if oscillation of a function is zero then the function has a right limit. I was given this question:
$$\omega_f(I)=sup_{x,y\in I}{|f(x)-f(y)|}$$
True/False:
$$\lim_{\epsilon\rightarrow0}{\omega_f((x_0,x_0+\epsilon))}=0 \iff \lim_{x\rightarrow x_0^+}{f(x)} \in \mathbb{R}$$
I think I proved the direction from right to left:
if $\lim_{x\rightarrow x_0^+}{f(x)} =L$ then
$$\forall \epsilon_0 \exists \delta_0 : x_0<x<x_0+\delta_0 \rightarrow |f(x)-L|<\epsilon_0$$
Therefore, choose $\epsilon_0/2, \forall x,y\in(x_0,x_0+\delta_0) \rightarrow |f(x)-f(y)|<\epsilon_0 \rightarrow \omega_f(I)\le\epsilon_0$ (triangle inequality) therefore when $\epsilon\le\delta_0$
$\rightarrow \lim_{\epsilon\rightarrow0}{\omega_f((x_0,x_0+\epsilon))}=0$
I tried doing the second direction but I can't determine what limit the function would even have, I also couldn't think of any counterexamples, any help would be greatly appreciated!
 A: thanks to @Jose27 .
Proof from left to right:
Look at two sequences $z_n, y_n \rightarrow x_0^+$.
$$\forall \epsilon>0 \exists N_z \forall n\ge N_z:x_0<z_n<x_0+\epsilon$$
 And: $$\forall \epsilon>0 \exists N_y \forall n\ge N_y:x_0<y_n<x_0+\epsilon$$
We know that  $$(*) \forall\epsilon_0>0 \exists \delta_0:|\epsilon|<\delta_0\rightarrow sup_{x,y\in(x_0,x_0+\epsilon)}{|f(x)-f(y)|}<\epsilon_0$$
From a certain point $N_z:x_0<z_n<x_0+\epsilon$ therefore $$\forall \epsilon_0>0 \exists N : n\ge N:\forall k>0 |f(z_n)-f(z_{n+k})|<\epsilon_0$$,Because $N$ is the point from which $z_n\in(x_0,x_0+\delta_0) $ where $\delta_0  $ comes from $(*)$ .   Therefore $f(z_n)$ converges and we do the same for $f(y_n)$. 
We know  from (*) that $\forall \epsilon>0 z_n,y_n \in (x_0,x_0+\epsilon)$ from a certain point. 
So know $$\forall \epsilon > 0 \exists N :\forall n \ge N : |f(z_n)-f(y_n)|<\epsilon_0$$ 
So: $$|f(z_n)-f(y_n)| \rightarrow 0$$ Therefore $$\lim_{n\rightarrow \infty}{f(z_n)}=\lim_{n\rightarrow\infty}{f(y_n)}$$
Therefore $\lim_{x\rightarrow x_0^+}{f(x)}\in \mathbb{R}$ 
