Proving the existence of a one-sided limit of a monotonically decreasing function If the function $f(x)$ is monotonically decreasing over $(a,b)$ and $x_0\in (a,b)$, I have to prove that there exist a left-hand limit $f(x_0-0)$ and that $f(x_0-0) \ge f(x_0)$.
I think i have to prove that the limit is equal to the inf
 A: I assume that $f(x)$ is piecewise continuous, thus it is made up of many continuous segments (such that there are as many jump discontinuities as you like, but in between are continuously decreasing functions). Clearly, the limit exists along all continuous segments, then the only question is what about the jump discontinuities? Prove one exists and then they all must exist.

The picture shows the proof, for every positive $\delta$, there exists a unique positive $\varepsilon$ such that$f(c - \delta) = f(c) + \varepsilon$, the left-handed values. This is proved from the definition of monotonically decreasing: for every $x < y$, $f(x) > f(y)$ (the definition of $f(x) > f(y)$ means there exists a positive $\varepsilon$ such that $f(x) = f(y) + \varepsilon$ and $x < y$ means there exists a $\delta$ such that $x = y - \delta$).
This effectively provides the $(\varepsilon, \delta)$ definition of a limit (for a left sided limit). For every $\varepsilon > 0$, there exists a $\delta_0 > 0$ such that $f(c - \delta_0) = L + \varepsilon$. Then for every $0 < \delta < \delta_0$, we have that $f(c - \delta_0) > f(c - \delta) > f(c)$ because:
\begin{align}
\delta < \delta_0 &&\\
-\delta_0 < -\delta && \text{switch sides}\\
c - \delta_0 < c - \delta && \text{Add $c$ to each side} \\
f(c - \delta_0) > f(c - \delta) && \text{definition of monotonicly decreasing} \\
L + \varepsilon > f(c - \delta) && \text{$f(c - \delta_0) = L + \varepsilon$}\\
f(c - \delta) - L < \varepsilon && \text{q.e.d.}
\end{align}
...now this would not strictly be true for a monotonically non-increasing function--but, again, you'd just take out the special case of a flat segment to prove the limit still exists there.
Edit (after a while)
Looking back, there is a problem with the above, it's possible that no value of $x$ exists such that $f(x) = L + \varepsilon$ (just look at the huge gap in my picture) and thus it's incorrect to assume that for all $\varepsilon$ there exists a $\delta_0 > 0$ such that $f(c - \delta_0) = L + \varepsilon$. By the way, this shows that the left-handed limit cannot be the lower value at the jump discontinuity in my picture because there is a gap between $f_{\text{top}}$ and $f_\text{bottom}$, $\Delta f$, where if I choose $0 < \varepsilon < \Delta f$, the smallest possible $f(x)$ (to the left) I can find is $f_\text{top}$ and $f_\text{top} - L = \Delta f > \varepsilon$.
My original existence proof still works in the neighborhood of the jump discontinuity. In that case we would assume there is some $\zeta > 0$ such that on the interval $x \in [c - \zeta, c)$, $f(x)$ is continuous and thus $f(c - \zeta) = L + \varepsilon_0$ such that $\varepsilon_0 > 0$ (whatever that value is). From there we can show that for any $0 < \varepsilon \leq \varepsilon_0$, there does exist a value $f(c - \delta_0) = L + \varepsilon$ and then we can just choose $0 < \delta < \delta_0$ then through the definition of monotonicly decreasing, $f(c - \delta) < f(c - \delta_0) = L + \varepsilon$.
Once that's established, for all other $\varepsilon > \varepsilon_0$, we can just choose $\delta < \zeta$. From there we have $f(c - \delta) < L + \varepsilon_0 < L + \varepsilon$.
The last part missing from the above also requires monotonicity, we need to show that for all $x \in (c - \delta, c)$, $f(x) - L < \varepsilon$. That's pretty easy, because the above range means $c - \delta < x < c$, we have all $x > c - \delta$ for all $x$ in this range thus, $f(x > c - \delta) < f(c - \delta)$ by the definition of monotonicly decreasing (the same logic can be applied to the proof requiring continuity around the neighborhood of the jump discontinuity).
One last Note
If the function is completely discontinuous (or partially around $x_0$), then in fact the left-handed limit does not exist! For the exact reasons that you can say the lower value at the jump discontinuity is definitely not the value of the left-handed limit.
A: Let $\{x_n\}$ be a sequence increasing to $x_0$. Let $a=\inf\{f(x):x>x_0\}$. If we show that $\{f(x_n)\}$ converges to $a$ it will follow that $f(x_0-0)$ exists and equals $a$. Since $f(x)\geq f(x_0)$ whenever $x<x_0$ we get $f(x_0)\geq a$ and hence $f(x_0)\geq f(x_0-0)$  as required. Now let $\epsilon >0$. By definition of $a$ there exists $x<x_0$ such that $a+\epsilon >f(x)$. For $n$ sufficiently large we have $x<x_n<x_0$. Hence $f(x_n)\leq f(x) <a+\epsilon$. On the other hand $f(x_n)\geq a$ by definition of $a$. This proves that $\{f(x_n)\}$ converges to $a$.
