Past exam of the mathematics course of the University of Cambridge / 2019 / ib / Paper 3 / 2E / b

The function $x\mapsto x^2$ is not uniformly continuous on $\mathbb{R}$. Take $x_n=n$ and $y_n=n+1/n$. Then $|x_n-y_n|\to 0$, whereasThis contradicts the sequential criterion for uniform continuity.

The square-root function is uniformly continuous on $[0,\infty )$. For $x,y\ge 0$,so $|\sqrt{x}-\sqrt{y}|\le \sqrt{|x-y|}$. Given $\epsilon >0$, choosing $\delta ={\epsilon }^2$ proves uniform continuity.

The function $x\mapsto \cos (1/x)$ is uniformly continuous on $[1,\infty )$. Its derivative satisfiesso the mean value theorem givesThus the function is Lipschitz, which implies uniform continuity.