User:Keith.U/Sandbox

Theorem
Let $\ln: \R_{>0}$ denote the real natural logarithm.

Then:
 * $\displaystyle \forall x \in \R_{>0}: \ln x = \lim_{h \mathop \to 0} \frac {x^h - 1} h$

Proof
First, we show that:
 * $\displaystyle \forall x \in \R_{>0} : \lim_{h \mathop \to 0^+} \frac {x^h - 1} h$ exists.

Fix $x \in \R_{>0}$.

Let $x > 1$.

From Power Function on Base Greater than One is Strictly Convex, $x^h$ is strictly convex.

Thus:

Further, $0 < \dfrac{1}{x} < 1$, so:

Hence $\dfrac {x^h - 1} h$ is strictly increasing on $\R \setminus \left\{{ 0 }\right\}$.

Next:

and:

So $\dfrac {x^h - 1} h$ is strictly positive on $\R \setminus \left\{{ 0 }\right\}$.

In particular:
 * $ \dfrac {x^h - 1} h$ is bounded below and increasing on $\left({0, \,.\,.\, \to}\right)$
 * $ \dfrac {x^h - 1} h$ is bounded above (by $\displaystyle \inf_{h \mathop > 0} \frac {x^h - 1} h$) and increasing on $\left({\gets, \,.\,.\, 0}\right)$

So from Limit of Monotone Function, $\displaystyle \lim_{h \mathop \to 0^+} \frac {x^h - 1} h$ and $\displaystyle \lim_{h \mathop \to 0^-} \frac {x^h - 1} h$ exist.

Further:

where $\left\langle{ n \left({x^{1 / n} - 1 }\right\rangle_{n \mathop \in \N}$ is now a real sequence.

Similarly:

Thus, for $x > 1$:

So from Limit iff Limits from Left and Right, for $x > 1$:
 * $\displaystyle \lim_{h \mathop \to 0} \frac {x^h - 1} h = \ln x$

Suppose instead that $0 < x < 1$.

From Ordering of Reciprocals:
 * $\dfrac 1 x > 1$

Thus, from above:

Hence the result.