Derivative of Logarithm at One/Proof 2

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Theorem

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

Then:

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

$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle \lim_{x \to 0} \frac {\ln \left({1 + x}\right)} {x}$	$=$	$\displaystyle \lim_{x \to 0} \frac {\ln \left({1 + x}\right) - \ln 1}{x}$	$\displaystyle $	$\displaystyle $	$\displaystyle $	subtract $\ln 1 = 0$ from the numerator, from Logarithm of 1 is 0
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle \left.{\dfrac {\mathrm d} {\mathrm dx} \ln x}\right \vert_{x=1}$	$\displaystyle $	$\displaystyle $	$\displaystyle $	definition of derivative at a point
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle \frac 1 1$	$\displaystyle $	$\displaystyle $	$\displaystyle $	Derivative of Natural Logarithm Function
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle 1$	$\displaystyle $	$\displaystyle $	$\displaystyle $

$\blacksquare$

\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \lim_{x \to 0} \frac {\ln \left({1 + x}\right)} {x}\)	\(=\)	\(\displaystyle \lim_{x \to 0} \frac {\ln \left({1 + x}\right) - \ln 1}{x}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	subtract $\ln 1 = 0$ from the numerator, from Logarithm of 1 is 0
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left.{\dfrac {\mathrm d} {\mathrm dx} \ln x}\right \vert_{x=1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	definition of derivative at a point
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac 1 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Derivative of Natural Logarithm Function
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)