Derivative of Logarithm at One/Proof 3

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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

Note that this proof does not presuppose Derivative of Natural Logarithm Function.

\(\ds \lim_{x \mathop \to 0} \frac {\map \ln {1 + x} } x\) \(=\) \(\ds \lim_{n \mathop \to \infty} \frac {\map \ln {1 + \frac 1 n} } {\frac 1 n}\)
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} n \, \map \ln {1 + \frac 1 n}\)
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} \map \ln {\paren {1 + \frac 1 n}^n}\)
\(\ds \) \(=\) \(\ds \ln e\) Definition of Euler's Number as Limit of Sequence
\(\ds \) \(=\) \(\ds 1\) Natural Logarithm of e is 1

$\blacksquare$