Derivative of Logarithm at One
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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 1
L'Hôpital's Rule gives:
- $\ds \lim_{x \mathop \to c} \frac {\map f x} {\map g x} = \lim_{x \mathop \to c} \frac {\map {f'} x} {\map {g'} x}$
(provided the appropriate conditions are fulfilled).
Here we have:
\(\ds \map \ln {1 + 0}\) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds \map {D_x} {\map \ln {1 + x} }\) | \(=\) | \(\ds \dfrac 1 {1 + x}\) | Chain Rule for Derivatives | |||||||||||
\(\ds \map {D_x} x\) | \(=\) | \(\ds 1\) | Derivative of Identity Function |
Thus:
- $\ds \lim_{x \mathop \to 0} \frac {\map \ln {1 + x} } x = \lim_{x \mathop \to 0} \frac {\paren {1 + x}^{-1} } 1 = \frac 1 {1 + 0} = 1$
$\blacksquare$
Proof 2
\(\ds \lim_{x \mathop \to 0} \frac {\map \ln {1 + x} } x\) | \(=\) | \(\ds \lim_{x \mathop \to 0} \frac {\map \ln {1 + x} - \ln 1} x\) | subtract $\ln 1 = 0$ from the numerator, from Logarithm of 1 is 0 | |||||||||||
\(\ds \) | \(=\) | \(\ds \intlimits {\dfrac {\d} {\d x} \ln x} {x \mathop = 1} {}\) | Definition of Derivative of Real Function at Point | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 1\) | Derivative of Natural Logarithm Function | |||||||||||
\(\ds \) | \(=\) | \(\ds 1\) |
$\blacksquare$
Proof 3
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 |
$\blacksquare$