Limit of Monotone Real Function/Increasing

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Theorem

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:

$\displaystyle \lim_{x \mathop \to b^-} f \left({x}\right) = L$

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


Corollary

Let $f$ be a real function which is 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 $a < x < \xi < y < b$.


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 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.

$\blacksquare$


Sources