Is an increasing, bounded and continuous function on $[a,+\infty)$ uniformly continuous? Supose $f$ is increasing, bounded and continuous on $[a,+\infty)$. 
Is  $f$ uniformly continuous ?
I think yes. how to prove that?
My idea is to show there exists $X$ , $f$ is uniformly continuous on $[X,+\infty)$.
How to fix such $X$?
 A: Since $f$ is increasing and bounded, there exists $l\in\mathbb R$ such that $\lim_{x\to+\infty}f(x)=l$. Fix $\varepsilon>0$; there is $t_0>a$ such that if $x\geq t_0$ then $|f(x)-l|\leq\varepsilon/2$. Using the continuity of $f$, we get that $f$ is uniformly continuous on $[a,t_0+1]$, so there is a $\delta\in (0,1)$ such that if $a\leq x,y \leq t_0+1$ and $|x-y|\leq \delta$ then $|f(x)-f(y)|\leq\varepsilon/2$. Now, let $x,y\geq a$ such that $|x-y|\leq \delta$. If $x,y\in [a,t_0]$ we have $|f(x)-f(y)|\leq\varepsilon$; if $x, y>t_0$ then $|f(x)-f(y)|\leq |f(x)-l|+|f(y)-l|\leq \varepsilon$ and if $x\leq t_0$ and $y>t_0$ then $y\in [a,t_0+1]$ so $|f(x)-f(y)|\leq\varepsilon$.
A: Yes, it is.  Let $\epsilon > 0$   Since $f$ is bounded and increasing, $\lim_{x\to\infty }f(x)$ exists; let us denote this by $M$.  Choose $R\ge a$ so that $f(R) > M - \epsilon//2$.  Since $f$ is continuous on $[a, R]$ is is uniformly continuous there.  Chose $\delta > 0$
so $|x - y|< \delta \implies |f(x - f(y)| < \epsilon/2$.  
I is an easy matter now to argue that on the entire line if $|x - y| < \delta\implies |f(x)-f(y)|< \epsilon.$
