If the function $\sqrt{x} \sin x$ is uniformly continuous on $\mathbb{R}$ First confusion is $\sqrt x$ has a domain of $[0, \infty)$, how we can say anything on $\mathbb{R}$?
If for $[0, \infty)$ than it is a continuous function on $(0, \infty)$ but what about the point zero will it be uniformly continuous there are not?

Hint:
Take $x_n = n\pi$ and $y_n = n\pi + \frac{1}{\sqrt{n}}$ and show that $|y_n-x_n| \to 0$ but $|f(y_n) - f(x_n)|\not\to 0$ as $n \to \infty$.
