1550251515-Classical_Complex_Analysis__Gonzalez_

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Functions. Limits and Continuity. Arcs and Curves 133

Note in the definition above that the domain D can be any infinite set of
points having at least an accumulation point, and that a need not belong
to D, so that f(a) may or may not be defined. Also, it should be noted
that a function may fail to have a limit at an accumulation point of its

d()_m<'!-~· For instance, if we define f(z) = z/z for z =f 0, this function


does not have a limit as z _, 0. In fact, if we let z = _re;o, we find that

f(z) = e^2 io, Jf(z)J,;;;,: 1; so the image of N~(O) by f is the unit circle, which

is not contained in N<(L) if E < 1, no matter what value is assumed for L.

If the limit of a function at a point exists, it is unique. This is suggested


by the wording of the definition, and it is a consequence of the fact that f is

supposed to be a single-valued function, as shown in the following theorem.

Theorem 3.1 If a function has a limit at a point, then this limit is unique.


Proof Suppose that limz-->a f(z) = L1 and limz-->a f(z) = L2, with L1 =f
L 2. Choose e <^1 / 2 IL 1 -L 2 1. Then N€(L 1 )nN<(L 2 ) = 0. By the definition

of limit there exists 81 > 0 such that

f(z) E N<(L1) whenever z E N~ 1 (a) n D

Also, there exists 82 > 0 such that

Hence if 8 = min( 81 , 82 ) for any z such that

z E NHa) nD


we must have


and

which is impossible since the neighborhoods N<(L 1 ) and N<(L 2 ) are disjoint


and f is single-valued.

Example Let f(z) = (z^2 -1)/(z -1), D = <C - {1}. Then limz__, 1 f(z) =



  1. In fact, for z =f l, f(z) = z + 1, and for any given E > 0 we have


lf(z) - 21 = lz - ll < E whenever 0 < lz -ll < 8, with 8 = E.

3.6 Finite Limit at Infinity

Definition 3.2 Let f be as in Definition 3.1, and suppose that oo is an

accumulation point of D. The number L ( L =f oo) is said to be the limit of

f(z) as z approaches oo [or the limit of f(z) at oo] if for every E > 0 there

is a positive number M (depending on E) such that

lf(z)-LlMand z ED

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