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 |
$\blacksquare$