# Multiply both sides of an equation by infinitesimal quantity

The following linear equation

$$-x \frac{a}{b^2} + c \frac{1}{b^2} = 0 \tag{1}\label{eq1}$$

is such that $$a, b, c \in \mathbb{R}$$. It must be evaluated for $$b \to 0$$.

1. It can be solved immediately by multiplying both sides by $$b^2$$:

$$-xa + c = 0$$ $$x = \frac{c}{a} \tag{2}\label{eq2}$$

1. Otherwise,

$$x \frac{a}{b^2} = c \frac{1}{b^2}$$ $$x = \frac{b^2}{b^2} \frac{c}{a} = \frac{c}{a} \tag{3}\label{eq3}$$

If both sides of an equation are multiplied by a non-zero quantity, the new equation is equivalent to the original one, as stated in a well-known property. But is this property is still valid if the quantity is infinitesimal, as here? Why?

My attempt:

In the procedure 2, leading to $$\eqref{eq3}$$, the fraction $$b^2 / b^2$$ is unitary: it's easier to accept that it's correct, regardless of the value of $$b^2$$.

However, with equations like $$\eqref{eq1}$$ (and more complex cases), it's a common practice to immediately follow the procedure 1. I wonder if it's still correct and why.

One way of looking at this is to consider $$b = \frac 1t, t \to \infty$$. The equation then resolves to $$t^2(-xa+c) = 0$$ Since $$t^2$$ tends to infinity and the RHS is zero, we must have $$-xa + c = 0 \implies x = \frac ca$$.
Another way of looking at it would be that an infinitesimal is a finite number $$\epsilon$$ such that $$\epsilon > 0$$ and $$\epsilon \to 0$$. Thus, the infinitesimal is not exactly zero, but is very, very close to zero and hence multiplying both sides by said infinitesimal does not change the equation.