Let $f(x)= x\sin(1/x)$ if $x>0$ and $0$ if $x=0 $ .I want to show that there isn't any sequence of polynomials converging to $f$ uniformly on $[1,\infty)$
I have to show for any sequence of polynomials $P_n(x),$ $\sup_{x\in[1,\infty)}|P_n(x)−f(x)|\nrightarrow0$?
$\lim_{x\rightarrow \infty}f(x) =1$ Is this correct? Where do I go from here?
For $x>0$ let $y=1/x.$ Then $f(x)=\frac {\sin y}{y}.$ Since $y\to 0^+$ as $x\to \infty,$ and $\lim_{y\to 0^+}\frac {\sin y}{y}=1,$ we have $\lim_{x\to \infty}f(x)=1.$
If $p(x)$ is a non-constant polynomial then as $x\to \infty$ we have $|p(x)|\to \infty$ and $f(x)\to 1$ so $\sup_{x\geq 0}|p(x)-f(x)|=\infty.$
If $p(x)$ is a constant $K$ then $$\sup_{x\geq 0}|p(x)-f(x)|\geq \max (|K-f(0)|,\lim_{x\to \infty}|K-f(x)| )=$$ $$=\max (|K|,|K-1|)\geq 1/2.$$