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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.

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Yes, this is valid.

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.

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