Q:

Foundations of Differential Calculus

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 $2x+\omega$".

A:

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".

