Derivative of Logarithm at One

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

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

Proof
L'Hôpital's rule gives $$\lim_{x \to c} \frac {f \left({x}\right)} {g \left({x}\right)} = \lim_{x \to c} \frac {f^{\prime} \left({x}\right)} {g^{\prime} \left({x}\right)}$$ (provided the appropriate conditions are fulfilled).

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

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