Complex Analysis - Maths KU

2.6 Limits and Continuity 119
See (5) in Section 1.6 and Problems 25 and 26 in Exercises 1.6. Because
z is not allowed to take on the value 1 + √ 3i in the limit, we can cancel
the common factor in the numerator and denominator of the rational
function. That is,
lim
z→1+ √ z
3i
2 − 2z +4
z − 1 − √ = lim
3i z→1+ √ 3i
= lim
z→1+ √ 3i
� z − 1+ √ 3i �� z − 1 − √ 3i �
z − 1 − √ 3i
�
z − 1+ √ �
3i .
ByTheorem 2.2(ii) and the limits in (15) and (16), we then have
�
z − 1+ √ �
3i =1+ √ 3i − 1+ √ 3i =2 √ 3i.
lim
z→1+ √ 3i
Therefore, lim
z→1+ √ z
3i
2 − 2z +4
z − 1 − √ 3i =2√3i. In Section 3.1 we will calculate the limit in part (b) of Example 4 in a
different manner.
2.6.2 Continuity
Continuity of Real Functions Recall that if the limit of a real
function f as x approaches the point x0 exists and agrees with the value of
the function f at x0, then we saythat f is continuous at the point x0. In
symbols, this definition is given by:
Continuity of a Real Function f(x)
A function f is continuous at a point x0 if lim f(x) =f(x0). (17)
x→x0
Observe that in order for the equation lim f(x) =f(x0) in (17) to be sat-
x→x0
isfied, three things must be true. The limit lim f(x) must exist, f must be
x→x0
defined at x0, and these two values must be equal. If anyone of these three
conditions fail, then f cannot be continuous at x0. For example, the function
⎧
⎨ x
f(x) =
⎩
2 , x < 0
,
x − 1, x ≥ 0
illustrated in Figure 2.52 is not continuous at the point x = 0 since lim
x→0 f(x)
x
does not exist. On a similar note, even though lim
x→1
2 − 1
= 2, the function
x − 1
f(x) = x2 − 1
x − 1
is not continuous at x = 1 because f(1) is not defined.