Let the function $f(x)$ is monotonically decreasing over $(a,b)$ and $x_0∈(a,b)$, I have to prove that there exist a one-sided limit $f(x_0−0)$ and that $f(x_0−0)≥f(x_0)$.
Whole proof will be very appreciated because I'm trying to understand all of this.
I'm guessing that $f(x_0-0)$ means $\lim_{x\to{x_0}^-}f(x)$.
Let $l=\inf\{f(x)\,|\,x\in(a,x_0)\}$. Note the, for each $x\in(a,x_0)$, $f(x)\geqslant f(x_0)$, since $f$ is decreasing. So, the set $\{f(x)\,|\,x\in(a,x_0)\}$ has a lower bound (namely, $f(x_0)$) and the definition of $l$ makes sense; besides, $l\geqslant f(x_0)$. Now, given $\varepsilon>0$, since $l+\varepsilon>l$, there is a $x_1\in(a,x_0)$ such that $f(x_1)<l+\varepsilon$ and, since $f$ is decreasing, the equality $f(x)<l+\varepsilon$ holds for every $x\in(x_1,x_0)$. So, if $\delta=x_0-x_1$, this proves that\begin{align}|x-x_0|<\delta\wedge x<x_0&\implies l\leqslant f(x)<l+\varepsilon\\&\implies\bigl|f(x)-l\bigr|<\varepsilon.\end{align}So, by definition, $\lim_{x\to{x_0}^-}f(x)=l$.