1550251515-Classical_Complex_Analysis__Gonzalez_

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fact, we have

so that

lim R = 0

16.zl->O IAzl

Chapter6

and lim ..!..__ = 0

6.z->0 Az

With the "little-oh" notation the result above can be expressed briefly by
writing e: = o(IAzl), or c = o(Az). Conversely, if u can be written in the
form (6.4-2) with c^1 = e:/IAzl -t 0 as IAzl -t 0, it can also be expressed
as in (6.4-1) by taking

and

A differentiable function at a point ( x, y) is clearly continuous at that

point since (6.4-2) implies that &u -t 0 as Ax, Ay -t 0. However, a

continuous function at ( x, y) is not necessarily differentiable at the same

point, as the following example shows.

Let

xy

u( x, y) = --;=========;=

yx2 + y2

for (x,y)-:/: (0,0)

and u(O, 0) = 0. This function is continuous at (0, 0), since for lxl :5 8,

IYI :5 8 we have

lu(x,y)-u(O,O)I= lxllYI :5lxl:58

vx2 + y2

which can be made less than a given p > 0 by choosing 0 < 8 < p.

The given function vanishes on the coordinate axes, i.e., u(x,O) = 0 for

all x, and u(O,y) = 0 for ally. Hence we obtain

Ux(O, 0) = 0 and uy(O,O) = 0

Although the partial derivatives exist at the origin, the function is not

differentiable there since

AxAy

Au= u(Ax, .6.y)-u(O, 0) = --;:::===

.JAx^2 + Ay^2

and. the principal part of Au is O·Ax+O·Ay = O, so that the residual part is

· AxAy

c= -;::=====

vAx2 + Ay2