Page 149 - DMTH401_REAL ANALYSIS

Unit 10: Limit of a Function

This completes the proof of the theorem. You may note the theorem is true even when a = ±. Notes
You may also see that while proving (iv), we have proved that if

1 1
lim g(x) = B  0, then lim = .
x a x g(x) B
a
Before we solve some examples, we prove two more theorems.
Theorem 5: Let f and g be defined in the domain D and let f(x)  g(x) for all x in D. Then if lim f(x)
x a
and lim g(x) exist,
x a
lim f(x)  lim g(x).
x n x a
Proof: Let lim f(x) = A, lim g(x) = B. If possible, let A > B.
x a x a
A B
-
for  = , there exist  ,  > 0 such that
2 1 2
0 < |x – a| <   |f(x) – A| < E
1
and 0 < |x – a| <   |g(x) – B| < E.
2
If 6 = min. ( ,  ), then for 0 < |x – a| < 6, g(x)  ]B – , B + [ and f(x)  ] A – , A +  [. But B +  =
1 2
+
A B
A – E = . Therefore g(x) < f(x) for 0 < |x – a|< 6 which contradicts the given hypothesis.
2
Thus A  B.
Theorem 6: Let S and T be non-empty subsets of the real set R, and let f: S  T be a function of
S onto T. Let g: U  R be a function whose domain U C R contains T. Let us assume that lim f(x)
x a
exists and is equal to b and lim g(y) exists and is equal to c. Then lim g(f(x)) exists and is equal
x b x a
to c.
Proof: Since lim g(y) = c, given a number E > 0, there exists a number  > 0 such that
x b 0
0 < |y – b| <   |g(y) – c| < .
0
Since lim f(x) = b, corresponding to  > 0, there exists  > 0 such that
x a 0
0 < |x – a| <   |f(x) – b| <  .
0
Hence, taking y = f(x) and combining the two we get that for
0 < |x – a| < , |g(f(x)) – c| = |g(y) – c| < 
(since |f(x) – b| <  ).
0
This completes the proof of the theorem. Finally we give one more result without proof.

Result: If lim f(x) = A, A > 0 and lim g(x) = B where A and B are finite real numbers then
x a x a
B
g(x)
lim f(x) = A .
x a
Now we discuss some examples. You will see how the above results help us in reducing the
problem of finding limit of complicated functions to that of finding limits of simple functions.
+
(2x 7)(3x 11)(4x 5)
-
+
Example: Find lim
3
x  4x + x - 1
Solution:
+
+
(2x 7)(3x 11)(4x 5)
-
lim 3
x  4x + x - 1
LOVELY PROFESSIONAL UNIVERSITY 143