Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics II / 4 / b

For any $x,y\in \mathbb{R}$, rationalizing the difference of square roots givessince $f(x),f(y)\ge 1$. Given $\epsilon >0$, use uniform continuity of $f$ with tolerance $2\epsilon$ to obtain $\delta >0$. Whenever $|x-y|<\delta$, the displayed estimate yields $|g(x)-g(y)|<\epsilon$. Thus $g$ has uniform continuity on $\mathbb{R}$. The square root function is in fact Lipschitz continuous with constant $1/2$ on $[1,\infty )$.

TakeBoth functions have values in $[1,\infty )$. The reverse triangle inequality gives $|g(x)-g(y)|\le |x-y|$, so $g$ is Lipschitz continuous and hence has uniform continuity.

For $x_n=n$ and $y_n=n+1/n$, the distances satisfy $|x_n-y_n|=1/n\to 0$, whereasIn particular tolerance $\epsilon =1$ defeats every proposed $\delta$: choose $n$ with $1/n<\delta$. Thus $f$ does not have uniform continuity on $\mathbb{R}$.