I know that there are questions about it but I want to have a clue before I look at the answers.

So can someone give me a clue to how can I prove that: 

$$
f(x) = \sqrt{x}\sin \frac{1}{x}
$$

at $(0,\infty)$

Has uniform continuity?

Note: I didn't learn about differentiability yet. 

\*EDIT: 

Using daruma answer:

1. We will extend the continuity of $f$ to $[0, \infty]$, but showing that $\displaystyle \lim\_{x \to 0}f(x) = f(0) = 0$ so $f$ is also continous at $0$:

We will use hiene and look at the function as multiplication of 2 series. 

$f$ is built from a serie that goes to $0$ at $x \to 0$: $\sqrt{x}$ multiply by a bounded serie $\displaystyle \sin \frac{1}{x}$ so the serie goes to $0$ so is the function $f$.

So the function is continuous at $[0, \infty)$

2. Showing that $\lim\_{x \to \infty}f(x) = 0$ that will allow us to say, that $\forall \epsilon > 0, \forall x,y > R: |f(x)|, |(f(y)| < \frac{\varepsilon}{4}$, therfore, $\forall \delta > 0,$ at $[R, \infty)$, if $|x-y| < \delta$: $|f(x) - f(y)| < \varepsilon$.

Therfore, $f$ has uniform continuity at $[R, \infty)$

Now, by Cantor theorom, that if $f$ is continious at a closed interval $[a,b]$ it is also uniformly continious at that interval we conclude that $f$ is uniformly continious at $[0,R]$ as a multiplication of continious functions. 

Making a union of the two intervals, we get that $f$ has uniform continuity at $[0 ,\infty)$, particularly, at $(0, \infty)$, as needed.

I would like to hear if you have notes or corrections if i didnt get something right.

Thanks all for the help.