# Derivative of Logarithm at One/Proof 2

## Theorem

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

Then:

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

## Proof

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

$\blacksquare$