Natural Logarithm as Derivative of Exponential at Zero

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

Let $x > 1$.

From Power Function on Strictly Positive Base is Convex, $x^h$ is convex.

Thus for $0 < h < s$:

Further, $0 < \dfrac 1 x < 1$.

So, for $h < s < 0 \iff 0 < -s < -h$:

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

Next:

So $\dfrac {x^h - 1} h$ is strictly positive on $\R_{>0}$.

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

So from Limit of Increasing 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 $\sequence {n \paren {x^{1 / n} - 1 } }_{n \mathop \in \N}$ 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.