User:Keith.U/Sandbox

Theorem
Let $\ln x$ be the natural logarithm function.

Then:
 * $D_x \left({\ln x}\right) = \dfrac 1 x$

Proof
This proof assumes the definition of the natural logarithm as the limit of a sequence of functions.

Let $\left\langle{ f_n }\right\rangle$ be the sequence of mappings $f_n : \R_{>0} \to \R$ defined as:
 * $f_n \left({ x }\right) = n \left({ \sqrt[n]{ x } - 1 }\right)$

Fix $x_0 \in \R_{>0}$.

Pick $k \in \N : x_0 \in J := \left[{ \dfrac{1}{k}, \,.\,.\, k }\right]$.

From definition of bounded interval, $J$ is bounded.

From Derivative of Nth Root and Combination Theorem for Sequences:
 * $\forall n \in \N \forall x \in J : D_x f_n \left({ x }\right) = \dfrac{ \sqrt[n]{x} }{ x }$

In particular,
 * $\forall n, f_n$ is differentiable on $J$

From Defining Sequence of Natural Logarithm is Convergent, $\left\langle{ f_n \left({ x_0 }\right) }\right\rangle$ is convergent.

Lemma
From the lemma, $\left\langle{ f_n' }\right\rangle$  converges uniformly to $\dfrac{1}{x}$ on $J$.

From Derivative of Uniformly Convergent Sequence of Differentiable Functions, $f' \left({ x }\right) = \dfrac{1}{x}$ on $J$

In particular, $f' \left({ x_0 }\right) = \dfrac{1}{x_0}$

Hence the result.