Infinitesimals

In the Differential Calculus we are usually concerned with the derivative, that is, with the ratio of the differentials $dy$ and $dx$. In some applications it is also useful to consider $dx$ as an infinitesimal (see §3.3), that is, as a variable whose values remain numerically small, and which, at some stage of the investigation, approaches the limit zero. Then by (9.2), and item 2 in §3.8, $dy$ is also an infinitesimal.

In problems where several infinitesimals enter we often make use of the following

Theorem 9.3.1   In problems involving the limit of the ratio of two infinitesimals, either infinitesimal may be replaced by an infinitesimal so related to it that the limit of their ratio is unity.

Proof: Let $\alpha$, $\beta$, $\alpha'$, $\beta'$ be infinitesimals so related that

$$\lim \frac{\alpha'}{\alpha} = 1,\ \ \ \lim \frac{\beta'}{\beta} = 1.$$

We have

$$\frac{\alpha}{\beta} = \frac{\alpha'}{\beta'} \cdot \frac{\alpha}{\alpha'} \cdot \frac{\beta'}{\beta}$$

identically, and

$$\lim \frac{\alpha}{\beta} = \lim \frac{\alpha '}{\beta '} \cdot ... ...cdot \lim \frac{\beta '}{\beta} =\lim \frac{\alpha'}{\beta'} \cdot 1 \cdot 1 ,$$

by Theorem 3.8.2. Therefore,

$$\lim \frac{\alpha}{\beta} = \lim \frac{\alpha'}{\beta'}.$$

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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 $\frac{\Delta x}{dx} = 1$. That is, by the above theorem, 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 $\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = f'(x)$ we may write $\frac{\Delta y}{\Delta x} = f'(x) + \epsilon$, where $\epsilon$ is an infinitesimal which approaches zero when $\Delta x \to 0$.

Clearing of fractions, remembering that $\Delta x \to dx$, $\Delta y= f'(x)dx + \epsilon \cdot \Delta x$, or $\Delta y= dy + \epsilon \cdot \Delta x$, by (9.2). Dividing both sides by $\Delta y$, $1 = \frac{dy}{\Delta y} + \epsilon \cdot \frac{\Delta x}{\Delta y}$, or $\frac{dy}{\Delta y} = 1 - \epsilon \cdot \frac{\Delta x}{\Delta y}$. Therefore, $\lim_{\Delta x \to 0} \frac{dy}{\Delta y} = 1$, and hence $\lim_{\Delta x \to 0} \frac{\Delta y}{dy} = 1$. That is, by the above theorem, In the limit of the ratio of $\Delta y$ and a second infinitesimal, $\Delta y$ may be replaced by $dy$.