Uniform convergence of a sequence of polynomials to a specified function

Question

Suppose we have $f: [1, \infty) \to \mathbb{C}$ is continuous and $\lim_{x\to\infty}f(x)$ exists. Is it true that there exists a sequence of polynomials $p_n$ such that $p_n(\frac{1}{x}) \to f(x)$ uniformly on $[1, \infty)$?

Should the Weierstrass's Approximation Theorem be used here? If so, how? I don't know where the $\frac{1}{x}$ comes from...

Thanks!

Answer 1

Let $g(x)=f(\frac 1x)$, then $g$ is defined on $(0, 1]$. Since $\lim_{x\to\infty}f(x)=L$ exists, we can define $g(0)=L$ and make $g$ a continuous function on $[0, 1]$. Apply Weierstrass to $g$, get $p_n(x)\to g(x)$, then $p_n(\frac 1x)\to g(\frac 1x)=f(x)$.