# Derivative of Logarithm at One

## Theorem

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

Then:

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

## Proof 1

L'Hôpital's Rule gives:

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

(provided the appropriate conditions are fulfilled).

Here we have:

 $\ds \map \ln {1 + 0}$ $=$ $\ds 0$ $\ds \map {D_x} {\map \ln {1 + x} }$ $=$ $\ds \dfrac 1 {1 + x}$ Chain Rule for Derivatives $\ds \map {D_x} x$ $=$ $\ds 1$ Derivative of Identity Function

Thus:

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

$\blacksquare$

## Proof 2

 $\ds \lim_{x \mathop \to 0} \frac {\map \ln {1 + x} } x$ $=$ $\ds \lim_{x \mathop \to 0} \frac {\map \ln {1 + x} - \ln 1} x$ subtract $\ln 1 = 0$ from the numerator, from Logarithm of 1 is 0 $\ds$ $=$ $\ds \intlimits {\dfrac {\d} {\d x} \ln x} {x \mathop = 1} {}$ Definition of Derivative of Real Function at Point $\ds$ $=$ $\ds \frac 1 1$ Derivative of Natural Logarithm Function $\ds$ $=$ $\ds 1$

$\blacksquare$

## Proof 3

Note that this proof does not presuppose Derivative of Natural Logarithm Function.

 $\ds \lim_{x \mathop \to 0} \frac {\map \ln {1 + x} } x$ $=$ $\ds \lim_{n \mathop \to \infty} \frac {\map \ln {1 + \frac 1 n} } {\frac 1 n}$ $\ds$ $=$ $\ds \lim_{n \mathop \to \infty} n \, \map \ln {1 + \frac 1 n}$ $\ds$ $=$ $\ds \lim_{n \mathop \to \infty} \map \ln {\paren {1 + \frac 1 n}^n}$ $\ds$ $=$ $\ds \ln e$ Definition of Euler's Number as Limit of Sequence $\ds$ $=$ $\ds 1$ Natural Logarithm of e is 1

$\blacksquare$