Is $\sin{(\log{x})}$ uniformly continuous on $(0,\infty)$? Is $\sin{(\log{x})}$ uniformly continuous on $(0,\infty)$? 
Let $x,y \in (0,\infty)$.
$$
|f(x)-f(y)| = |\sin{\log{x}} - \sin{\log{y}}| = \left|2 \cos{\frac{\log{xy}}{2}}\sin{\log{\frac{x}{y}}{2}} \right| \leq 2 \left|\sin{\frac{\log{\frac{x}{y}}}{2}} \right| \leq \left|\log{\frac{x}{y}} \right| = |\log{x} - \log{y}| 
$$
If we consider the interval is $(1,\infty)$, then it is easy to finish the above inequality.
But now the interval is $(0,\infty)$, I guess it is not uniformly continuous and now I don't have any idea to show it is not uniformly continuous.
Can anyone give a favor or some hints? Thank you!

No it is not uniformly continuous because $\log x$ is not uniformly continuous on $(0,1)$. 
If we take the sequence of numbers $(e^{-n})$ then $|e^{-n} -e^{-n-1}|$ tends to zero but $|\log(e^{-n})-\log(e^{-n-1})|=1$
