Page 137 - DMTH401_REAL ANALYSIS

Unit 10: Limit of a Function

Notes
Figure 10.1

Example: Let a function f: R  R be defined as

f(x) = x , " x  R.
2
Find its limits when x  2.

Solution: By intuition, it follows that
2
Lim f(x) = Lim x = 4.
x 2 x 2
In other words, we have to show that for a given E > 0, there exists a  > 0 such that
0 < |x – 2| <   |f(x) – 4| < .
Suppose that an  > 0 is fixed. Then consider the quantity |f(x) – 4|, which we can write as

|f(x) – 4| = |x – 4| = |(x – 2)(x + 2)|.
2
Note that the term |x – 2| is exactly the same that appears in the 6–inequality in the definition.
Therefore, this term should be less than . In other words,

|x – 2| < 
 2 –  < x < 2 + 

 x ] 2 – , 2 + [.
We restrict  to a value 2 so that x lies in the interval ] 2 – , 2 + [  ] 0, 4[. Accordingly, then
|x + 2| < . Thus, if   2, then

|x – 2| < 2  0 < |x + 2| < ,
and further that

|x – 2| <   2  |x + 2| |x – 2| <  |x – 2| < 6.
If 6 is small then so is 6. In fact it can be made less than  by choosing  suitably. Let us, therefore,
select 6 such that  = min.(2, /6). Then
0 < |x – 2| <   |f(x) – 4| <  |x – 2| < .   6. / = .

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