Uniform Continuity of function $f(x)=\sqrt x\sin\sqrt x$ on $[0,\infty)$

Question

I wanted to show uniform continuity of function $f(x)=\sqrt{x} \sin\sqrt{x}$ over $[0,\infty)$. I used all method know to me that continuity extension, definition, bounded derivative test, but not able to through.

Also thought for reverse that is not continuous but not able to find required equivalent sequence just like checking uct of $x \sin x $. Any Help will be appreciated.

Answer 1

Hint. Note that $f$ is differentiable (also at $x=0$) and the derivative $$f'(x)=\frac{\sin(\sqrt{x})}{2\sqrt{x}}+\frac{\cos(\sqrt{x})}{2}$$ is bounded in $[0,+\infty)$ .

Answer 2

Use that $f$ is uniformly continuous on $[0,1]$ (as continuous functions on compact sets are uniformly continuous) and that $f$ is uniformly continuous on $\mathbb{R}\setminus [0,1]$ as $f$ has bounded derivative there. Now combine those two results to deduce that $f$ is uniformly continuous on the enire real line.