uniform continuity of $f$ on $(a,b) \implies \exists \lim_{x \to a^+} f(x)$? Does $\lim_{x \to a^+} f(x)$ necessarily exist for all $f$ uniformly continuous on $(a,b)$?
This has no relation whatsoever to homework. It occurred to me after looking at another question (which asked whether or not we must have $\lim_{x \to a^+} f(x) < \infty$ under the added hypothesis that $f'$ is bounded on $(a,b)$). 
 A: Yes. Let $a_n\rightarrow a$ from the right. Consider the sequence $x_n=f(a_n)$. By uniform continuity of $f$, it follows that for $\epsilon>0$ there exists $\delta>0$ such that $\forall x,y\in(a,b)[|x-y|<\delta \implies |f(x)-f(y)|<\epsilon]$. Since $\{a_n\}_{n]\in Z^+}$ is Cauchy, choose $N$ such that $\forall m,n\geq N[|a_n-a_m|<\delta]$, it follows that $\forall m,n\geq N[|x_n-x_m|<\epsilon]$. $\{x_n\}_{n\in Z^+}$ is cauchy hence convergent.
As David Mitra noted, it should be shown that for any two sequences $(a_n)_{n\in Z^+},(b_n)_{n\in Z^+}$ in $(a,b)$ that converge to $a$ we have $lim_{n\rightarrow\infty}f(a_n)=lim_{n\rightarrow\infty}f(b_n)$. I will show that $lim_{n\rightarrow\infty}|f(a_n)-f(b_n)|=0$.
Proof: Since $a_n-b_n\rightarrow 0$, therefore one can choose $N$ such that $\forall n\geq N[|a_n-b_n|<\delta]$. By the uniform continuity of $f$, we know that $\forall n\geq N[|f(a_n)-f(b_n)|<\epsilon]$. Thus $lim_{n\rightarrow\infty}|f(a_n)-f(b_n)|<\epsilon$. Since $\epsilon$ is arbitrary, it follows that $lim_{n\rightarrow\infty}|f(a_n)-f(b_n)|=0$.
