Stone-Weierstrass theorem with $p(1/x)$

Question

I am trying to prove that for a continuous $f\colon[1,\infty)\rightarrow\mathbb R$ and $f(x)\to a$ as $x\to\infty$ it could be approximated by $g(x)=p(1/x)$ where $p$ is a polynomial.

Answer 1

Hint. If $f: [1,\infty)\to\mathbb R$ is continuous, and $g(x)=f(1/x)$, then $g$ is continuous in $(0,1]$. If $f$ also has a (finite) limit, as $x\to\infty$, then $g$ is continuous at $0$, and altogether it is continuous in the whole closed interval $[0,1]$.

If now $p$ is a polynomial, such that $$ \max_{x\in [0,1]}\lvert g(x)-p(x)\rvert<\varepsilon, $$
then $$ \sup_{x\in [1,\infty)}\lvert f(x)-p(1/x)\rvert<\varepsilon. $$ Therefore, $f$ can be approximated by polynomials of the form $p(1/x)$, by virtue of Stone-Weirestrass Theorem.