Page 141 - DMTH401_REAL ANALYSIS

Unit 10: Limit of a Function

Notes
Example: Consider the function I defined by

2
x - 1
f(x) = , x  R, x  1
x 1
-
Find its limit as x  1.
Solution: Note that f(x) is not defined at x = 1. (Why?).
2
x - 1
For any x  1, f(x) = = x + 1.
-
x 1
lim f(x) = lim (x + 1) = 2
x 1 + x 1 +
-
-
lim f(x) = lim (x + 1) = 2
-
-
x 1 - x 1 -
Since lim f(x) = lim f(x), by Theorem 2, lim f(x) = 2,
-
-
-
x 1 + x 1 - x 1
lim f(x) = 2 can be seen by  –  definition as follows:
-
x 1
Corresponding to any number E > 0, we can choose 8 =  itself. Then, it is clear that
0 < |x – 1| <  =  
2
x - 1
|f(x) – 2| = - 2 = |x + 1 – 2| = |x – 1| < .
x 1
-
From Theorem 2, it follows that f(l+) and f(l–) also exist and are both equal to 2.
Example: Let f: R  R be defined as
ì |x|, x  0
f(x) = í
î 3, x » 0.
Find its limit when x  0.

Solution: You are familiar with the graph of f as given in Unit 4. It is easy to see that lim f(x) = 0
-
x 0
= f(0+) = f(0–). The fact that f(0) = 3 has neither any bearing on the existence of the limit of f(x) as
x tends to 0 nor on the value of the lim f(x).
x 0
-
Example: Define f on the whole of the real L in  as follows:
ì 1 if x > 0
ï
f(x) = í 0 if x = 0
ï - 1 if x < 0.
î
Find its limit when x tends to 0.
Solution: Since f(x) = 1 for all x > 0,
f(0+) = lim f(x) = + 1.
0
x +
Similarly f(0–) = –1. Since lim f(x)  lim f(x), lim f(x) does not exist.
x 1+ x 1– x 0
Now, we give another proof using  – S definition.

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