Show that $f$ is uniformly continuous if limit exists 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
 A: 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 $\forall x,y\in (0,1]\; (|x-y|<\delta \implies |f(x)-f(y)|<\epsilon)$ .
A: You want to show the direction $f$ uniformly continuous on $(0,1]\Longrightarrow$ the $\lim_{x\to0^+}f(x)$ exists. 
Here are some hints:


*

*Let $(x_n)_{n\in\mathbb N}$ be a sequence on $(0,1]$ such that $x_n\to 0$. Show that $(f(x_n))_{n\in\mathbb N}$ is Cauchy and therefore convergent. Here use the uniformly continuity of $f$. 

*If $(x_n)_{n\in\mathbb N},(y_n)_{n\in\mathbb N}$ are two sequences on $(0,1]$ such that $x_n\to0, \ y_n\to 0$, then by considering the sequence $(z_n)_{n\in\mathbb N}$ defined as $z_{2k}=x_k$ and $z_{2k+1}=y_k \ \forall k\in\mathbb N$, show that $\lim_{n\to\infty}f(x_n)=\lim_{n\to\infty}f(y_n)$.
(The sequences $f(x_n),f(y_n),f(z_n)$ converge and $f(x_n),f(y_n)$ are subsequences of $f(z_n)$).

*Conclude that the $\lim_{x\to0^+}f(x)$ exists. 

