Page 145 - DMTH401_REAL ANALYSIS

Unit 10: Limit of a Function

We have already stated that if a function t is define a by f(x) = 1/x, x  0, then the limits f(0+) and Notes
f(0–) and lim f(x) do not exist. It simply means that these limits do not exist as real numbers. In
x 0
-
other words, there is no (finite) real number A such that f (0+) = A f(0–) = A, or lim f(x) = A.
x 0
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10.2 Sequential Limits
In Unit 5, you studied the notion of the limit of a sequence. You also know that a sequence is also
a function but a special type of function. What is special about a sequence? Do you remember it?
Recall it from Unit 5. Naturally, you would like to know the relationship of a sequence and an
arbitrary real function in terms of their limit concepts. Both require us to find a fixed number A
as a first step. Both assume a small positive number  as a test for closeness. For functions we
need a positive number  corresponding to the given positive number E and for sequences we
need a positive integer m which depends on . So, then what is the difference between the two
notions? The only difference is in their domains in the sense that the domain of a sequence is the
set of natural numbers whereas the domain of an arbitrary function is any subset of the set of
real numbers. In the case of a sequence, there are natural numbers only which exceed any choice
of m. But for a function with a domain as an arbitrary set of real numbers, this is not necessary
the case. Thus in a way, the notion of the limit of a function at infinity is a generalization of that
of limit of a sequence.
Let us now, therefore, examine the connection between the limit of a function and the limit of a
sequence called the sequential limit. We state and prove the following theorem for this purpose:
Theorem 3: Let a function f be defined in a neighbourhood of a point ‘a’ except possibly at ‘a’.
Then f(x) tends to a limit A as x tends to ‘a’ if and only if for every sequence (x ), x  a for any
n n
natural number n, converging to ‘a’, f(x ) converges to A.
n
Proof: Let, lim f(x) = A. Then for a number E > 0, there exists a 6 > 0 such that for 0 < |x – a| < 6 we
x a
-
have
|f(x) – A| < 
Let (x,) be a sequence (x  a for any n  N) such that (x ) converges to a i.e. x  a.
n n n
Then corresponding to  > 0, there exists a natural number m such that for all n  m
|x – a| < .
n
Consequently, we have

|f(X ) – A| < , " n  m.
n
This implies that f(x ) converges to A.
n
Conversely, let f(x ) converge to A for every sequence x which converges to a, x  a for any n.
n n n
Suppose lim f(x)  A.
x a
-
Then there exists at least one , say  =  such that for any  > 0 we have an x such that
0 
0 < |x – a| < 

and
|f(x ) – A|   .
 0
1
Let  = , n = 1, 2, 3......
n

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