Page 147 - DMTH401_REAL ANALYSIS

Unit 10: Limit of a Function

Definition 5: Algebraic Operations on Functions Notes
Let f and g be two functions with domain D  R. Then the sum, difference, product, quotient of
f and g denoted by f + g, f – g, fg, f/g are functions with domain D defined by

(f + g) (x) = f(x) + g(x)
(f – g) (x) = f(x) – g(x)
(fg) (x) = f(x). g(x)
(f/g) (x) = f(x)/g(x)

provided in the last case g(x)  0 for all x in D.
Now we prove the theorem.

Theorem 4: If lim f(x) = A and lim g(x) = B, where A and B are real numbers,
x a x a
(1) lim (f + g) (x) = A + B = lim f(x) + lim g(x),
x a x a x a
(ii) lim (f – g) (x) = A – B = lim f(x) – lim g(x),
x a x a x a
(iii) lim (f  g) (x) = A  B = lim f(x)  lim g(x),
x g x a x a
lim f(x)
f x a
(iv) If further lim g(x)  0, then lim f/g(x) exists and lim (x) = A/B = .
x a x g x g lim g(x)
g
x a
Proof: Since lim f(x) = A and lim g(x) = B, corresponding to a number  > 0. There exist numbers
x a x a
 > 0 and  > 0 such that
1 2
0 < |x – a| <   |f(x) – A| < /2 (1)
1
0 < |x – a| <   |g(x) – B| < /2 (2)
2
Let  = minimum ( ,  ). Then from (1) and (2) we have that
1 2
0 < |x – a| <   |f(x) + g(x) – (A + B)|  |f(x) – A| + |g(x) – B|
< /2 + /2 = .
Which shows that lim (f + g) (x) = lim f(x) + g(x) = A + B
x a x a
This proves part (i).
The proof of (ii) is exactly similar. Try it yourself.
(iii) |f(x) g(x) – AB| = |(f(x) – A) g(x) + A (g(x) – B)|

 |f(x) – A| |g(x)| + |A|. |(g(x) – B)|. (3)
Since lim g(x) = B corresponding to 1, there exists a number  > 0
x a 0
such that

0 < |x – a| <   |g(x) – B| < 1.
0
which implies that |g(x)  |g(x) – B| + |B|  1 + |B| = K (say) (4)
Since f(x) = A, corresponding to  < 0, there exists a number  > 0 such that number  < 0 such that
1 1
0 < |x – a| <   |f(x) – A| < /2K (5)
1

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