The limit at infinity of a function with derivative in $L^2$

Question

Suppose that $$\tag{1}f(x),f'(x),xf(x)\in L^{2}(\mathbb{R}),$$ then we have $$\tag{2}\frac{d}{dx}(x|f(x)|^{2})\in L^{1}(\mathbb{R}).$$ My question is can we get the following limit, and how $$\lim_{|x|\rightarrow\infty}x|f(x)|^{2}=0.$$

Progress: From $(1)$ we have $|x|^{1/2}f(x)\in L^{2}(\mathbb{R})$ and $f(x)\in L^{1}(\mathbb{R})$ by Cauchy-Schwarz inequality.

Answer 1

I don't see why $f$ would be in $L^1$...

EDIT. The hypothesis $f\in L^2$ is useless. Indeed, $f$ is continuous; then $xf\in L^2\implies f\in L^2$.

$xf,f'\in L^2 \implies \int_0^{+\infty} xff'=[xf^2/2]_0^{+\infty}-1/2\int_0^{+\infty} f^2\in\mathbb{R}$; since $f^2\in L^1$, $\lim_{+\infty}xf^2$ exists and is $0$ (because $xf^2\in L^1$). In the same way, $\lim_{-\infty}xf^2=0$.