# Intuition for nonstandard analysis from limits conception

I'm trying to gain an intuition for the use of nonstandard analysis over the limit approach.

Traditionally the motivation for derivatives is that the derivative of a point $$(x,f(x))$$ for some function $$f$$ is the slope of the line connecting that point to some nearby point, $$(x+\alpha, f(x+\alpha)), \alpha > 0$$, and making $$\alpha$$ smaller and smaller to the limit of $$x$$, hence getting a tangent line.

So maybe this means (I know its wrong but follow my train of thought) that the derivative of a point $$x$$ of some function $$D: \mathbb{Z}\rightarrow \mathbb{Z}, \text{ where } \mathbb{Z} \text{ is the integers}$$, as the slope of the line connecting $$x$$ to $$x+min(\mathbb{Z^+})$$, where of course $$min(\mathbb{Z^+})$$ = 1. So the derivative of such a function $$D$$ is $$D' = \frac{D(x+1) - D(x)}{1}.$$ For example, if $$D(x) = x^2$$ on the naturals then $$D' = (x+1)^2 - x^2 = 2x+1$$. Obviously if this same function was defined on the reals then $$D' = 2x$$.

So at this point it seems like more generally, the derivative for a function $$G: \mathbb{F}\rightarrow \mathbb{F}, \text{ where } \mathbb{F} \text{ is some field/ring/group}$$, is

$$G' = \frac{G(x+dx) - D(x)}{dx}, dx = min(\mathbb{F^+})$$ (again I know this isn't true, but follow my intuition). In words, my thinking so far is that the derivative at $$x$$ is to take the slope of the line from $$x$$ to $$x+dx$$ where $$dx$$ is the smallest nonzero element in the field or ring.

I think this until I consider a function $$L: \mathbb{R} \rightarrow \mathbb{R}$$, defined on the reals. I quickly realize that there is no minimal non-zero element to do my derivative calculation as described above.

BUT, I decide this isn't a problem if I assume there is some larger field to which the reals is a subset, $$\mathbb{R} \subset \mathbb{F_a}$$, and I assume this larger field $$F_a$$ contains an element $$dx \not\in \mathbb{R}$$ that is smaller than any non-zero element in $$\mathbb{R}$$, allowing me to proceed with my derivative calculation as above, as long as I remember that $$dx$$ is not a real number anymore.

So for the function $$f(x) = x^2$$ defined over the reals, I now calculate $$f' = \frac{(x+dx)^2 - f(x)}{dx} = \frac{x^2 - 2x(dx) - (dx)^2 - x^2}{dx} = \frac{2x(dx) + (dx)^2}{dx} = \frac{dx(2x+dx)}{dx} = 2x+dx$$

So now I have a result that is some number with a real part ($$2x$$) and a non-real part ($$dx$$), but given that the non-real part is smaller than any non-zero real number, if I just ignore that term and consider only the real-part, I will still have the correct slope (in the limit), so I just take $$2x$$ as my final derivative.

Is that a reasonable way to think about it?

• Aside from the minimum nonzero part, the early users of calculus thought of $d\!x$ as an infinitesimal without fully developing a theory of infinitesimal numbers as an extension of the reals. – Somos Nov 25 '18 at 13:45
• From the point of view of the internal set theory, there is no extension, just a distinction of standard and non-standard elements of $\Bbb R$ with some non-standard axioms. – LutzL Dec 7 '18 at 20:51