Page 148 - DMTH401_REAL ANALYSIS

Real Analysis

Notes Since lim g(x) = B, corresponding to a number  > 0, there exists a number  > 0 such that
x a 2
E
0 < |x – a| <   |f(x) – B| < (6)
2 2( A + 1)
Let  = min (a ,  ,  ). Then using (4), (5) and (6) in (3), we have for 0 < |x – a| < ,
1 1 2
|f(x) g(x) – AB|  |f(x) – A| |g(x)| + |A| |g(x) – B|
 |f(x) – A|. K + |A| |(g(x) – B)|

 
< . K + |A| < /2 + /2 = . (7)
2K 2( A + 1)
Therefore, lim g(x) = AB i.e. lim (fg) (x) = AB = lim f(x). lim g(x), which proves part (iii) of the
x a x a x a x a
theorem.
(iv) First we show that g does not vanish in a neighbourhood of a.

B
lim g(x) = B and B  0. Therefore |B| > 0. Then corresponding to we have a number > 0 such
x a 2
B
that for 0 < |x – a| < , |g(x) – B| < .
2
Now by triangle inequality, we have

B
||g(x)| – |B||  |g(x) – B| < .
2
B B
i.e., |B| – < |g(x)| < |B| + . (8)
2 2
B
In other words, 0 < |x – a| <   |g(x)| > .
2
Again since lim g(x) = B, for a given number  > 0, we have a number  > 0 such that 0 < |x – a|
x a
<  implies that
2
B 
|g(x) – B| < .
2
Let 6 = min (, p). Then if 0 < |x – a| < , from (7) and (8) we have
2
-
-
1 1 B g(x) 2 B g(x) 2 B 
- = < < 2 = .
g(x) B g(x) B B 2 2 B
1 1
This proves that lim = .
x g(x) B
a
Now by part (iii) of this theorem, we get that
f(x) 1 1
lim = lim f(x). = lim f(x). lim
x g(x) x a g(x) x a x g(x)
a
a
1
= A. = A/B.
B
lim f(x)
æ f ö x a
i.e., lim ç ÷ (x) = A/B = .
g
x è ø lim g(x)
a
x a

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