calculus - Proving uniform continuity of trigonometric function arctan

In my assignment I have to prove that the following function is uniformly continuous:

$$f(x) = \arctan x \cdot \sin\left(\frac{1} {x}\right) $$ in the open interval $(0,\infty) $.

I thought I'd prove the one side limits $ x\to 0^+$ and $ x\to \infty^-$.

$$\lim_{x\to\infty^-}=\frac{\pi} {2}\cdot0=0.$$

For the second limit I thought using the squeeze theorem. Please let me know if I made a mistake:

$$0\le \arctan x \cdot \sin \left(\frac{1} {x} \right)\le \arctan x .$$

Therefore, from the squeeze theorem:

$$\lim_{x\to 0^+} \arctan x \cdot\sin\left(\frac{1} {x}\right) =0$$

Since $\arctan$ and $\sin$ are continuous always when it is defined, we can say that the function is uniformly continuous, since the one side limits exists.

I have only one concern here: zero is not defined here, in the open interval from the beginning. Is it a problem to use the squeeze theorem here? If it is, thought I'd use $\frac {x} {x+1} $ instead of $0$.

Thanks,

Alan