Is $|x|^{\frac12} \sin (x)$ uniformly continuous on $\mathbb R$?

I spent almost two hours on this question but didn't work out yet.

Other two similar questions are to judge $x\sin(x),\ x\sin(x^2)$.

Former one let $x_n=2n\pi+\dfrac1n,\ y_n=2n\pi$, latter one let $x_n=\sqrt{2n\pi+\dfrac{\pi}{2}},\ y_n=\sqrt{2n\pi}$, both are not uniformly continuous.

About this one, its plot is similar to above two, so I guess it is not uniformly continuous, either. On the other hand, I think $x^{\frac12}$ can change the property of the function, it may be uniformly continuous. So I am confused now, please help me out. Thanks!

Hint. For $f(x)=|x|^{\frac12} \sin (x)$, consider the sequences $x_n:=2n\pi+\frac{1}{\sqrt{n}}$ and $y_n:=2n\pi$ for $n\geq 1$. We have that $x_n-y_n\to 0$ as $n\to \infty$. What about the following limit $$\lim_{n\to +\infty }\left(f(x_n)-f(y_n)\right)=?$$ What may we conclude?
P.S. More generally for $f(x)=|x|^{a} \sin (x^b)$ with $a,b>0$, take $$x_n:=\left(2n\pi+\frac{1}{n^{a/b}}\right)^{1/b}\quad\text{and}\quad y_n:=(2n\pi)^{1/b}.$$