prove $\sqrt{x}\sin \left(\frac{1}{x}\right)$ is uniformly continuous in $(0,+\infty)$

The function is defined in $(0,+\infty)$

I set $\epsilon > 0$ so there exist $\delta > 0$ such that for every $x_{1}, x_{2}$ such that $|x_1-x_2| < \delta$ it exist that: $|f(x_1) - f(x_2)| < L$

but as I got $|\sqrt{x_1}\sin(\frac{1}{x_1})-\sqrt{x_2}\sin(\frac{1}{x_2})|$ I could not find a way to prove it… I tried to multiply by its conjugate but it got me nowhere.

• Why the -1? This seems like a suitable question for this site. – Mark Fischler Sep 10 '17 at 16:17
• Something minor, is the function defined to be $0$ at $x=0$? – velut luna Sep 10 '17 at 16:28
• actually it is not that minor as I forgot to point out that $f$ is defined in $(0,\infty)$. now fixed – LosLas Sep 10 '17 at 16:33

Define$$\begin{array}{rccc}f\colon&[0,+\infty)&\longrightarrow&\mathbb R\\&x&\mapsto&\begin{cases}\sqrt x\sin\left(\frac1x\right)&\text{ if }x\neq0\\0&\text{ otherwise.}\end{cases}\end{array}$$I will prove that it is uniformly continuous. It is clear that it is continuous. Therefore, the restriction of $$f$$ to $$[0,1]$$ is uniformly continuous (this is a standard Real Analysis theorem). The restriction of $$f$$ to $$[1,+\infty)$$ is uniformly continuous because it is continuous and $$\lim_{x\to+\infty}f(x)=0$$; this equality comes from$$\lim_{x\to+\infty}\sqrt x\sin\left(\frac1x\right)=\lim_{x\to0^+}\frac{\sin x}{\sqrt x}=\lim_{x\to0^+}\sqrt{x}\frac{\sin x}x=0\times1=0.$$It is easy to deduce from this that $$f$$ is uniformly continuous.
• I think $x$ cannot be negative. – velut luna Sep 10 '17 at 16:36