Past exam of the mathematics course of the University of Cambridge / 2017 / ib / Paper 1 / 11G / iii / Solution

True. Define $c_n(x)=\max (-n,\min (x,n))$ and $f_n=f\circ c_n$. The map $c_n$ is Lipschitz continuous with constant one. The restriction of $f$ to $[-n,n]$ is uniformly continuous by the Heine-Cantor theorem; its modulus of continuity combined with $|c_n(x)-c_n(y)|\le |x-y|$ proves uniform continuity of $f_n$ on all of $\mathbb{R}$. For each fixed $x$, once $n\ge |x|$ we have $f_n(x)=f(x)$. HenceThis pointwise approximation by uniformly continuous functions even gives eventual equality on every fixed bounded interval. It need not give uniform convergence on the whole line, as the bounded counterexample in part (i) already demonstrates.