Natural Logarithm as Derivative of Exponential at Zero

From ProofWiki
Jump to navigation Jump to search

Theorem

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 $0 < h < s$:

\(\ds \frac {x^h - a^0} {h - 0}\) \(\le\) \(\, \ds \frac {x^s - a^0} {s - 0} \, \) \(\ds \) Definition of Convex Real Function
\(\ds \leadsto \ \ \) \(\ds \frac {x^h - 1} h\) \(\le\) \(\, \ds \frac {x^s - 1} s \, \) \(\ds \)


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

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

\(\ds \frac {\paren {\frac 1 x}^{-s} - \paren {\frac 1 x}^0} {-s - 0}\) \(\le\) \(\, \ds \frac {\paren {\frac 1 x}^{-h} - \paren {\frac 1 x}^0} {-h - 0} \, \) \(\ds \) Power Function on Strictly Positive Base is Convex
\(\ds \leadsto \ \ \) \(\ds \frac {x^s - 1} {-s}\) \(\le\) \(\, \ds \frac {x^h - 1} {-h} \, \) \(\ds \) Exponent Combination Laws: Negative Power
\(\ds \leadsto \ \ \) \(\ds \frac {x^h - 1} h\) \(\le\) \(\, \ds \frac {x^s - 1} s \, \) \(\ds \) Order of Real Numbers is Dual of Order of their Negatives


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


Next:

\(\ds h\) \(>\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds x^h\) \(>\) \(\ds 1\) Power Function on Base Greater than One is Strictly Increasing
\(\ds \leadsto \ \ \) \(\ds x^h - 1\) \(>\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \frac {x^h - 1} h\) \(>\) \(\ds 0\) Order of Real Numbers is Dual of Order of their Negatives

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:

\(\ds \lim_{h \mathop \to 0^+} \frac {x^h - 1} h\) \(=\) \(\ds \lim_{h \mathop \to \infty} h \paren {x^{1 / h} - 1}\) Limit of Composite Function
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} n \paren {x^{1 / n} - 1}\) Limit of Sequence is Limit of Real Function
\(\ds \) \(=\) \(\ds \ln x\) Sequential definition of natural logarithm

where $\sequence {n \paren {x^{1 / n} - 1 } }_{n \mathop \in \N}$ is now a real sequence.


Similarly:

\(\ds \lim_{h \mathop \to 0^-} \frac {x^h - 1} h\) \(=\) \(\ds \lim_{h \mathop \to 0^+} -\frac {x^{-h} - 1} h\) Limit of Composite Function
\(\ds \) \(=\) \(\ds -\lim_{h \mathop \to \infty} h \paren {x^{-1 / h} - 1}\) Limit of Composite Function
\(\ds \) \(=\) \(\ds -\lim_{h \mathop \to \infty} h \paren {\paren {\frac 1 x}^{1 / h} - 1}\) Exponent Combination Laws: Negative Power
\(\ds \) \(=\) \(\ds -\lim_{n \mathop \to \infty} n \paren {\paren {\frac 1 x}^{1 / n} - 1}\) Limit of Sequence is Limit of Real Function
\(\ds \) \(=\) \(\ds -\ln \frac 1 x\) Definition 3 of Natural Logarithm
\(\ds \) \(=\) \(\ds \ln x\) Logarithm of Reciprocal


Thus, for $x > 1$:

\(\ds \lim_{h \mathop \to 0^-} \frac {x^h - 1} h\) \(=\) \(\ds \ln x\)
\(\ds \) \(=\) \(\ds \lim_{h \mathop \to 0^+} \frac {x^h - 1} h\)

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:

\(\ds \ln \frac 1 x\) \(=\) \(\ds \lim_{h \mathop \to 0} \frac {\paren {\frac 1 x}^h - 1} h\)
\(\ds \leadsto \ \ \) \(\ds \ln x\) \(=\) \(\ds -\lim_{h \mathop \to 0} \frac {\paren {\frac 1 x}^h - 1} h\) Logarithm of Reciprocal
\(\ds \) \(=\) \(\ds -\lim_{h \mathop \to 0} \frac {\paren {\frac 1 x}^{-h} - 1} {-h}\) Limit of Composite Function
\(\ds \) \(=\) \(\ds \lim_{h \mathop \to 0} \frac {\paren {\frac 1 x}^{-h} - 1} h\)
\(\ds \) \(=\) \(\ds \lim_{h \mathop \to 0} \frac {x^h - 1} h\) Exponent Combination Laws: Negative Power


Hence the result.

$\blacksquare$

Retrieved from "https://proofwiki.org/w/index.php?title=Natural_Logarithm_as_Derivative_of_Exponential_at_Zero&oldid=524761"