Asymptotics of uniformly continuous functions over $\mathbb R$ For $\alpha >1$, $x^\alpha$ is not uniformly continous over $\mathbb R$ , essentially because $\forall \delta>0, \lim_n (n+\delta)^{\alpha}-n^{\alpha}=\infty$
I don't have a clear intuition about the behavior of uniformly continuous functions over an unbounded interval. 

Is is safe to say that if $f$ is uniformly continuous over $\mathbb R$ then $\displaystyle \forall \alpha >1, f=o_{\infty}(x^{\alpha})$ ?

 A: If $f :[1, \infty) \to \mathbb{R}$ is uniformly continuous over $\mathbb{R}$, then we must have $|f(x)| = O(x)$ and, hence, $|f(x)| = o(x^\alpha)$ for $\alpha > 1$ as $x \to \infty$.
There exists $\delta > 0$ such that $|f(y) - f(x)| < 1 $ if $|y-x| \leqslant \delta.$
For any $x > 1$ there is an integer $n  = \lfloor (x-1)/ \delta\rfloor$ such that $x = 1 + n\delta + c$ where $0 \leqslant c < \delta$. 
Then $f(x) - f(1)$ telescopes into $n+1$ increments with absolute difference not exceeding $1$:
$$|f(x)| = |f(1) + [f(1+\delta) - f(1)] + \ldots  + [f(1 + n\delta) - f(1 +(n-1)\delta)] + [f(x) - f(1 + n\delta)]| \\ \leqslant |f(1)| + n + 1.$$
Consequently, we can find an upper bound for $|f(x)|/x$ that is independent of $n$:
$$\frac{|f(x)|}{x} \leqslant \frac{|f(1)| + n + 1 }{1 + n\delta + c} \\ = \frac{|f(1)|/n + 1 + 1/n }{1/n + \delta + c/n} \\ \leqslant \frac{|f(1)| + 2}{\delta} \\ = C.$$
If $\alpha > 1$ then as $x \to \infty ,$
$$\frac{|f(x)|}{x^\alpha} = \frac{|f(x)|}{x}\frac1{x^{\alpha -1 }} \leqslant \frac{C}{x^{\alpha - 1}} \to 0,$$
and $|f(x)| = o_\infty(x^\alpha)$.
