Let $f:(0,\infty)\to(0,\infty)$ be uniformly continuous function, is the following statement true? 
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.

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