# Prove quantitatively that for infinitesimally small Deltax, (Deltax)/x ~~ Delta(lnx)?

## I actually have probably proved this, but I think I did it qualitatively. Not sure if it's what my book is looking for... For some infinitesimally small $\Delta x$, supposedly, $\frac{\Delta x}{x} \approx \Delta \ln x$. But if $\Delta x$ is small, then $\Delta x = \mathrm{dx}$, the differential change in $x$. That is, $\frac{1}{x} \mathrm{dx} = d \left(\ln x\right)$. Integrating both sides: $\int \frac{1}{x} \mathrm{dx} = \int d \left(\ln x\right) \mathrm{dx}$ The integral of a derivative cancels out to give: $\int \frac{1}{x} \mathrm{dx} = \textcolor{b l u e}{\ln | x | + C}$ which we know to be true from calculus.

Aug 23, 2016

slightly different way of looking at it, but same idea.

#### Explanation:

${\lim}_{\Delta x \to 0} \frac{\Delta \left(\ln x\right)}{\Delta x} = \frac{d \left(\ln x\right)}{\mathrm{dx}} = \frac{1}{x}$.

$\therefore \frac{\Delta \left(\ln x\right)}{\Delta x} \approx \frac{1}{x}$

And so $\Delta \left(\ln x\right) \approx \frac{\Delta x}{x}$

or you could go more formal and write it as

${\lim}_{\Delta x \to 0} \frac{\Delta \left(\ln x\right)}{\Delta x} = {\lim}_{\Delta x \to 0} \frac{\ln \left(x + \Delta x\right) - \ln x}{\Delta x}$

...and complete the derivation of the derivative of ln x from first principles.

So
$= {\lim}_{\Delta x \to 0} \frac{1}{\Delta x} \ln \left(\frac{x + \Delta x}{x}\right)$

$= {\lim}_{\Delta x \to 0} \frac{1}{\Delta x} \ln \left(1 + \frac{\Delta x}{x}\right)$

$y = \frac{\Delta x}{x}$

$= {\lim}_{y \to 0} \frac{1}{y x} \ln \left(1 + y\right)$

$= {\lim}_{y \to 0} \frac{1}{x} \ln {\left(1 + y\right)}^{\frac{1}{y}}$

$= \frac{1}{x} \ln \left(e\right) = \frac{1}{x}$