$dx$ isn't a nilsquare infinitesimal because $(dx)^2$ isn't zero. $\varepsilon$ is a nilsquare infinitesimal in this case. I wrote that there is not only point $(dx,0)$ but there is also point $(\varepsilon, 0)$ which coincide with $(dx,0)$.

Apolinar's proof works for $dx$. In this case $dx$ isn't nilsquare infinitesimal.

@Mike_bb, saying that "Apolinar's proof works for dx" does not say much about what Apolinar's theoretical framework is. Is it classical analysis?

Yes. It's classical analysis. @Mikhail Katz

@Mike_bb, if he is working in the framework of classical analysis, he is just being sloppy. A reputable publisher would not publish a book like this in mathematics.

I made mistake. It's not classical analysis. The book is called "Calculus with infinitesimals". @Mikhail Katz

@Mike_bb, good to know, but does he use classical logic or intuitionistic logic?

Classical logic. @Mikhail Katz

OK, see my comment above.

Ok. But what's wrong in Apolinar's approach? I read about infinitesimals in this book. I checked what is written there and I didn't find errors. @Mikhail Katz

@Mike_bb, if he is using classical logic, then the equality $dx^2 + dx^4=dx^2$ can only be true if $dx=0$.

Why? When two or more infinitesimals of different order are added, the infinitesimal of the least order is obtained. If this is not enough to understand we can rewrite this expression as $(dx)^2 + (dx)^4 = (1+(dx)^2) (dx)^2$ , $(dx)^2$ in brackets can be neglected and result is $(dx)^2$ . There is theorem in the book "A Treatise on Infinitesimal Calculus: Differential calculus" Bartholomew Price (page 23). I can share link to this book. @Mikhail Katz

@Mike_bb, as discussed in an earlier post, Price is not rigorous. The quantity $1+dx^2$ is just not equal to $1$! To give a rigorous presentation of this, one needs to use the standard part.

The standard part is used when we talk about Nonstandard analysis. Isn't it? A few days ago I could explain what Price meant. I can try to post explanation in an earlier post. @Mikhail Katz

@Mike_bb, nonstandard analysis is the best approach we have today of accounting for infinitesimal analysis in accordance with modern standards. If $1+dx^2=1$ then by canceling the $1$, we get $dx^2=0$, which is impossible in classical logic. So there is a problem here!

To avoid this we can write $0+(dx)^2=0$. If we agree that $2xdx+(dx)^2 = 2xdx$ and $x=0$ then we get $0 + (dx)^2 = 0$ and it correctly works. @Mikhail Katz

@Mike_bb, the usual laws of elementary algebra show that $0+dx^2=0$ is the same as $dx^2=0$. You can't get out of a mathematical problem by finessing notation :-)

I agree that this isn't rigorous but it works. @Mikhail Katz

@Mike_bb, some of the things that seemed to work all along, led mathematicians into serious contradictions in the second half of the 19th century. If it weren't for the contradictions, we could happily continue setting $dx$ equal zero, as Price does :-)

Using infinitesimal quantities is more natural to me. When I first saw definition of instantaneous speed as derivative I asked myself: how is it possible that there is speed at the point? Point has zero length. But infinitesimals explain "how". If infinitesimals didn't exist we couldn't say about speed "at the point". @Mikhail Katz

@Mike, I cannot be accused of not sharing your enthusiasm for infinitesimals :-) see e.g., u.math.biu.ac.il/~katzmik/infinitesimals.html However, one still needs an approach that meets modern requirements of a mathematical theory.