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.