Since the right-hand member is an infinitesimal by (1), p. 19, we have, from the converse theorem on p. 18,
or,
|
|
which was to be proved.
Proof of Theorem II. Let
and
be the variables,
and
their respective limits, and
and
infinitesimals; then
and |
 |
and
|
Multiplying, |
 |
Multiplying,
|
or,
(B)

Since the right-hand member is an infinitesimal by (1) and (2), p. 19, we have, as before,
which was to be proved.
Proof of Theorem III. Using the same notation as before,
or,
(C)

Here again the right-hand member is an infinitesimal by (4), p. 19, if
; hence
which was to be proved.
It is evident that if any of the variables be replaced by constants, our reasoning still holds, and the above theorems are true.
21. Special limiting values. The following examples are of special importance in the study of the Calculus. In the following examples
and
.