uniform continuity in $(0,\infty)$ Prove uniform continuity over $(0,\infty)$ using epsilon-delta of the function
$$f(x)=\sqrt{x}\sin(1/x)$$
I know that I start with: For every $\varepsilon>0$ there is a $\delta>0$ such that for
any $x_1$, $x_2$ where $$|x_1-x_2|<\delta$$ then $$|f(x_1)-f(x_2)|<\varepsilon$$
But I can't manage the expression of $f(x)$.
Thank you for helping.

Note that
$$\left| \sqrt{x} \sin (1/x)- \sqrt{y} \sin (1/y) \right|\\ \leqslant \sqrt{x}\left|  \sin (1/x)-  \sin (1/y) \right|+|\sqrt{x}-\sqrt{y}||\sin(1/y)| \\\leqslant \sqrt{x}\left|  \sin (1/x)-  \sin (1/y) \right|+|\sqrt{x}-\sqrt{y}|.$$
Using $\displaystyle \sin a - \sin b = 2 \sin \left(\frac{a-b}{2}\right)\cos \left(\frac{a+b}{2}\right),$ we have
$$\left| \sqrt{x} \sin (1/x)- \sqrt{y} \sin (1/y) \right| \\ \leqslant 2\sqrt{x}\left|  \sin \left(\frac1{2x}-\frac1{2y}\right) \right|\left|  \cos \left(\frac1{2x}+\frac1{2y}\right) \right|+|\sqrt{x}-\sqrt{y}| \\ \leqslant  2\sqrt{x}\left|  \sin \left(\frac1{2x}-\frac1{2y}\right) \right|  +|\sqrt{x}-\sqrt{y}| \\ \leqslant2\sqrt{x}\left| \frac1{2x}-\frac1{2y} \right|  +|\sqrt{x}-\sqrt{y}| \\ \leqslant \frac{\left| x-y \right|}{\sqrt{x}|y|}  +|\sqrt{x}-\sqrt{y}| \\ \leqslant  \frac{\left| x-y \right|}{\sqrt{x}|y|}  +\frac{|x-y|}{{\sqrt{x}+\sqrt{y}}} .$$
Hence, on the interval $[a,\infty)$ we have
$$\left| \sqrt{x} \sin (1/x)- \sqrt{y} \sin (1/y) \right| \leqslant \left(\frac1{a^{3/2}} + \frac1{2a^{1/2}}\right)|x - y| < \epsilon,$$
when
$$ |x-y| < \delta = \epsilon \left( \frac1{a^{3/2}} + \frac1{2a^{1/2}} \right)^{-1}.$$
Therefore $f$ is uniformly continuous on $[a, \infty)$ for any $a > 0$.
Note that $f$ is continuous on $(0,a]$ and $\lim_{x \to 0+}f(x) = 0.$ Then $f$ is continuously extendible and uniformly continuous on the compact interval $[0,a]$ as well.
