Prove that $\sqrt{x}{\sin \frac1x}$ is uniformly continuous at $(0, \infty)$ I need to prove that $f(x) =\sqrt{x}{\sin \frac1x}$ is uniformly continuous at $(0, \infty )$.
I managed to show that:
$-\sqrt{x} <= f(x) <= \sqrt{x}$, 
$\lim_{x\to0}\sqrt{x} = \lim_{x\to0}-\sqrt{x} = 0$, 
and so $\lim_{x\to0}f(x) = 0$
meaning that: $\lim_{x\to0^+}f(x) = 0$.
I also showed: $\lim_{x\to\infty}f(x) = 0$.
How should I continue? Will showing the $f$ is continuous at $(0,1)$ be enough?
If so how can I prove that it is continuous at $(0,1)$?
 A: We know that $f(x):=\sqrt{x}\sin\left(\frac{1}{x}\right)$ is continuous on the closed and bounded interval $[0,1]$ - this is easy to see, as you have proven that the limit as $x\rightarrow 0$ exists and is equal to $0$, therefore the function can be extended to a continuous function on $[0,1]$ and this well known result still holds.
We now look at the interval $[1,\infty)$.
Calculating the first derivative, we have:
$$f'(x) = \frac{\sin\left(\frac{1}{x}\right)}{2\sqrt{x}}-\frac{\cos\left(\frac{1}{x}\right)}{x^{\frac{3}{2}}}$$
Note that $f'(1)=\frac{\sin(1)-2\cos(1)}{2}\approx -0.1196$ and that $\lim\limits_{x\rightarrow\infty}f'(x)=0$. 
Since $f'(x)$ is continuous on $[1,\infty)$, the fact that these two limits are finite implies that there must exist an upper bound on $|f'(x)|$ for $x\in[1,\infty)$.
Since the derivative is bounded, $f(x)$ is Lipschitz-continuous on $[1,\infty)$, which implies uniform continuity.
As $f(x)$ is uniformly continuous on $[0,1]$ and $[1,\infty)$, it is therefore also uniformly continuous on $[0,\infty)$ (see Paramand Singh's excellent answer on this question as to why the uniform continuity of a function on two sets $A$ and $B$ implies the uniform continuity on $A\cup B$ if $A$, $B$ and $A\cup B$ are intervals) and any subset of this set - thus it is certainly uniformly continuous on $(0,\infty)$.
A: Denote your function as $f(x)$
First $\lim_{x\rightarrow 0} \sqrt{x}sin{\frac{1}{x}}=0$,add a definition when x=$0$ let your function equals $0$ when $x=0$,so your function is uniformly continous on $(0,a]$,$a>0$.
$f^{'}(x)=\frac{1}{2\sqrt{x}}sin{\frac{1}{x}}-cos{\frac{1}{x}}x^{-\frac{3}{2}}$
is bounded on $[a,\infty)$ with upper bound $|f^{'}(x)|\le \frac{1}{2\sqrt{a}}+a^{-\frac{3}{2}}$,so f continous uniformly on $R_{+}$
