Let $f:[1,\infty]\longrightarrow \mathbb{R}$ a continous function such that $\text{lim}_{x\rightarrow \infty}f(x)$ exists and let $\epsilon >0 $. Show that exists a polynomial $p$ such that $\text{sup}_{x\in[1,\infty]}\{|f(x)-p\left(\frac{1}{x}\right)|\}<\epsilon$.
By the Weierstrass aproximation theorem exists $p$ polynomial such that $\text{sup}_{x\in[1,\infty]}\{|f(x)-p(x))|\}<\epsilon$. But I don't know how to use the exitence of $\text{lim}_{x\rightarrow \infty}f(x)$ to conclude that $\text{sup}_{x\in[1,\infty]}\{|f(x)-p\left(\frac{1}{x}\right)|\}<\epsilon$. Any suggestions
Hint: Start by showing that the function $g : [0,1] \to \mathbb{R}$ defined by $$g(y) = \begin{cases}f(\tfrac{1}{y}) & y \in (0,1] \\ \lim\limits_{x \to \infty}f(x) & y = 0\end{cases}$$ is continuous on $[0,1]$. Then use the Weierstrass aproximation theorem on $g(y)$ to find a polynomial $p$ such that ....
Can you finish the proof from here?