An introduction to the study of the elements of the differential and integral calculus. From the German of the late Axel Harnack, With the permission of the author.

~ 52.

Examples.

89

3. z =- -- is discontinuous at all points of the right line x = 0
X
while y is finite, and is quite indeterminate at the point x =- 0 y = 0.
4. The function z = sin (4tan-~ -~) being defined for x = 0') to
be zero for all values of y including y = 0, is a discontinuous function
il the point x = 0, y = 0, although for a constant y it is a continuous
function of x and for a constant x a continuous function of y. But if
we put y = ax the function sin (4tan-la) can take all possible values
between the limits -+- 1 and - 1 as we approach the point x == 0, y = 0
in all possible directions, whereas it should there be zero: or, forming the
differences for the neighbourhood of this point, no value can be chosen for
k independently of O h, nor for h independently of ',,l which will make
sill (4tan- -— ) < 6: the criterion of continuity therefore is not
satisfied.
5. If we form the function - =- xay-i, where a and 3 are positive
and f < a, and for all values of x replace it for y = 0 by the value
z = 0O then it is a discontinuous function at x ==0, y = 0, although
when we put y = — ax it is a continuous function: C o- x-ia-il of the
variable x and so is continuous on every direction proceeding from
the origin. For here too it is not possible to find a finite value of h
independent of dk, for which we have
(+ h>)a (~+ nho)- < 6.
In a domain, in which the criterion of continuity holds without
exception for every point, including its limits, f(x, y) is a uniformly continuous function of both variables, i.e. a value
can be assigned for h and one for k, which, whatever be the values
of x and y, are sufficient to satisfy the inequality
f(x + oh, y + ~ )- f(x, y) < a, (o< <, O< < ).
For if it were assumed that such minima values could not be assigned
to h and k, there should be points in the domain, in whose immediate
neighbourhood the criterion of continuity could only be satisfied by
h and k ultimately falling below any assignable value. That this is
impossible is seen as follows: Suppose x1, Y, to be such a point, and
determine ih and k for a point x1 - E, y1 - E' arbitrarily near it, so that
abs [f(x, - + ~ OQh, Yw - E'-~ rc) - f(x, - e, y, - )] < 6.
Now if the only way in which this inequality can continue to hold,
when E and 8' converge to zero, be by h and k also falling below any
assignable value, then - E +- 0h and - '+- k will always remain
less than zero, so that the point xl, y, is never reached in this process.
*) Thome: Einleitung in die Theorie der bestimmten Integrale p. 31.