Proving that $f$ is uniformly continuous on $[0,\infty)$ Prove: If $f$ is uniformly continuous on $[0,b]$ for all $b>0$ and $\lim_{x\to \infty}f(x)=L$, then $f$ is uniformly continuous on $[0,\infty)$.
I intend to answer that problem. Can you please look at my proof? 
 A: Let $\epsilon>0$. Then there exists $M>0$ such that for all $x>M$,
$$|f(x)-L|<\frac{\epsilon}{2}.$$
Thus, 
$$x,y\in (M,\infty)\quad\implies\quad |f(x)-f(y)|\leq |f(x)-L|+|f(y)-L|  <\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.$$
This means that
$$x,y\in (M,\infty)\quad\implies\quad |f(x)-f(y)|<\epsilon.\quad \quad(1)$$
Using the assumption, it follows that $f$ is uniformly continuous on $[0,M+1]$. Then, there exists $\Delta>0$ such that for all $x,y\in [0,M+1]$,
$$|x-y|<\Delta\quad\implies\quad|f(x)-f(y)|<\epsilon.\quad\qquad  (2)$$
We take 
$$\delta=\min\left\{\frac{1}{2},\Delta\right\}.$$ 
Let $x,y\in [0,\infty)$ such that $|x-y|<\delta$. Then $|x-y|<\Delta$ and $|x-y|<\frac{1}{2}$. Without loss of generality, assume that $x\leq y$. We consider the following cases:
Case 1. Suppose $y\leq M+1$. Then $x,y\in[0,M+1]$ so that by using $(2)$, we get $|f(x)-f(y)|<\epsilon$. 
Case 2. Suppose $y>M+1$. The subtracting by $\frac{1}{2}$ to both sides, we get
$$y-\frac{1}{2}>M+\frac{1}{2}.\qquad (3)$$
Since $|x-y|<\frac{1}{2}$, we get $-\frac{1}{2}<x-y<\frac{1}{2}$ implying 
$$y-\frac{1}{2}<x.\qquad\quad (4)$$
Combining $(3)$ and $(4)$, we get
$$x>y-\frac{1}{2}>M+\frac{1}{2}>M.$$
Thus, $x,y\in(M,\infty)$ and using $(1)$, $|f(x)-f(y)|<\epsilon$.
In either case, we have shown that for any $x,y\in[0,\infty)$,
$$|x-y|<\delta\quad\implies\quad|f(x)-f(y)|<\epsilon.$$ This proves that $f$ is uniformly continuous on $[0,\infty)$.
