Limit of Monotone Real Function/Decreasing

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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 $b$.


Corollary

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

If $\xi \in \left({a \,.\,.\, b}\right)$, then:

$f \left({\xi^-}\right)$ and $f \left({\xi^+}\right)$ both exist

and:

$f \left({x}\right) \ge f \left({\xi^-}\right) \ge f \left({\xi}\right) \ge f \left({\xi^+}\right) \ge f \left({y}\right)$

provided that $a < x < \xi < y < b$.


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.

$\blacksquare$


Sources