Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/157

We have	${\frac {\alpha }{\beta }}$	$={\frac {\alpha '}{\beta '}}\cdot {\frac {alpha}{\alpha '}}\cdot {\frac {\beta '}{\beta }}$ identically,
and	$\lim {\frac {\alpha }{\beta }}$	$=\lim {\frac {\alpha '}{\beta '}}\cdot \lim {\frac {\alpha }{\alpha '}}\cdot \lim {\frac {\beta '}{\beta }}$	Th. II, §20
		$=\lim {\alpha '}{\beta '}\cdot 1\cdot 1$.	By (C)
(D)	∴ $\lim {\frac {\alpha }{\beta }}$	$=\lim {\frac {\alpha '}{\beta '}}$	Q.E.D

Now let us apply this theorem to the two following important limits.

For the independent variable x, we know from the previous section that $\Delta x$ and dx are identical.

Hence their ratio is unity, and also limit ${\tfrac {\Delta x}{dx}}=1$. That is, by the above theorem,

(E)	In the limit of the ratio of $\Delta x$ and a second infinitesimal, $\Delta x$ may be replaced by dx.

On the contrary it was shown that, for the dependent variable y, $\Delta y$ and dy are in general unequal. But we shall now show, however, that in this case also

$\lim {\frac {\Delta y}{dy}}=1$.

Since $\lim _{\Delta x=0}{\tfrac {\Delta y}{\Delta x}}=f'(x)$ we may write

${\frac {\Delta y}{\Delta x}}=f'(x)+\epsilon$,

where $\epsilon$ is an infinitesimal which approaches zero when $\Delta x{\dot {=}}0$.

Clearing of fractions, remembering that $\Delta x{\dot {=}}dx$,

	$\Delta y$	$=f'(x)dx+\epsilon \cdot \Delta x$,
or	$\Delta y$	$=dy+\epsilon \cdot \Delta x$,	(B), §89

Dividing both sides by $\Delta y$,

	$1$	$={\frac {dy}{\Delta y}}+\epsilon \cdot {\frac {\Delta x}{\Delta y}}$,
or	${\frac {dy}{\Delta y}}$	$=1-\epsilon \cdot {\frac {\Delta x}{\Delta y}}$.
		∴ $\lim _{\Delta x\to 0}{\frac {dy}{\Delta y}}=1$,

and hence $\lim _{\Delta x\to 0}{\tfrac {\Delta y}{dy}}=1$. That is, by the above theorem,

(F)	In the limit of the ratio of $\Delta y$ and a second infinitesimal, $\Delta y$ may be replaced by dy.