varepsilon. \] Therefore \(f\) is uniformly continuous on \([0,\infty)\).