Problem with infinitesimals and approximations I'm reading about infitesimals and I came accros the following calculation.
Suppose f is differentiable at point a.
$Δf - dy = Δf - f'(a)Δx = (\frac{Δy}{Δx} - f'(a))\cdot Δx = ε\cdot Δx$
So, as Δx$\rightarrow 0$ it is that $\frac{Δy}{Δx}\rightarrow f'(a)$ and therefore ε$\rightarrow 0$(based on the textbook)
But the definition of $\frac{Δy}{Δx}$ as Δx approaches 0 is exactly f'(a) so why ε$\rightarrow 0$ and not ε=$0$.
The textbook im using is Thoma's calculus 14 edition. This can be found in chapter 3.9.
 A: It's because although $\Delta x \to 0$ causes $\frac{\Delta y}{\Delta x} \to f'(a)$, note it's a limiting case, where the fraction normally never actually reaches the integral value, although it will become arbitrarily close as $\Delta x$ becomes small enough. In particular, you have by definition of derivatives that
$$\lim_{\Delta x \to 0}\frac{\Delta y}{\Delta x} = f'(a) \tag{1}\label{eq1A}$$
Note with limits that you have the variable approach the limiting value, but never actually get there since, in some cases, the expression being checked doesn't even have a value there (e.g., with $\frac{\Delta y}{\Delta x}$, you get $\frac{0}{0}$). Thus, you can't have $\Delta x$ actually be $0$, although it can become arbitrarily close. As such, in general, you have, from what you wrote
$$\frac{\Delta y}{\Delta x} - f'(a) = \epsilon \neq 0 \tag{2}\label{eq2A}$$
As you noted in your comment below, $\epsilon$ is not a constant but, rather, it depends on $\Delta x$. As such, this is why you have $\epsilon \to 0$ instead of $\epsilon = 0$.
