A new paper by David Sherry and Mikhail G. Katz. Abstract:

Leibniz entertained various conceptions of infinitesimals, considering them sometimes as ideal things and other times as fictions. But in both cases, he compares infinitesimals favorably to imaginary roots. We agree with the majority of commentators that Leibniz's infinitesimals are fictions rather than ideal things. However, we dispute their opinion that Leibniz's infinitesimals are best understood as logical fictions, eliminable by paraphrase. This so-called syncategorematic conception of infinitesimals is present in Leibniz's texts, but there is an alternative, formalist account of infinitesimals there too. We argue that the formalist account makes better sense of the analogy with imaginary roots and fits better with Leibniz's deepest philosophical convictions. The formalist conception supports the claim of Robinson and others that the philosophical foundations of nonstandard analysis and Leibniz's calculus are cut from the same cloth.

It seems very interesting blog specialized in infinitesimal!

Even if it is a fiction, I thought mathematicians sought consistency. Can you imagine lim (x -> ∞) 1/x = 0 now. But if we write it also as infinitesimal δ or some, many result would be changed.

lim (x -> ∞) 1/x = 0, lim (x -> ∞) 1/2x = 0, lim (x -> ∞) 1/x^2 = 0

So far, the equations are equally zero. But if you also write is as infinitesimal and categorize different types of infinitesimal, you can discover something new.

For instance,

lim (x -> ∞) 1/x = δ(1/x, ∞), lim (x -> ∞) 1/2x = δ(1/2x, ∞), lim (x -> ∞) 1/x^2 = δ(1/x^2, ∞)

Also, we already know multiple types of infinity such as ℵ, ℵ_0. In that case, these might be expressed as

lim (x -> ℵ) 1/x

lim (x -> ℵ_0) 1/x

But in the current calculus notation, both are equally approximated as 0. Is the gap such small enough to regard as nothing between them? In my imaginary notation, it can be like

lim (x -> ℵ) 1/x = δ(1/x, ℵ) ... (1)

lim (x -> ℵ_0) 1/x = = δ(1/x_0, ℵ) ... (2)

I think (1) - (2) is not zero and very big. Perhaps, you can invent another good expression to solve this problem.

there was mistake in the last part.

lim (x -> ℵ_0) 1/x = = δ(1/x_0, ℵ) ... (2)

---> lim (x -> ℵ_0) 1/x = = δ(1/x, ℵ_0) ... (2)

I mistakenly put _0 for the x, it shows this notation is too complicated to use for the daily calculation.