Limit of Monotone Real Function/Decreasing

Theorem
Let $f$ be a real function which is decreasing and bounded below on the open interval $\left({a \,.\,.\, b}\right)$.

Let the infimum of $f$ on $\left({a \,.\,.\, b}\right)$ be $l$.

Then:
 * $\displaystyle \lim_{x \to b^-} f \left({x}\right) = l$

where $\displaystyle \lim_{x \to b^-} f \left({x}\right)$ is the limit of $f$ from the left at $b$.

Proof
Let $\epsilon > 0$.

We have to find a value of $\delta > 0$ such that $\forall x: b - \delta < x < b: \left|{f \left({x}\right) - l}\right| < \epsilon$.

That is, that $l - \epsilon < f \left({x}\right) < l + \epsilon$.

As $l$ is a lower bound for $f$ on $\left({a \,.\,.\, b}\right)$, $l - \epsilon < f \left({x}\right)$ automatically happens.

Since $l + \epsilon$ is not a lower bound for $f$ on $\left({a \,.\,.\, b}\right)$, $\exists y \in \left({a \,.\,.\, b}\right): f \left({y}\right) < l + \epsilon$.

But $f$ decreases on $\left({a \,.\,.\, b}\right)$.

So:
 * $\forall x: y < x < b: f \left({x}\right) \le f \left({y}\right) < l + \epsilon$

We choose $\delta = y - a$ and hence the result.