Natural Logarithm as Derivative of Exponential at Zero

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

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

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

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

Begin by supposing $x > 1$.

From Exponential with 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 $h \to 0^{+}$.

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

Further:

Thus:

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

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

Similarly:

Thus, for $x > 1$:
 * $\displaystyle \lim_{ h \to 0^{-} } \frac{ x^{h} - 1 }{ h } = \ln x = \lim_{ h \to 0^{+} }\frac{ x^{h} - 1 }{ h }$

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

Suppose instead that $0 < x < 1$.

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

Thus, from above:

Hence the result.