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CHAPTER 03
THEORY OF LIMITS
FORMAL DEFINITION OF A LIMIT
In the previous chapter we provided an intuitive definition of a limit, and we stated, but
did not prove, a number of useful limit theorems
...
The intuitive definition earlier introduced is
correct
...
What do we mean
with close?
In everyday life the use of “close” is relative, but in mathematics we need a more precise
description
...

Definition 3
...

(ii)
A neighborhood of a point c is an open interval (a,b) such that c  (a,b)
...
They
will almost always represent small positive quantities
...
2
Suppose that the function f is defined near a point c (i
...
in a neighborhood of a point c)
...

The inequality 0 < |x - c| that appears in the definition means “x is not c”
...
The two inequalities
combined as a single statement 0 < |x - c| <  describes the open interval
( c - , c + ) from which c is deleted
...

The formal definition of a limit requires that for every open interval about L,
34

(L - , L + ), no matter how small, we can find an interval ( c - , c +) about c, of
whose function values lie within that interval about L
...
No matter how small  is chosen,  can
be made small enough so that f(x) lies within a distance  from L
...

y

L+
 L
L-

x
0

c- cc+

Definition 3
...


Definition 3
...

Examples:
1
...

Now,
|(3x-1)-5|
= |3x-1-5|
= |3x- 6| = |3(x- 2)|
= |3 ||x-2| = 3 |x-2|

35

|(3x-1)-5|
= 3 |x-2| <  if
|x-2| < /3
Thus we can take  = /3
...
Any smaller positive  will also work
...

(iii) The value of  depends on the value chosen for 
...
It does not
mean the two statements are equal, and should not be used in the place
of “=”
...


Show that lim x  4  1
x -3

Solution:
Given any  > 0, we must find a  > 0 such that if 0 < |x+3| < , then
| |x+4|-1| < 
...
From the properties of absolute values it follows that
|x+4| - |1| ≤ | |x+4|-1| < 
Hence
|x+4| - |1| < 
Or
|x+4| <  + 1
From the properties of absolute values follow:
- ( + 1) < x + 4 < 

-  -1
< x+4 < 

-  -2
< x+3 < 

-  -2
< x+3 < 

-  -2
< x+3 < 

|x + 3| <  + 2
Therefore we can choose  =  + 2
3
...

Now, we know (x3 -1) = (x-1)(x2+x+1)
Hence, if
|x3 - 1| < 

|x-1||x2+x+1| < 
(i)
We observe that the expression |x-1| we are looking for appears on the left hand
side, but we also have the expression |x2+x+1| for which we must find a
replacement
...

If
|x-1| < 1
Then
-1
0
-2 < 0 < x < 2

-2 < x < 2

|x| < 2
Therefore
|x2+x+1| ≤ |x2 | + |x |+ |1| < 22 + 2 + 1 = 7
From expression (i) follows that |x-1||x2+x+1| < 7 |x-1| < 

|x-1| < 
 = min{1, 

Therefore

7

7

}

EXERCISES
Use the formal definition of a limit to show that
lim (x  6)  5
1
...


x 5

3
...


x 1

lim (x 2  2x  1)  2

5
...


x 2

lim 2x  2  6

7
...


x 5

lim 1  5x  4

9
...


x 3

2 x 3  5x 2  2 x  5
lim
7
x 1
x2 1

11
...

x0

37

THEOREMS ON LIMITS

Theorem 3
...

x c

x c

Proof:
We use a method of proof called “Proof by contradiction”
...

Assume M ≠ L then

Since

LM
2

> 0
...


2

lim f(x)  M , for each  =

LM

x c

if 0 < |x-c| < 2 , then |f(x) - M| <  =

LM
2

2

> 0 there exists a 2 > 0 such that


...
It follows that for each   0 , there exists   0 such that
|L - M|

= |[L – f(x)] + [f(x) – M]|
≤

|-[f(x) – L]|+|f(x)-M|

=

|f(x) - L|+|f(x)-M|

<
=

LM
2

+

LM
2

|L - M|

The conclusion |L - M| < |L - M| is a contradiction
...


Remark:
Since the definition states “………for each  > 0 there exists a  > 0…
...


Theorem 3
...
For each


> 0 there exists a  > 0 such that if 0 < |x - c| <  , then
a
|f(x) - L| <


a

a



|a||f(x) - L| < |a|



| |a|f(x) - |a|L| < 



| af(x) - aL| < 

Therefore
lim[a f(x)] =

a lim f(x) = aL
x c

xc

Theorem 3
...

2

39

Also since lim g(x)  L 2 , then for each
x c

0 < |x - c| < 2, implies |g(x) – L2| <


> 0 there exists 2 > 0 such that
2



...
For each  > 0, there exists  > 0 such that
0 < |x - c| < 



|f(x)+g(x) - L1 - L2|
=

|(f(x)- L1 ) + (g(x) - L2 )|

≤

|f(x) – L1|+|g(x) – L2|

<



+
2
2

=



Therefore,
lim[f(x)  g(x)]

= lim f(x) + lim g(x)

x c

xc

=

xc

L1 + L2

Theorem 3
...
g(x) ]
xc

= lim f(x)
...
L2

Proof
We must show that for each  > 0 there exists a  > 0 such that if
0 < |x - c| <  then |f(x)
...
L2)| < 
Since lim f(x)  L1 , for each 1 > 0 there exists a 1 > 0 such that
x c

0 < |x - c| < 1

 |f(x) – L1| < 1

But,
|f(x)| - |L1|

≤ |f(x) – L1 │ < 1

Hence,
|f(x)|

< 1 + |L1 |……………………………………………………
...
g(x) - (L1
...
lim g(x)

lim[f(x)
...
L2

Theorem 3
...


1
1

x c g(x)
L2

We first show that lim

We must show that for a given  > 0, there exists  > 0 such that when
0 < |x - c|<  , then

1
1

g(x) L 2

 

41

But,

1
1

g(x) L 2

L 2  g(x)
L 2 g(x)

=

<

 …………………………
...
½ |L2|

=

½ (L2)2

Taking reciprocals, we obtain

1
2
 2
L 2 g(x)
L2
Similarly, there is a 2 such that if 0 < |x - c| < 2 , then from (A)
| L2 – g(x)| <  |L2 g(x)| < 
...
Then for 0 < |x - c| < ,

L 2  g(x)
L 2 g(x)

=

1

...

L22 2

= 

1
1

...
lim
x
c
xc g(x)
xc g(x)

lim

42

=

L1
...

It is used to confirm the limit of a function via comparison with two other functions, of
which the limit is either known or can be easily computed
...
9 (The Pinching Theorem)
(Also known as the Squeeze Theorem or the Sandwich Theorem)
Let p > 0
...

Since lim h(x)  L , for each  > 0, there exists 1 > 0 such that if 0 < |x - c| < 1
x c

then | h(x) – L | < 
...


Therefore
-  < g(x) - L < 


L -  < g(x) < L + 

Let  = min{p, 1 ,2 }
...

x c

43

y

g
f

L

h
x

c

INFINITE LIMITS AND LIMITS AT INFINITY
Consider the function defined by f(x) =

1
; x ≠ 0 and given by the following graph and
x2

table:
Fig
...
We say that as x → 0, f(x) becomes infinite and, adopting the limit
notation for convenience, we write
1
lim 2  
x 0 x
On the other hand, look at the graph of f(x) =

44

1
, x ≠ 0
...
B
y

y=

1
,x≠0
x

x
0,01
0,001
0,0001
-0,01
-0,001
-0,0001

x

As x approaches zero from the right,

f(x)
100
1000
10000
-100
-1000
-10000

1
becomes positively infinite
...
Symbolically we write
x
1
1
and
lim  
lim  
x 0 x
x 0 x
1
Now let us again examine the function f(x) =
, x ≠ 0, as x becomes infinite in both a
x
positive and negative sense
...
001

...
00001

...
001
-
...
00001
-
...

Like-wise, as x decreases without bound through negative values, the corresponding
values of f(x) also approach zero
...
For example consider

45

x 2  3x  5
x  x 2  x  1
It is clear that as x →∞, both numerator and denominator become infinite
...
Since x→ ∞, we are concerned only with those values of x which are
very large
...

A frequently used “gimmick” is to divide both the numerator and denominator by
the power of x which is the largest in either the numerator or denominator
...
Thus
3 5
x 2 (1   2 )
2
x  3x  5
x x
=
lim
lim 2
x 
x  x  x  1
1
1
x 2 (1   2 )
x x
1
1
lim 1  3 lim  5 lim 2
x 
x  x
x  x
=
1
1
lim 1  lim  lim 2
x 
x  x
x  x
1
1
As x→ ∞,
approaches 0
...

x


x
x
2
x  3x  5
1  3(0)  5(0)
1
Thus, lim 2
=
=
= 1
x  x  x  1
1 0  0
1
lim

Examples

8x  2x  3
x  2x 3  3x  1
2

1
...


x2 1  3 x2 1

lim

x  4

x4 1  5 x4 1

=

=

lim

x  4

lim

x 

x 2 (1 

1
x2

)  3 x 3 ( x1 

1
x3

)

x 4 (1 

1
x4

)  5 x 5 ( 1x 

1
x5

)

x 1

1
x2

x 3

1
x



1
x3

x 4 1

1
x4

x 5

1
x



1
x5

=

lim

=

1 0
1 0

x  4

1

1
x2

3

1
x



1
x3

1

1
x4

5

1
x



1
x5

= 1
...

lim 2
x  5x  3x
1
2
...

4
...

6
...


x 2 1

lim

x 

4x 2  x
x3 1
lim
x  (x  1)(x 2  x  1)

lim

x 

x

3

5x

2

2
3

2

1


lim  x  x 


x  1

lim 4x  x  1  2x
2

x 

2

x 

2

8
...

x    x  1 x 2  x  1

9
...

c) Prove that your guess is correct
...


Repeat question 9 for f ( x)  3x2  8x  6  3x2  3x  1

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Title: THE THEORY OF LIMITS
Description: DIFFERENT APPROACHES TO DEFFERENTIATING DIFFERENT LIMIT AND DERIVATIVES

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