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
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:

Hence $\dfrac {x^h - 1} h$ is strictly increasing.

Further:

Hence, $\dfrac {x^h - 1} h$ is decreasing and bounded below as $h \mathop \to 0^+$.

So by Limit of Monotone Function:
 * $\displaystyle \lim_{h \mathop \to 0^+} \frac {x^h - 1} h$ exists.

Further:

Thus:

Hence $\dfrac{x^h - 1} h$ is increasing and bounded above as $h \to 0^-$.

So by Limit of Monotone Function:
 * $\displaystyle \lim_{h \mathop \to 0^-} \frac {x^h - 1} h$ exists.

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.