How is it possible that $dx = \varepsilon$?

Question

Let's consider graph of function $y=x^2$ ("Calculus with infinitesimals" Efraín Soto Apolinar, see picture below).

We have the point $(dx, (dx)^2)$ which coincide with point $(dx, 0)$ and $dx$ isn't nilsquare infinitesimal. But if we know that $(dx)^2 = 0$ then we can conclude that $dx$ must be nilsquare infinitesimal $\varepsilon$. Thus the point $(dx,0)$ coincide with the point $(\varepsilon, 0)$.

I'm confused.

How is it possible that $dx = \varepsilon$ ?

Thanks.

Answer 1

Yes, $dx$ is a nilsquare infinitesimal. As mentioned in my answer here, one prefers to use the $dx$ notation when discussing derivatives or related notions such as maxima and minima, as Bell does here.

It is not clear to me what the theoretical background is for Apolinar's remarks. Is he working in Synthetic Differential geometry? The way he passes from $dr^2=dx^2$ to $dr=dx$ certainly looks suspicious.