Why is it possible to do the following approximation?
$$\frac{\mathrm{d} x}{x+\mathrm{d} x} \approx \frac{\mathrm{d}x}{x} $$
Why does that $\mathrm{d} x$ in the denominator "count" less than the one in the numerator?
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1$\begingroup$ What is ${\rm d}x$ here? Usually it's used to denote a very small quantity (${\rm d}x \ll x$). Then the approximation makes sense as $x+{\rm d}x \approx x$. $\endgroup$– WintherCommented Sep 14, 2016 at 14:42
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$\begingroup$ If $\mathrm dx$ is infinitesimal and $x$ is not $\approx 0$, we have $x+\mathrm dx\approx x\not\approx 0$ $\endgroup$– Hagen von EitzenCommented Sep 14, 2016 at 14:42
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4 Answers
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The $dx$ in the denominator is dwarfed by the $x$. The difference between $\frac{0.000001}{1.000001}$ and $\frac{0.000001}{1}$ is miniscule.
answered Sep 14, 2016 at 14:42
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It's $\displaystyle \frac{dx}{x+dx}=\sum\limits_{k=0}^\infty (-1)^k (\frac{dx}{x})^{k+1}$.
Because of $\displaystyle (\frac{dx}{x})^{k+1}<<\frac{dx}{x}$ for $k\in\mathbb{N}$ you can approximate $\displaystyle \frac{dx}{x+dx}\approx \frac{dx}{x}$.
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This is only valid as long as $x\not\approx 0$.
We have $x+\mathrm dx\approx x$ because $\mathrm dx\approx 0$ and addition is continuous. Then provided $x\not\approx 0$, the claim follows by continutity of division.
answered Sep 14, 2016 at 14:45
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You are allowed to approx when terms are added as the difference between the results will be negligible however you cannot do the same with multiplication or division $10000+0.0001=10000.0001$ which is quite close however $10000*0.0001$ is definitely nowhere near $10000 $
