Page 146 - DMTH401_REAL ANALYSIS

Real Analysis

Notes We get a sequence (x,) such that x, = x where 6 = 1/n and

1
0 < |x – a| < for n = 1, 2,.....
n n
and
|f(x ) – A|   .
n 0
0 < |x – a|  x,  a for any n.
n
1 1
Since  0 and |x – a| < , it follows that x,  a.
n n n
But |f(x ) – A |    f(x )  A i.e. f(x ) does not tend to A.
n 0 n n
Therefore x  a " a and x tends to a as n tends to  whereas f(X ) does not converge to A,
n n n
contradicting our hypothesis. This completes the proof of the theorem.
You may note that the above theorem is true even when either a or A is infinite or both a and A
are infinite (i.e. + or –).
By applying this theorem, we can decide about the existence or non-existence of limit of a
function at a point. Consider the following examples:

ì 0 if x is rational
Example: Let f(x) = í
î 1 if x is irrational
Show that an point a in the real line R lim f(x) exists.
x R
Solution: Consider any point ‘a’ of the real line. Let (p ) be a sequence of rational numbers
n
converging to the point ‘a’. Since p is a rational number, f(P ) = 0 for all n and consequently
n n
lim f(P ) = 0, Now, consider a sequence (q) of irrational numbers converging to ‘a’. Since q, is an
n
irrational number, f(q ) = 1 for all n and consequently lim f(q ) = 1. So for two sequences (p ) and
n n n
(q,) converging to ‘a’; sequences (f(p )) and (f(q )) do not converge to the same limit. Therefore
n n
lim f(x) cannot exist for if it exists and is equal to A, then both (f(p )) and (f(q )) would have
-
x a n n
converged to the same limit A.
Example: Show that for the function f: R  R defined by f(x) = x Q x  R, lim f(x) exists
-
for every a  R. x p
Solution: Consider any point a  R. Let (x,) be a sequence of points of R converging to ‘a’. Then
f(x ) = x, and consequently lim f(x ) = lim (x ) = a. So for every sequence <x > converging to ‘a’
n n n n
(f(X )) converges to ‘a’. So by Theorem 3, lim f(x) = a. Consequently lim f(x) exists for every a  R.
n
-
x a x a
-
Task Show that lim 2 = 2 by proving that for any sequence (x ), x  1, converging to 1,
x
n n
x 1
2 converges to 2.
xn
10.3 Algebra of Limits
We discussed the algebra of limits of sequences. In this section, we apply the same algebraic
operations to limits of functions. This will enable us to solve the problem of finding limits of
functions. In other words we discuss limits of sum, difference, product and quotient of functions.

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