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:

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


Proof 1

L'Hôpital's Rule gives:

$\displaystyle \lim_{x \mathop \to c} \frac {f \left({x}\right)} {g \left({x}\right)} = \lim_{x \mathop \to c} \frac {f' \left({x}\right)} {g' \left({x}\right)}$

(provided the appropriate conditions are fulfilled).


Here we have:

$\ln \left({1 + 0}\right) = 0$
$D_x \left({\ln \left({1 + x}\right)}\right) = \dfrac 1 {1 + x}$ from the Chain Rule
$D_x \left({x}\right) = 1$ from Derivative of Identity Function.


Thus:

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

$\blacksquare$


Proof 2

\(\displaystyle \lim_{x \mathop \to 0} \frac {\map \ln {1 + x} } x\) \(=\) \(\displaystyle \lim_{x \mathop \to 0} \frac {\map \ln {1 + x} - \ln 1} x\) 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
\(\displaystyle \) \(=\) \(\displaystyle 1\)

$\blacksquare$


Proof 3

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

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

$\blacksquare$