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.

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

  • $\begingroup$ I think $x$ cannot be negative. $\endgroup$ – velut luna Sep 10 '17 at 16:36
  • $\begingroup$ @velutluna Thanks. I will edit my answer. $\endgroup$ – José Carlos Santos Sep 10 '17 at 16:36
  • $\begingroup$ I made the mistake of not writing its possible values in the first place. $\endgroup$ – LosLas Sep 10 '17 at 16:37
  • $\begingroup$ @LosLas I know. I wrote my answer before you corrected that, but it's OK. One thing, though: anything that's relevant for your question should be put within the body of the question, no matter what the title says. $\endgroup$ – José Carlos Santos Sep 10 '17 at 16:39
  • $\begingroup$ OK. got it. thank you so much for your rapid answer $\endgroup$ – LosLas Sep 10 '17 at 16:43

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