2.3 • LIMITS AND CONTINUITY 71Solution If x = r cos 8 and y = r sin 8 , then
2r^3 cos^3 8u(x, y) = 2 2 2. 2 = 2rcos^3 8.
r COS 8+r Sill 8
Because .J (x -0)
2
- (y - 0)
2
= r and because lcos^3 Bl < 1,
lu(x, y)- 01 =2r icos^3 el < € whenever o < vx^2 + y^2 = r < ~·
Hence, for any c > 0, Inequality (2-15) is satisfied for o = ~;that is, u (x , y) has
the limit uo = 0 as (x, y) approaches (O, O).The value uo of the limit must not depend on how (x, y) approaches (xo, Yo),
sou (x , y) must approach the value uo when (x, y) approaches (xo, Yo) along any
curve that ends at the point (xo, Yo). Conversely, if we can find two curves C1
and C 2 that end at (xo, Yo) along which u (x, y) approaches two distinct values
u 1 and u 2 , then u(x, y) does not have a limit as (x , y) approaches (xo, Yo).
- EXAMPLE 2.15 Show that the function u(x, y) = x•".+\ 2 does not have a
limit as (x, y) approaches (0, 0).
Solution If we let (x, y) approach (O, O) along the x-axis, then
lim u (x, 0) = lim (x)(O) = O.(x,0)-(0.0) (x,o}-(O,O) x2 + Q2
But if we let (x, y) approach (0, O) along the line y = x, then
lim u (x x) = lim (x)(x) = -
2(^1).
(x,z)-(0,0) ' (x,x}-(0,0} x2 + x2
Because the value of the limit differs depending on how (x, y) approaches (O, O),
we conclude that u (x, y) does not have a limit as (x, y) approaches (O, O).
Let f (z) be a complex function of the complex variable z that is defined for
all values of z in some neighborhood of z 0 , except perhaps at the point zo. We
say that f has the limit wo as z approaches zo, provided the value f (z) can be
made as close as we want to the value w 0 by ta.king z to be sufficiently close to
zo. When this happens we write
lim f (z) = wo.
z -z o
The distance between t he points z and zo can be expressed by lz - z-0!, so
we can give a precise definition similar to the one for a function of two variables.