Solution
= Solution
\b[True.] Define $h(u)=g(1/u)$ for $0<u\leq1$ and $h(0)=0$. The hypothesis at infinity makes $h$ <continuous> on $[0,1]$. The <Weierstrass approximation theorem> gives <polynomials> $R_n$ with $\sup_{0\leq u\leq1}|R_n(u)-h(u)|\to0$. Since $t\mapsto1/t$ maps $[1,\infty)$ onto $(0,1]$,
$$
\boxed{\sup_{t\geq1}|R_n(1/t)-g(t)|\longrightarrow0.}
$$