Derivative of Logarithm at One/Proof 2

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

\(\ds \lim_{x \mathop \to 0} \frac {\map \ln {1 + x} } x\)

\(=\)

\(\ds \lim_{x \mathop \to 0} \frac {\map \ln {1 + x} - \ln 1} x\)

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

\(\ds \)

\(=\)

\(\ds 1\)

$\blacksquare$