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



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,

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 $$\Delta x \dot= dx$$,

Dividing both sides by $$\Delta y$$,

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