Prove that $f$ is uniformly continuous iff there exist sequences $a_n,b_n$ such that if $\lim a_n=\lim b_n \implies \lim f(a_n)=\lim f(b_n)$ 
Suppose $f$ is a real-valued function, and $a_n,b_n$ are real sequences.
Prove that $f$ is uniformly continuous $\iff (\lim a_n=\lim b_n \implies  \lim f(a_n)=\lim f(b_n))$




*

*Suppose $f$ is uniformly continuous. Then, in particular, $f$ is continuous. Suppose $a_n\to x$ as $n\to\infty$. We know by continuity of $f$ that: $$\lim a_n=x\implies \lim f(a_n)=f(x)$$ We know that $\lim a_n=\lim b_n=x$ so that also: $$\lim b_n=x\implies \lim f(b_n)=f(x)$$
Together this gives that if $f$ is uniformly continuous, that $\lim f(a_n)=\lim f(b_n)$.



*

*Suppose $f$ is not uniformly continuous. We then know that:
$$\forall_{\delta>0}\exists_{\epsilon>0}:|x-a|<\delta \not \Rightarrow |f(x)-f(a)|<\epsilon$$
Suppose we know that $\lim a_n=a=\lim b_n=b$, we want to show that $\lim f(a_n)\neq f(b_n)$. 
As the sequences are arbitrary in $\mathbb{R}$, we know that:
$a-b=0\implies|a-b|<\delta$. 
As $f$ is not uniformly continuous, we know that for a particular $\delta>0$ we pick, the following holds:
$$|a-b|<\delta \not \Rightarrow |f(a)-f(b)|<\epsilon \ \ \ (*)$$
Now suppose that $\lim f(a_n)=\lim f(b_n)$ so that $f(a)=f(b)$. Then for any $\epsilon>0$:
$$|a-b|=0<\delta\implies|f(a)-f(b)|=0<\epsilon$$
This is contradictory with the statement $(*)$; non-uniform continuity. Thus: $f(a)$ cannot equal $f(b)$ if $a=b$ so that if $f$ is not uniformly continuous, we know that: $$\lim a_n = \lim b_n \not \Rightarrow \lim f(a_n) = \lim f(b_n)$$ $\tag*{$\Box$}$ 

$\impliedby$ fails. On $\mathbb R,$ let $f(x) = x^2.$ Suppose $\lim a_n = \lim b_n.$ Denote the common limit by $L.$ Then by continuity of $f,\lim f(a_n) = f(L) = \lim f(b_n).$ But $f$ is not uniformly continuous on $\mathbb R.$
