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 $$\lim_{x \to b^-} f \left({x}\right) = L$$, where $$\lim_{x \to b^-} f \left({x}\right)$$ 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 $$\lim_{x \to a^+} f \left({x}\right) = l$$, where $$\lim_{x \to a^+} f \left({x}\right)$$ 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.