# Limit of Monotone Real Function/Increasing

## Theorem

Let $f$ be a real function which is increasing and bounded above on the open interval $\openint a b$.

Let the supremum of $f$ on $\openint a b$ be $L$.

Then:

$\displaystyle \lim_{x \mathop \to b^-} \map f x = L$

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

### Corollary

Let $f$ be a real function which is increasing on the open interval $\openint a b$.

If $\xi \in \openint a b$, then:

$\map f {\xi^-}$ and $\map f {\xi^+}$ both exist

and:

$\map f x \le \map f {\xi^-} \le \map f \xi \le \map f {\xi^+} \le \map f y$

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

## Proof

Let $\epsilon > 0$.

We have to find a value of $\delta > 0$ such that $\forall x: b - \delta < x < b: \size {\map f x - L} < \epsilon$.

That is, that $L - \epsilon < \map f x < L + \epsilon$.

As $L$ is an upper bound for $f$ on $\openint a b$, $\map f x < L + \epsilon$ automatically happens.

Since $L - \epsilon$ is not an upper bound for $f$ on $\openint a b$, it follows that:

$\exists y \in \openint a b: \map f y > L - \epsilon$

But $f$ increases on $\openint a b$.

So:

$\forall x: y < x < b: L - \epsilon < \map f y \le \map f x$

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

$\blacksquare$