Show that $f$ is uniformly continuous if limit exists

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.

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 $\forall x,y\in (0,1]\; (|x-y|<\delta \implies |f(x)-f(y)|<\epsilon)$ .