Can $\sin(1/x)$ be approximated pointwise by polynomials over $(0,\infty)$ Can the function $f(x)=\sin(1/x)$ on $(0,\infty)$ be approximated by a sequence of polynomials pointwise on the domain?I am sure that uniform approximation is not possible because $\lim_{x\to 0+}\sin(1/x)$ does not exist.But is there a possibility of pointwise approximation by a polynomial sequence?
[Note: I am an undergraduate student and the only thing that I can use is Weierstrass polynomial approximation and any other independent idea,but I know nothing of approximation theory,so I am expecting some elementary answer.]
 A: I feel like you can do $\sup_{x\in[1/n,n]}|f(x)-p_{n}(x)|<1/n$ by applying Weierstrass on each $[1/n,n]$.
A: Short answer: Yes, because any continuous function on a compact interval can be approximated arbitrarily sharply by a polynomial. So, at step $n$ of your approximating sequence, consider the compact subset $[1/n, n]$ of $[0, \infty)$, and find (by Weierstrass approximation) a polynomial $P_n(x)$ which is at distance $\leq 1/n$ from your $\sin(1/x)$ function, uniformly on $[1/n, n]$. Then for any fixed point $x$ of $(0, \infty)$, $P_n(x)$ will converge to $\sin(1/x)$, since for $n$ large enough you will always have $x \in [1/n, n]$, and therefore $\left|\sin(1/x) - P_n(x)\right| \leq 1/n \stackrel{n \to \infty}\to 0$.
Actually there is nothing specific to $\sin(1/x)$ in this argument: it works the same for any function that is continuous on any open interval of $\mathbf{R}$.
A: Sure, but the result doesn't mean much.
Let $I_n = [{1 \over n},n]$, this is compact and we can choose a polynomial $p_n$ such that
$\sup_{x \in I_n}|p_n(x)-f(x) | < {1 \over n}$.
Then for any fixed $x>0$ we have $p_n(x) \to f(x)$.
