Page 138 - DMTH401_REAL ANALYSIS

Real Analysis

Notes This completes the solution.
Note that the first step is to manipulate the term |f(x) – A| by using algebra. The second step is
to use a suitable strategy to manipulate |f(x) – A| into the form
|x – a| (trash)
where the ‘trash’ is some expression which has the property that: it is bounded provided that 6
is sufficiently small. Why we use the term ‘trash, for the expression as a multiple of |x – a|? The
reason is that once we know that it is bounded, we can replace it by a number and forget about
it.
The number 6 arose by virtue of this ‘trash’. If you take 6  3 (instead of 6  2), you can still show
that 6 will be replaced by 7. In that case you can set  as
 = min.(3, /7)

and the proof will be complete. Thus, there is nothing special about 6. The only thing is that such
a number (whether 6 or 7) has to be evaluated by the restriction placed on 6.

Finally, note that in general, 6 will depend upon .

Task For a function f: R  R defined by f(x) = x , find its limit when x tends to 1 by the
2
 –  approach.
In Unit 5, we proved that a convergent sequence cannot have more than one limit. In the same
way, a function cannot have more than one limit at a single point of its domain. We prove it in
the following theorem:

Theorem 1: If lim f(x) = A, lim f(x) = B, then A = B.
-
x a
-
x a
Proof: In short, we have to show that if lim f(x) has two values say A and B, then A = B. Since
x a
-
lim f(x) = A, lim f(x) = B, given a number E > 0, there exists numbers  ,  > 0 such that
-
x a x a 1 2
-
|f(x) – A| < /2 whenever 0 < |x – a| < 
1
and
|f(x) – B| < /2 whenever 0 < |x – a| <  .
2
If we take  equal to minimum of  and  , then we have
1 2
|f(x) – A| < /2 and |f(x) – B| < /2 whenever 0 < |x – a| < .
Choose an x such that 0 < |X – a| < . Then
0 0
|A – B| = (A – f(x ) + f(x ) – B|  (A – f(x )| + |f(x ) – B|
0 0 0 0
< /2 + /2 = .
E is arbitrary while A and B are fixed. Hence |A – B| is less than every positive number  which
implies that |A – B| = 0 and hence A = B. (For otherwise, if A  B then A – B = C  0 (say). We can
choose  < |C| which will be a contradiction to the fact that |A – B| <  for every  > 0.)
In the example considered before defining limit of a function, we allowed x to assume values
both greater and smaller than 1. If we consider values of x greater than 1 that is on the right of 1,
we see that values of f(x) approaches 5. We say that f(x) tends to 5 as x tends to 1 from the right.

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