= Solution
\b[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 $f_n(x)=f(x)$. Hence
$$
\boxed{f_n\longrightarrow f\text{ pointwise on }\mathbb R}.
$$
This <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.
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