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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!

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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 $$x_n:=\left(2n\pi+\frac{1}{n^{a/b}}\right)^{1/b}\quad\text{and}\quad y_n:=(2n\pi)^{1/b}.$$

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  • $\begingroup$ Thank you so much! I tried this sequences yesterday but got nothing, it must be too late that I was so stupid. Now it is obvious, and for general, +1! $\endgroup$ – Ferry Tau Oct 16 '17 at 3:02

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