xo, yo, dxo, dy。, for x, y, dx, dy. Proceeding to the limit we have

xo, yo, dx, dy, being finally substituted for x, y, dx, dy.

Since the ratio of de to dy is undefined, it appears that the value of (x, y) is generally indeterminate suppose however that, when xx, and y = y, either

df

dx

=

be simultaneously zero, we must proceed to second differentials,

an expression generally indeterminate by reason of the indefiniteness of the ratio of de to dy. If all the partial differentials of ƒ and F of the second order are zero, we must proceed to the third order, and so on.

The extension of the preceding considerations to indeterminate functions of any number of independent variables is obvious. We have considered only the case of indetermination of the form the application of the method, however, to that of the form∞

0

∞

may be established just as in the instance

where a is an arbitrary quantity. Thus p(x, y) may have any

The partial differentials of the first order being zero, we must proceed to differentials of the second order.

a being an arbitrary quantity. The value of p(x, y) is therefore indeterminate, within certain limits; its greatest and least values corresponding to the least positive and least negative values of a + +1. Suppose that

then

1

a

hence + 2 and 2 are the least positive and negative values of

ẞ or a + 1. It appears therefore that (~,, y.) may have any

α

is satisfied identically by a certain value x, of x, whatever be the value of y. The function y will for this value of x appear to be indeterminate.

Differentiating the proposed equation, we get

But since, when x = x ̧, ƒ(x, y) has a constant value zero for all values of y whatever, it follows that in this case also

The value of y, must be determined from the equation (3). In case the equation (3) be satisfied identically for all values of y, we must, the function

df

dx

now occupying the place of the original

function f, proceed to determine y, from the equation

and so on, until the indeterminateness is eradicated.

Ex. 1. Suppose that

f(x, y) = mx2 - x + log (1 + xy) = 0, x = 0.

H

ƒ (x, y) = (y2 – 1) x2 − y {log (1 + x)}2 = 0, x = 0.

or

F(x, y) = (y2 − 1) (x2 + x) − y log (1 + x) = 0 :

but this equation is identically satisfied by x = 0: we must therefore differentiate again with regard to x.