L'Hopital's rule with dual numbers

Question

Background:

For the dual numbers, we extend the reals with an additional unit vector $\epsilon$ subject to the constraint that $\epsilon^2 = 0$. We can write dual numbers as $x_0 + x_1 \epsilon$ for $x_0,x_1 \in \mathbb{R}$.

We have a rule for multiplication, $$ (x_0 + x_1 \epsilon)(y_0 + y_1 \epsilon) = x_0 y_0 + x_1 y_0 \epsilon + x_0 y_1 \epsilon + x_1 y_1 \epsilon^2 = x_0 y_0 + (x_1 y_0 + x_0 y_1)\epsilon, $$ as well as division, $$ \frac{x_0 + x_1 \epsilon}{y_0 + y_1 \epsilon} = \frac{x_0 + x_1 \epsilon}{y_0 + y_1 \epsilon} \left(\frac{y_0 - y_1 \epsilon}{y_0 - y_1 \epsilon}\right) = \frac{x_0 y_0 + (x_1 y_0 - x_0 y_1)\epsilon}{y_0^2}. $$ It's clear that if $y_0=0$, division is undefined for all values of $y_1$. That is, numbers of the form $x_1 \epsilon$ are zero divisors. Furthermore, we have the exact Taylor series $$ f(x_0 + x_1\epsilon) = f(x_0) + x_1 f'(x_0) \epsilon. $$

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My question is: Is L'Hopital's rule $$\lim_{x\rightarrow x_0} \frac{f(x_0)}{g(x_0)} = \lim_{x\rightarrow x_0} \frac{f'(x_0)}{g'(x_0)}$$ when $f(x_0)=g(x_0)=0$ just the addition of the rule $$ \frac{x_1 \epsilon}{y_1 \epsilon} = \frac{x_1}{y_1}? $$ That is, we can write that, if $f(x_0)=g(x_0)=0$, then L'Hopital's rule can be implemented by making the simplification $$ \frac{f(x_0 + x_1 \epsilon)}{g(x_0 + x_1 \epsilon)} = \frac{x_1 f'(x_0) \epsilon}{x_1 g'(x_0) \epsilon} = \frac{f'(x_0)}{g'(x_0)}? $$

Basically, is L'Hopital's rule equivalent to saying we can "cancel" the unit $\epsilon$ when the number is pure imaginary? What are the conditions necessary for making this "cancellation"?

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The case where the numerator and denominator are different variables seem to make this less than trivial: if $f(x_0) = g(y_0) = 0$, then the simplistic version of the rule would say that you should assign the value $$ \frac{f(x_0 + x_1 \epsilon)}{g(y_0 + y_1 \epsilon)} = \frac{x_1 f'(x_0) \epsilon}{y_1 g'(y_0) \epsilon} = \frac{x_1 f'(x_0)}{y_1 g'(y_0)} $$ which does give a well-defined procedure for defining the value of the ratio, but it doesn't necessarily correspond to a limit in $x$ and $y$ (since it's not obvious what order the limits on $x$ and $y$ should be taken). It does, however, correspond to the limit along the path $\big(x(t),y(t)\big) = \big(x_0 + x_1 t, y_0 + y_1 t\big)$ such that $$ \lim_{t\rightarrow 0} \frac{f\big(x(t)\big)}{g\big(y(t)\big)} = \lim_{t\rightarrow 0}\frac{f'\big(x(t)\big)x'(t)}{g'\big(y(t)\big)y'(t)} = \frac{x_1 f'(x_0)}{y_1 g'(y_0)}. $$ In the context of the dual numbers, is there a "good" reason for choosing this path? It does seem to have the advantage that it reduces to L'Hopital's rule for $y=x$.

Answer 1

Non-standard analysis is, at least in calculus, about replacing limits with algebraic operations. The dual numbers, and in generalization, truncated Taylor series, are one step towards this, doing "infinitesimal" calculus without using infinitesimals, and represent about what Leibniz understood as or how he used infinitesimals. (Eliminating infinitesimals from common use was a thing of the second half of the 19th century, introducing modern infinitesimals happened in the second half of the 20th century.)

So like in algebra where one algebraic number stands for all its conjugates, here one infinitesimal stands for all, provided the functions that they are applied to are "standard", not constructed with infinitesimals (like polynomials with some coefficients infinitesimal). The $ϵ$ stands for this idealized infinitesimal, usually $ϵ^2$ is just a much smaller infinitesimal, but for basic calculus situations indeed practically $ϵ^2=0$.

You could also think about $ϵ$ being a very very small number so that any normal $C^1$ function restricted to $[x_0-ϵ,x_0+ϵ]$ is essentially linear and the quotient $\frac{f(x_0+t)}{g(x_0+t)}$, $0<|t|\le ϵ$ is nearly constant up to infinitesimal variations.

In itself, the isolated division by the zero-divisor $ϵ$ is impossible. Cancellation has to be treated carefully, as much as the application of l'Hopitals rule.