prove that if f(x) is monotone increasing on an open interval (a,b) and x_0 is any point in (a,b), then lim_{x \to x_0+} f(x) exists and satisfies lim_{x to x_0+} f(x) >= f(x_0). (from the right side)

Is this proof correct? if not could someone please correct it/ explain what I'm doing wrong?

Fix $\displaystyle x_0$ in (a,b). First consider x in ($\displaystyle x_0$,b] then f(x)>= f($\displaystyle x_0$) so inf f(x) ($\displaystyle x_0$<x<b) >=f($\displaystyle x_0$)

Claim: lim f(x)=inf f(x) ($\displaystyle x_0$<=x<b) . Let S=inf f(x). Let $\displaystyle \epsilon$>0. Then there exists x$\displaystyle \epsilon$ in ($\displaystyle x_0$,b) such that f(x$\displaystyle \epsilon$)<S-$\displaystyle \epsilon$.

But if x is in (x$\displaystyle \epsilon$,$\displaystyle x_0$) then f(x$\displaystyle \epsilon$)>=f(x) so S<=f(x)<S-$\displaystyle \epsilon$

Hence |f(x)-s|<$\displaystyle \epsilon$ for all x in (x$\displaystyle \epsilon$,$\displaystyle x_0$).

therefore, lim f(x) (x->$\displaystyle x_0$+) exists and equals S since S>=f($\displaystyle x_0$) we get lim (x->$\displaystyle x_0$+) >= f($\displaystyle x_0$)