Let $f:(0,\infty) \rightarrow \mathbb{R}$ be uniformly continous .Then

$(1)\lim_{x\rightarrow 0+}f(x)$ and $\lim_{x\rightarrow \infty}f(x)$ exist

$(2)\lim_{x\rightarrow 0+}f(x)$ exist but $\lim_{x\rightarrow \infty}f(x)$ need not exist

$(3)\lim_{x\rightarrow 0+}f(x)$ need not exist but $\lim_{x\rightarrow \infty}f(x)$ exist

$(4)$neither $\lim_{x\rightarrow 0+}f(x)$ nor $\lim_{x\rightarrow \infty}f(x)$ exist

My attemp:

I show $\lim_{x\to 0+}f(x)$ exist .

Ok, let $\{x_n\}$ be a sequence in $(0,\infty)$ converging to $0$. Since uniformly continous functions maps Cauchy sequenced to Cauchy sequences, $\{f(x_n)\}$ is Cauchy and hence it is convergent . Since $x_n$ is arbitary sequence converging to $0$ , it follows by sequential criterion that $\lim_{x\to 0+} f(x)$ exist.Is this proof correct??

Now, if $f(x)=sinx$ , then it is Lipschitz and uniformly continous but $\lim_{x\to \infty}f(x)$ does not exist.

So correct option is $(2)$

Please give an example of a function satisfying the hypothesis and $\lim_{x\to \infty}f(x)$ does exist.I suppose it is bounded .

Thanks for your time.

  • 1
    $\begingroup$ Your proof is almost perfect! The only thing you have left to prove is that the limit of $(f(x_{n}))$ does not depend on the choice of the sequence $x_{n}\to 0$. Nevertheless, the correct option is the one you said. $\endgroup$ – Darth Lubinus May 27 at 19:34
  • 1
    $\begingroup$ As for the example you asked for, how about $f(x)=0$? $\endgroup$ – Darth Lubinus May 27 at 19:36
  • $\begingroup$ Thanks for the response. I didn't think of the constant function !! Sorry, but I am too curious not to ask for a non-constant function.I can ,kind of picture it in mind but cannot write the function explicitly. $\endgroup$ – user710290 May 28 at 6:28
  • $\begingroup$ As for the first part you said, I am trying to prove that: Let $f(x_n) \to l$ and $f(y_n)\to l'$ where $x_n,y_n \to 0$. Now $|l-l'|=|l-f(x_n)+f(y_n)-l'+f(x_n)-f(y_n)|\le |f(x_n)-l|+|f(y_n)-l'|+|f(x_n)-f(y_n)|$ .Now each term on right side can be made smaller than $\epsilon /3$ by a selecting $n\gt M$ , which is possible. Hence $0\le |l-l|\lt \epsilon $ .Thus $l=l'$. Please give a confirmation. $\endgroup$ – user710290 May 28 at 6:50
  • $\begingroup$ To me it seems correct. I'd simply clarify that since $f$ is uniformly continuous, and $|x_{n}-y_{n}|\to 0$, it's true that $|f(x_{n})-f(y_{n})|\to 0$ (come to think of it, that alone proves that $l=l'$, I believe). $\endgroup$ – Darth Lubinus May 28 at 7:47

Your Answer

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

Browse other questions tagged or ask your own question.