Derivative of Logarithm at One

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

Then:
 * $\displaystyle \lim_{x \to 0} \frac {\ln \left({1 + x}\right)} {x} = 1$

Proof
L'Hôpital's rule gives:
 * $\displaystyle \lim_{x \to c} \frac {f \left({x}\right)} {g \left({x}\right)} = \lim_{x \to c} \frac {f^{\prime} \left({x}\right)} {g^{\prime} \left({x}\right)}$

(provided the appropriate conditions are fulfilled).

Here we have:
 * $\ln \left({1 + 0}\right) = 0$
 * $D_x \left({\ln \left({1 + x}\right)}\right) = \dfrac 1 {1 + x}$ from the Chain Rule
 * $D_x x = 1$ from Differentiation of the Identity Function.

Thus:
 * $\displaystyle \lim_{x \to 0} \frac {\ln \left({1 + x}\right)} {x} = \lim_{x \to 0} \frac {\left({1 + x}\right)^{-1}} {1} = \frac 1 {1 + 0} = 1$