1
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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
5
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.