Theorem: Natural Logarithm as Derivative of Exponential at Zero

Let $\ln: \R_{>0}$ denote the real natural logarithm.
Then:
:$\ds \forall x \in \R_{>0}: \ln x = \lim_{h \mathop \to 0} \frac {x^h - 1} h$

Proof:
Fix $x \in \R_{>0}$.
Let $x > 1$.
From Power Function on Strictly Positive Base is Convex, $x^h$ is convex.
Thus for $0 < h < s$:
{{begin-eqn}}
{{eqn | l = \frac {x^h - a^0} {h - 0}
      | o = \le
      | m = \frac {x^s - a^0} {s - 0}
      | c = {{Defof|Convex Real Function}}
}}
{{eqn | ll= \leadsto
      | l = \frac {x^h - 1} h
      | o = \le
      | m = \frac {x^s - 1} s
}}
{{end-eqn}}
Further, $0 < \dfrac 1 x < 1$.
So, for $h < s < 0 \iff 0 < -s < -h$:
{{begin-eqn}}
{{eqn | l = \frac {\paren {\frac 1 x}^{-s} - \paren {\frac 1 x}^0} {-s - 0}
      | o = \le
      | m = \frac {\paren {\frac 1 x}^{-h} - \paren {\frac 1 x}^0} {-h - 0}
      | c = Power Function on Strictly Positive Base is Convex
}}
{{eqn | ll= \leadsto
      | l = \frac {x^s - 1} {-s}
      | o = \le
      | m = \frac {x^h - 1} {-h}
      | c = Exponent Combination Laws: Negative Power
}}
{{eqn | ll= \leadsto
      | l = \frac {x^h - 1} h
      | o = \le
      | m = \frac {x^s - 1} s
      | c = Order of Real Numbers is Dual of Order of their Negatives
}}
{{end-eqn}}
Hence $\dfrac {x^h - 1} h$ is increasing on $\R \setminus \set 0$.
Next:
{{begin-eqn}}
{{eqn | l = h
      | o = >
      | r = 0
}}
{{eqn | ll= \leadsto
      | l = x^h
      | o = >
      | r = 1
      | c = Power Function on Base Greater than One is Strictly Increasing
}}
{{eqn | ll= \leadsto
      | l = x^h - 1
      | o = >
      | r = 0
}}
{{eqn | ll= \leadsto
      | l = \frac {x^h - 1} h
      | o = >
      | r = 0
      | c = Order of Real Numbers is Dual of Order of their Negatives
}}
{{end-eqn}}
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 $\ds \inf_{h \mathop > 0} \frac {x^h - 1} h$) and increasing on $\openint \gets 0$
So from Limit of Increasing Function, $\ds \lim_{h \mathop \to 0^+} \frac {x^h - 1} h$ and $\ds \lim_{h \mathop \to 0^-} \frac {x^h - 1} h$ exist.
Further:
{{begin-eqn}}
{{eqn | l = \lim_{h \mathop \to 0^+} \frac {x^h - 1} h
      | r = \lim_{h \mathop \to \infty} h \paren {x^{1 / h} - 1}
      | c = Limit of Composite Function
}}
{{eqn | r = \lim_{n \mathop \to \infty} n \paren {x^{1 / n} - 1}
      | c = Limit of Sequence is Limit of Real Function
}}
{{eqn | r = \ln x
      | c = Sequential definition of natural logarithm
}}
{{end-eqn}}
where $\sequence {n \paren {x^{1 / n} - 1 } }_{n \mathop \in \N}$ is now a real sequence.
Similarly:
{{begin-eqn}}
{{eqn | l = \lim_{h \mathop \to 0^-} \frac {x^h - 1} h
      | r = \lim_{h \mathop \to 0^+} -\frac {x^{-h} - 1} h
      | c = Limit of Composite Function
}}
{{eqn | r = -\lim_{h \mathop \to \infty} h \paren {x^{-1 / h} - 1}
      | c = Limit of Composite Function
}}
{{eqn | r = -\lim_{h \mathop \to \infty} h \paren {\paren {\frac 1 x}^{1 / h} - 1}
      | c = Exponent Combination Laws: Negative Power
}}
{{eqn | r = -\lim_{n \mathop \to \infty} n \paren {\paren {\frac 1 x}^{1 / n} - 1}
      | c = Limit of Sequence is Limit of Real Function
}}
{{eqn | r = -\ln \frac 1 x
      | c = {{Defof|Natural Logarithm|subdef = Positive Real|index = 3}}
}}
{{eqn | r = \ln x
      | c = Logarithm of Reciprocal
}}
{{end-eqn}}
Thus, for $x > 1$:
{{begin-eqn}}
{{eqn | l = \lim_{h \mathop \to 0^-} \frac {x^h - 1} h
      | r = \ln x
      | c = 
}}
{{eqn | r = \lim_{h \mathop \to 0^+} \frac {x^h - 1} h
      | c = 
}}
{{end-eqn}}
So from Limit iff Limits from Left and Right, for $x > 1$:
:$\ds \lim_{h \mathop \to 0} \frac {x^h - 1} h = \ln x$
Suppose instead that $0 < x < 1$.
From Ordering of Reciprocals:
:$\dfrac 1 x > 1$
Thus, from above:
{{begin-eqn}}
{{eqn | l = \ln \frac 1 x
      | r = \lim_{h \mathop \to 0} \frac {\paren {\frac 1 x}^h - 1} h
}}
{{eqn | ll= \leadsto
      | l = \ln x
      | r = -\lim_{h \mathop \to 0} \frac {\paren {\frac 1 x}^h - 1} h
      | c = Logarithm of Reciprocal
}}
{{eqn | r = -\lim_{h \mathop \to 0} \frac {\paren {\frac 1 x}^{-h} - 1} {-h}
      | c = Limit of Composite Function
}}
{{eqn | r = \lim_{h \mathop \to 0} \frac {\paren {\frac 1 x}^{-h} - 1} h
}}
{{eqn | r = \lim_{h \mathop \to 0} \frac {x^h - 1} h
      | c = Exponent Combination Laws: Negative Power
}}
{{end-eqn}}
Hence the result.
{{qed}}
Category:Natural Logarithms
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