Let $f$ uniformly continuous function on $(a,b)$ then the limits $\lim_{x\to a^{+}}f(x)$ and $\lim_{x\to b^{-}}$ exists.
I read a lot of similar proofs of this kind problems, but I don't understand the intuition of these proof, and I don't have idea of how I do start.
For instance I tried to use sequence $\lbrace x_n \rbrace$ near of $a$ and $\lbrace y_n \rbrace$ near of $b$ but I don't get connect that ideas for complete the proof, too I try use the $\varepsilon-\delta$ definition, but I fail too.
Finally I try use the module of continuity of $f$ denoted by $w_{f}(\delta)$ and use the fact that uniform continuity implies $\lim_{\delta \to 0^{+}}w_{f}(\delta)=0$ but I failed too.
Any idea or intuition of how I should start.
Let $\epsilon>0$ given.
$f $ uniformly continuous at $ (a,b) \implies$
$$(\exists \eta>0)\;:\; (\forall(x,y)\in (a,b)^2)\;$$ $$|x-y|<\eta\implies |f(x)-f(y)|<\epsilon$$ Take now two sequences $ (u_n)$ and $(v_n) $ in $(a,b)$ such that $$\lim_{n\to+\infty}u_n=\lim_{n\to+\infty}v_n=a$$
then for $ n $ and $ m $ great enough
$$|u_n-u_m|<\eta \text{ , } |v_n-v_m|<\eta$$ and $$|u_n-v_n|<\eta$$ thus
$$|f(u_n)-f(u_m)|<\epsilon\text{ and }$$ $$ |f(v_n)-f(v_m)|<\epsilon$$ and $$|f(u_n)-f(v_n)|<\epsilon$$ hence $(f(u_n)) $ and $ (f(v_n)) $ are Cauchy and converge to the same limit $ L$
By the sequential characterization of the limit, we conclude that $$\lim_{x\to a^+}f(x)=L$$