Class notes Nov 1 st
Example 1
Some examples of limits of functions
. We show using the definition that lim( 2 ) 3 x

1 x
3  x

.
Let
 
0 . We need to find
 
0 such that for any :| 1|

we have | x
3 
2 x
  
.
We start by assuming that such given

can be found and try to find its value in terms of the

and the point where we compute the limit. We evaluate:
| x
3 
2 x
  x
3   x
  
1||
2 x x 3 |

|
2 x x

(1)
Now, we want to find an estimate for the expression multiplying on x . Since x is in a

that will not depend

-neighborhood of 1, and since we think of

as a small number, we can (and shall) assume that it is not greater than 1. This will imply that if
| x

1 , then 0 x 2 , and therefore x
2 x 3 2
2   
9 . Then we may rewrite (1)
| x
3 
2 x 3 | | x 1|| and, we see that if we choose
 

9
then | x
3
2 x x 3 | 9

2 x
  
 
,
and the proof is complete.
1
Example 2 . Let us show that lim sin x

0
does not exist. x
Observation : Notice that we are given a periodic function (the sine function in particular) and although the limit is computed at zero, the argument of the function is
1 x
, which makes the limit computation equivalent to looking at lim sin x
 x . As we have seen, this last limit does not exist.
We will give a direct proof. Using the sequential limit criterion, we will show that there exists a sequence ( x n
) : x n
Let us define number, sin
9, …), and
1 x n sin x n

 sin
2 n
 n

2
1 x n
 sin
. Obviously, lim n
 x n
 n

2
 n


0 for which
1 if n

4 k

1 (that is n is a multiple of 4 plus 1, such as: 1, 5,
 
1
 lim sin n

0 but
1 x n sin
does not exist.
1 x n
 sin n

2

0 if n is an even
if n

4 k

3 (such as 3, 7, 11,…). Since the sequence

 sin x
1 n


has three subsequences converging to three different numbers, -1, 0 and 1, it follows that the limit does not exist.

Below is the graph of the function in a neighborhood of zero.
Example 3 . We will show that with a small modification the function above has a limit at zero. More specifically we show that lim sin x

0
1 x

0 .
Here is the graph of the function ( )
 x sin
1 x
:
We will use the definition to prove the limit. Let
 
0 . Again, we need to find
 
0 such that for any :| |
 
we have x
1
| sin | x
 
. Assuming that such

can be found,

we have x
1 x
 x
1 x
|
  desired result.
. So we observe that taking
Some example on continuous functions
 
we obtain the
Example 4 . Exercise 4.3.10. (Abbott) So we have a function :

R , property (
 y )

( )

( ) (1). satisfying the a) If in the given equation we take from which it follows that f (0)

0
Let now x

,
  x . Then x y
.
(

0 x
, we get
)
 f (0) f

(0)

( ) f

(0 0) f (
 x )

, f (0) and
 f (0) , therefore f (
  
. b) Let us show that if f is continuous at zero, then it is continuous at every x

R .
Let x

R , and consider a sequence x n
 x . We want to show that n lim (
 n
)

( )
Notice that the last statement is equivalent to lim n

 f x n
 f x
 
0
.
. But
( n
)

( )

( n
 x ) , and if x n
 x , then y n
 x n
 x is a sequence converging to zero. But, by hypothesis, f is continuous at zero, therefore n
 f y n
)
 n
 f x n
 x )
 f (0)

0 , and we proved that f is continuous at x . c) Let k
 f (1) . If in (1) we let x y 1 we get f (1 1) f (2)

2 (1)

2 k that for some positive integer : ( )
 kn . Then
(
  f n
 f (1)
 kn
 
(

1) .
Since we proved that f ( x ) , it follows that f ( n )
  kn
. Assume
, so ( )
 kz for any integer, z .
Let now r q
, ,

Z . We have: ( )
 kp
 f q
 p q
) q f p
( ) q
, the last equality f p following from (1). Then ( ) q
 k p
, or ( )
 kr . q
Now, we proved that if f is continuous at zero it must be continuous everywhere.
Moreover, we saw in Exercise 4.3.8 that if two continuous functions agree on the set of rational numbers, then they must be identical. Since f is continuous and ( )
 kx for all x

Q , it follows that ( )
 kx , for all x

R .
Exercise 1 . Find all continuous functions on R satisfying (
 y )

( ) ( ) . Assume f (0)

0 .