Derivative of Logarithm at One/Proof 1

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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

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:

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

$\blacksquare$


Sources