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:

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


Proof

\(\displaystyle \lim_{x \mathop \to 0} \frac {\map \ln {1 + x} } x\) \(=\) \(\displaystyle \lim_{x \mathop \to 0} \frac {\map \ln {1 + x} - \ln 1} x\) subtract $\ln 1 = 0$ from the numerator, from Logarithm of 1 is 0
\(\displaystyle \) \(=\) \(\displaystyle \intlimits {\dfrac {\d} {\d x} \ln x} {x \mathop = 1} {}\) Definition of Derivative of Real Function at Point
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 1\) Derivative of Natural Logarithm Function
\(\displaystyle \) \(=\) \(\displaystyle 1\)

$\blacksquare$