Derivative of Logarithm at One/Proof 1

Theorem

Let $\ln x$ be the natural logarithm of $x$ for real $x$ where $x > 0$.

Then:

$\ds \lim_{x \mathop \to 0} \frac {\map \ln {1 + x} } x = 1$

$\ds \lim_{x \mathop \to c} \frac {\map f x} {\map g x} = \lim_{x \mathop \to c} \frac {\map {f'} x} {\map {g'} x}$

(provided the appropriate conditions are fulfilled).

Here we have:

$\ds \map \ln {1 + 0}$

$=$

$\ds 0$

$\ds \map {D_x} {\map \ln {1 + x} }$

$=$

$\ds \dfrac 1 {1 + x}$

$\ds \map {D_x} x$

$=$

$\ds 1$

Thus:

$\ds \lim_{x \mathop \to 0} \frac {\map \ln {1 + x} } x = \lim_{x \mathop \to 0} \frac {\paren {1 + x}^{-1} } 1 = \frac 1 {1 + 0} = 1$

$\blacksquare$