Show for a monotonic function $f$ on $(a,b)$ is $\omega_f(x_0)=|\lim\limits_{x \searrow x_0}f(x)-\lim\limits_{x \nearrow x_0}f(x)|$ Show for a monotonic function $f$ on $(a,b)$ is $\omega_f(x_0)=|\lim\limits_{x \searrow x_0}f(x)-\lim\limits_{x \nearrow x_0}f(x)|$

My attempt:
$\forall \delta >0$ we call
$\Omega_f(\mathcal{U}_\delta(x_0)\cap(a,b))$ the oscillation of $f$ in $\mathcal{U}_\delta(x_0)\cap(a,b)$ which means, the 
$\Omega_f(\mathcal{U}_\delta(x_0)\cap(a,b))=\sup f(\mathcal{U}_\delta(x_0)\cap(a,b))-\inf f (\mathcal{U}_\delta(x_0)\cap(a,b))$
$=\sup\lbrace |f(x)-f(y)||x,y \in \mathcal{U}_\delta(x_0)\cap(a,b) \rbrace$
The existence of $\sup f(\mathcal{U}_\delta(x_0)\cap(a,b))$ and $\inf f(\mathcal{U}_\delta(x_0)\cap(a,b))$ is obvious, since 
$\mathcal{U}_\delta(x_0)\cap(a,b)=\lbrace x\in (a,b)| x_0-\delta<x<x_0+\delta \rbrace$
and since $f$ is monotonic (lets say w.l.o.g) growing 
$f(x_0-\delta)\le f(x) \le f(x_0+\delta)$, $\forall x \in \mathcal{U}_\delta(x_0)\cap(a,b)$
$\omega_f(x_0):=\lim\limits_{\delta \searrow 0}\Omega_f(\mathcal{U}_\delta(x_0)\cap(a,b))$
So finally :

Given

$\omega_f(x_0):=\lim\limits_{\delta \searrow 0}\Omega_f(\mathcal{U}_\delta(x_0)\cap(a,b))=\lim\limits_{\delta \searrow 0}\sup\lbrace |f(x)-f(y)||x,y \in \mathcal{U}_\delta(x_0)\cap(a,b) \rbrace$

End of given

$=\lim\limits_{\delta \searrow 0}\left(\left|\lim\limits_{x \searrow x_0-\delta}f(x)-\lim\limits_{x \nearrow x_0+\delta}f(x)\right|\right)=|\lim\limits_{x \searrow x_0}f(x)-\lim\limits_{x \nearrow x_0}f(x)|$
Could someone verify my solution, or give critic ? :)
Thank you
 A: You have set this up correctly, but you have not actually proved the final conclusive statement.
If $\delta$ is sufficiently small we have $\bar{U}_\delta(x_0) \subset (a,b)$ -- that is the closed ball of sufficiently small radius is contained in the open interval -- and $f$ is defined on $\bar{U}_\delta(x_0)$.  Thus, for all $x,y \in \bar{U}_\delta(x_0)$ it follows that  $f(x_0-\delta) \leqslant f(x), f(y) \leqslant f(x_0 + \delta)$ and 
$$|f(x) - f(y)| \leqslant f(x_0+ \delta) - f(x_0-\delta)$$
Whence,
$$\sup_{x,y \in U_\delta(x_0) \cap (a,b)}|f(x) - f(y)| \leqslant f(x_0+ \delta) - f(x_0-\delta)$$
Since, $f$ is monotonic we have existence of the limits
$$\lim_{x \searrow x_0}f(x) = \lim_{\delta \to 0+}f(x_0 + \delta), \quad\lim_{x \nearrow x_0}f(x) = \lim_{\delta \to 0+}f(x_0 - \delta)$$
Thus,
$$\lim_{x \searrow x_0}f(x)- \lim_{x \nearrow x_0}f(x) \leqslant\\ \liminf_{\delta\to 0+}\sup_{x,y \in U_\delta(x_0) \cap (a,b)}|f(x) - f(y)|\leqslant \limsup_{\delta\to 0+}\sup_{x,y \in U_\delta(x_0) \cap (a,b)}|f(x) - f(y)|\\  \leqslant\lim_{x \searrow x_0}f(x)- \lim_{x \nearrow x_0}f(x) ,$$
which implies that the limit of $\displaystyle \sup_{x,y \in U_\delta(x_0) \cap (a,b)}|f(x) - f(y)|$ as $\delta \to 0+$ exists and 
$$\omega_f(x_0) = \lim_{\delta \to 0+}\sup_{x,y \in U_\delta(x_0) \cap (a,b)}|f(x) - f(y)|  \leqslant\lim_{x \searrow x_0}f(x)- \lim_{x \nearrow x_0}f(x)$$
