The Real And Complex Number Systems

xy
(b) fx, y
2
if x, y 0, 0, f0, 0 0.
xy 2 xy 2
Proof: 1. Since (x 0)
xy 2
limfx, y lim
0 for all y,
x0 x0
xy 2 x y 2
we have
2. Since (y 0)
we have
lim
y0
limfx, y 0.
x0
xy 2
limfx, y lim
0 for all x,
y0 y0
xy 2 x y 2
lim
x0
limfx, y 0.
y0
3. (x, y 0, 0) Letx r cos and y r sin , where 0 2, and note that
x, y 0, 0 r 0. Then
xy 2
fx, y
xy 2 x y 2
So,
r 4 cos 2 sin 2
r 4 cos 2 sin 2 r 2 2r 2 cossin
cos 2 sin 2 .
cos 2 sin 2 12cossin
r 2
0ifr 0
fx, y
1 if /4 or 5/4.
Hence, we know that the limit of fx, y does not exists as x, y 0, 0.
(c) fx, y 1 x sinxy if x 0, f0, y y.
Proof: 1. Since (x 0)
limfx, y lim 1
x0 x0 x sinxy y *
we have
2. Since (y 0)
we have
limfx, y
y0
lim
y0
lim
x0
limfx, y 0.
x0
1
lim y0 x sinxy 0ifx 0,
lim y0 y 0ifx 0,
limfx, y 0.
y0
3. (x, y 0, 0) Letx r cos and y r sin , where 0 2, and note that
x, y 0, 0 r 0. Then