Limit of Monotone Real Function

Increasing Function
Let $f$ be a real function which is increasing and bounded above on the open interval $\left({a \,.\,.\, b}\right)$.

Let the supremum 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 $b$.

Decreasing Function
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:
 * $\displaystyle \lim_{x \to a^+} f \left({x}\right) = l$

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

Corollary
Let $f$ be a real function which is increasing 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
 * $f \left({x}\right) \le f \left({\xi^-}\right) \le f \left({\xi}\right) \le f \left({\xi^+}\right) \le f \left({y}\right)$

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

A similar result applies for decreasing functions.

Proof for Increasing Function
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 an upper bound for $f$ on $\left({a \,.\,.\, b}\right)$, $f \left({x}\right) < L + \epsilon$ automatically happens.

Since $L - \epsilon$ is not an upper bound for $f$ on $\left({a \,.\,.\, b}\right)$, $\exists y \in \left({a \,.\,.\, b}\right): f \left({y}\right) > L - \epsilon$.

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

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

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

Proof for Decreasing Function
Let $\epsilon > 0$.

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

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

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

Proof of Corollary
$f$ is bounded above on $\left({a \,.\,.\, \xi}\right)$ by $f \left({\xi}\right)$.

By Limit of Increasing Function (proved above), the supremum is $f \left({\xi^-}\right)$.

So it follows that:
 * $\forall x \in \left({a \,.\,.\, \xi}\right): f \left({x}\right) \le f \left({\xi^-}\right) \le f \left({\xi}\right)$

A similar argument for $\left({\xi \,.\,.\, b}\right)$ holds for the other inequalities.

Likewise, a similar argument can be used to show the similar result for decreasing functions.