TITLE: Show that $f$ is uniformly continuous if limit exists
QUESTION [8 upvotes]: Let $f(x)$ be continuous on $(0,1]$. Show that $f$ is uniformly continuous IFF $\displaystyle \lim_{x\to0^+} f(x)$ exists.

Thoughts:
Backward Proof:
Let another function $\overline f(x)$ be continuous on $[0,1]$ which is equal to $f(x)$ plus the limit point. Thus the limit exists and it is uniformly continuous. So, as $f(x)$.
Forward Proof:
I Don't really have any idea...??
Please help guys

REPLY [4 votes]: A variation of the already-accepted answer: Show that if  $\lim_{x\to 0}f(x)$ does not exist then $f$ is not uniformly continuous: By the definition of $\lim_{x\to 0}f(x),$ if $\lim_{x\to 0}f(x)$ does  not exist then there exists $r>0$ such that for every $\delta\in (0,1)$ there exist $x,y\in (0,\delta)$ such that $|f(x)-f(y)|>r.$ Now let  $\epsilon =r.$  There cannot exist $\delta >0$ such that $\forall x,y\in (0,1]\; (|x-y|<\delta \implies |f(x)-f(y)|<\epsilon)$ .