Let $f:[1,\infty )\to \Bbb R$ be continuous, and suppose that $\lim \limits_{x \to \infty}f(x)$ exists. Use the Weierstrass approximation theorem to prove that $f$ can be uniformly approximated on $[1,\infty )$ by a function $g$ of the form $g(x)=p(\frac{1}{x})$, where $p$ is a polynomial.
By the assumption that $\lim \limits_{x \to \infty}f(x)=L<\infty$, given $\epsilon>0$ there exists $R>0$ such that $\forall x>R$, we have $|f(x)-L|<\varepsilon$. If we restricts $f$ to $[R,a]$ for any $a>R$, then by the Weierstrass theorem, there exists a sequence of polynomials $\{P_n\}$ converges uniformly to $f$ on $[R,a]$. We find that $P_n(x)=L$ on $[R,a]$, when $n$ is large enough. So, I wonder if we restrict $f$ to $[1,R]$, can we find a sequence of polynomials $\{p_n\}$ such that $\lim \limits_{n\to \infty} p_n(x)=f(x)$ uniformly on [1,R] and $p_n(x)=L$ on $[R,\infty)$ such that $p_n\to f$ uniformly on $[1,\infty)$?
I have not figured out how to find the function $g$ mentioned in the statement, I 'm not sure how to prove this statement. Appreciate any suggestion.
Define $\tilde{f}(x) = \begin{cases} f({1 \over x}),& x \in (0,1]\\ \lim_t f(t), & x=0 \end{cases}$. Note that $\tilde{f}$ is continuous on the compact interval $[0,1]$ hence for any $\epsilon>0$ there is some polynomial $p$ such that $\|\tilde{f}-p\|_\infty < \epsilon$.
For $t \ge 1$ we have $|\tilde{f}({ 1 \over t}) - p({ 1 \over t})|< \epsilon$ and since $\tilde{f}({1 \over t}) = f(t)$, we get the desired result.