Derivative of Logarithm at One/Proof 1

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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

L'Hôpital's Rule gives:

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

(provided the appropriate conditions are fulfilled).


Here we have:

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


Thus:

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

$\blacksquare$


Sources