Foundations of Differential Calculus

Question

In the preface of Foundations of Differential Calculus there's a section that says:

Thus, if the quantity $x$ is given an increment $\omega$, so that it becomes $x + \omega$, its square $x^2$ becomes $x^2 + 2x\omega + \omega^2$, and it takes the increment $2x\omega + \omega^2$. Hence, the increment of $x$ itself, which is $\omega$, has the ratio to the increment of the square, which is $2x\omega+\omega^2$, as $1$ to $2x+\omega$. This ratio reduces to $1$ to $2x$, at least when $\omega$ vanishes.

$f(x) = x^2$

$f(x+w) = (x+\omega)^2 = x^2 + 2x\omega + \omega^2$

$(x+\omega)^2 - x^2 = 2x\omega+\omega^2$

And then the part I don't understand where $\omega = 2x\omega+\omega^2$ goes to "as $1$ to $2x+\omega$".

Answer 1

The paragraph says that the increment of $x$ (which means $(x + \omega) - x$, i.e. $\omega$) and the corresponding increment of $f$ from $f(x) = x^2$ to $f(x + \omega) = x^2 + 2 \omega x + \omega ^2$ (which means $f(x + \omega) - f(x) = 2 \omega x + \omega ^2$) have the ratio

$$\frac {(x + \omega) - x} {f(x + \omega) - f(x)} = \frac {\omega} {2 \omega x + \omega ^2} = \frac 1 {2 x + \omega}$$

which, when $\omega \to 0$, is exactly $\frac 1 {2x}$.

This prepares you mentally for later considering the limit of the reversed fraction:

$$\lim \limits _{\omega \to 0} \frac {f(x + \omega) - f(x)} {(x + \omega) - x} = 2x ,$$

which is an illustration of the concept of "derivative".

Answer 2

The text shaded in khaki has numerator and denominator interchanged in its last part. Instead it should read as follows:

Thus, if the quantity $x$ is given an increment $ω$, so that it becomes $x + ω$. In this way the square $x^2$ becomes $x^2 + 2xω + ω^2$, so that the square takes the increment $2xω + ω^2$. Hence, the ratio of the increment of the square, which is $2xω+ω^2$, to the increment of $x$ itself, which is $ω$, is $2x+ω$. This ratio reduces to $2x$ when $ω$ is "negligible", or "goes to $0$".