# limit of $f:(0,\infty) \rightarrow \mathbb{R}$ as $x\to \infty$

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 .

• 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. – Darth Lubinus May 27 at 19:34
• As for the example you asked for, how about $f(x)=0$? – Darth Lubinus May 27 at 19:36
• 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. – user710290 May 28 at 6:50
• 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). – Darth Lubinus May 28 at 7:47