# Uniformly approximating $f\in \mathcal{C}([1,\infty))$ with polynomials where $\underset{x\rightarrow +\infty}{\lim} f(x)=a$

Suppose $$f\in \mathcal{C}([1,\infty))$$ and $$\underset{x\rightarrow +\infty}{\lim} f(x)=a$$. Show that $$f$$ can be uniformly approximated on $$[1,\infty)$$ by functions of the form $$g(x)=p(1/x)$$, where $$p$$ is a polynomial.

So I know I need to use the Weierstrass theorem, but I do not know if my approach is right. First, I know that Weierstrass is applied only on intervals of the form $$[a,b]$$. Would I be able to use it on the interval $$[a, a+n]$$ and show I can approximate $$f$$ for any $$n\in \mathbb{N}$$?

I am also thinking that the end goal here is for my polynomial, $$p_n$$, to equal $$a$$ at $$x=0$$? That way $$\underset{x\rightarrow +\infty}{\lim}p_n(1/x)=a$$.

If anyone had any hints for the problem and/or can point out if my thinking about this is on the right track, then I'd greatly appreciate it.

• Think about $q\colon[0,1]\to\mathbb{C}$, $$q(t)=\begin{cases}f(1/t)&t>0\\a&t=0\end{cases}$$ – user10354138 Nov 14 '18 at 4:10

Hint: Since $$f\in \mathcal{C}([1,\infty))$$ we have that $$f\circ 1/x\in \mathcal{C}((0,1])$$. Now, since $$\lim_{x\to\infty}f(x)=a$$, $$\lim_{x\to 0^+}f\circ 1/x=a$$. What does this then tell us about the function $$g:[0,1]\to\mathbb{R},\qquad g(x)=\begin{cases} f\circ 1/x & x\in (0,1] \\ a & x=0 \end{cases}?$$