Page 144 - DMTH401_REAL ANALYSIS

Real Analysis

Notes In this case also f(x) is unbounded and tends to – as x tends to a. You can give similar definitions
for f(a+) = +, f(a–) = +, f(a+) = –, f(a–) = –.

Now we define lim f(x) = .
x 
f is said to tend to  as x tends to  if given a number M > 0, there exists a number k > 0 such that
f(x) > M for x  k.

We may similarly define
lim f(x) = + , lim f(x) = –, lim f(x) = –.
x - x + x -
In all such cases we say that the function f becomes unbounded as x tends to + or – as the case
may be.

It is easy to see from the definition of limit of a function that the limit of a constant function at
any point in its domain is the constant itself. Similarly if lim f(x) = A, then lim cf(x) = cA for any
-
-
x a x a
constant c where c is a real number.
Example: Justify that

1
lim = .
x 2 (x 2- ) 2
-
Solution: You have to verify that corresponding to a given positive number M, there exists a
positive number 6, such that
1 > M whenever 0 < |x – 2| < 6.
(x 2- ) 2
Indeed for x  2,

1 > M  (x – 2) < 1
2
(x 2- ) 2 M
1
 |x – 2| < .
M
1
Take  = . Then you can see that
M
1
2 > M whenever 0 < |x – 2| < 6.
(x 2- )
Hence

1
lim 2 = .
x 2 (x 2- )
-

Task
1. Consider f(x) = |x|, x  R. Show that lim f(x) = +, and lim f(x) = +w and f (0 +) =
x + x -
f (0–) = 0 = f(0)
2. Let f(x) = –|x|, x  R. Prove that lim f(x) = – and lim f (x) = +  and f(0) = f (0+) =
x + x -
f(0–) = 0.

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