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Let $f:(0,\infty)\to(0,\infty)$ be uniformly continuous function. Does it imply $$\lim_{x\to\infty} {f(x+{1\over x})\over f(x)}=1\;?$$

By uniform continuity , for any $\epsilon>0$, $\exists\delta>0$ such that $|f(x)-f(y)|<\epsilon$ as $|x-y|<\delta$.
So, $|f(x+{1\over x})-f(x)|<\epsilon$ as $|{1\over x}|<\delta$ or $x>\delta$.
Thus, $$\left| {f(x+{1\over x})\over f(x)}-1\right|<{\epsilon\over f(x)},\ \forall x>\delta$$.
Now, by definition of range set, $f$ is bounded below by $0$, but can we get a non zero lower bound for function $f$ on $(\delta,\infty)$ i.e. can we get a $M>0$ such that $f(x)\ge M\ \forall x>\delta$? If yes then the proof can be completed easily then.
Thanks for assistance in advance.

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  • $\begingroup$ Please use \lim instead of lim: $ \lim_{x \to nifty} \neq lim_{x \to \infty}.$ $\endgroup$ – Digitalis Apr 13 '19 at 19:31
  • $\begingroup$ Yes, sorry for typing mistake but can you solve it? $\endgroup$ – Biswarup Saha Apr 13 '19 at 19:38
  • $\begingroup$ The range of the zero function is not $(0,\infty)$ though. $\endgroup$ – Mark Apr 13 '19 at 19:44
  • $\begingroup$ Ok, then you need to disprove the statement with a counter example $\endgroup$ – Biswarup Saha Apr 13 '19 at 19:44
  • $\begingroup$ The range of the function is given $(0,\infty)$ $\endgroup$ – Biswarup Saha Apr 13 '19 at 19:46
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The statement is not true. Any bounded, continuous function $f:(0,\infty) \to (0,\infty)$ where $f(x) \to 0$ as $x \to \infty$ is uniformly continuous.

Construct such a continuous function which is piecewise linear and where $f(x) = 1/n$ for $x = n $ and $f(x) = 2/n$ for $x = n + 1/n$ where $n \geqslant 2$ is an integer. The graph of the function looks like a sequence of declining sawtooth peaks.

Here we have $f(n+1/n)/f(n) \to 2$ as $n \to \infty$.

Thus, it is not the case that $f(x +1/x)/f(x) \to 1$ for $x \in (0,\infty)$ tending to $\infty$.

I'll leave any remaining details to you.

Addendum

$$f(x) = \begin{cases} 1/2,& 0 \leqslant x \leqslant 2\\1/n, &x = n, n\geqslant 2 \\ 2/n, & x = n + 1/n, n\geqslant 2\\ x - n + 1/n, & n < x < n +1/n, n \geqslant 2\\ 2/n + (1/n-2/n)(x - n - 1/n)/(1-1/n), &n + 1/n < x < n+1, n \geqslant 2 \end{cases}$$

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  • $\begingroup$ Can you please write the function properly, $f(x)=1/n\ \forall x\in(0,n), f(x)=2/n \ \forall x\in[n,n+1/n)$ Do you want to write like this? $\endgroup$ – Biswarup Saha Apr 14 '19 at 1:19
  • $\begingroup$ And one question more, $f:(0,\infty)\to(0,\infty), f(x)=x^2$, it is bounded near zero but not uniformly continuous. $\endgroup$ – Biswarup Saha Apr 14 '19 at 1:27
  • $\begingroup$ Bounded and continuous on $(0,\infty)$ along with $\lim_{x \to \infty} f(x) = 0$ are sufficient conditions for uniform continuity on $(0,\infty)$. You don't have a limit of $0$ as $x \to \infty$ for $f(x) = x^2$. $\endgroup$ – RRL Apr 14 '19 at 1:40
  • $\begingroup$ I changed the sentence to say bounded , continuous function with a limit of $0$ as $x \to \infty$. That must be UC. I just said bounded near zero originally, but that is redundant. Obviously $f(x) = 1/x$ is unbounded near zero only but not uniformly continuous. $\endgroup$ – RRL Apr 14 '19 at 1:42
  • $\begingroup$ Are you really unable to draw a graph as I described or write the function explicitly yourself. Take any constant for $f(0)$ with a straight line connecting to $f(2) = 1/2$, another line connecting to $f(2+1/2) = 2/2$, another straight line connecting to $f(3) = 1/3$, etc. $\endgroup$ – RRL Apr 14 '19 at 1:46

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