Calculus I-Limit of function, Slides of Advanced Calculus

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1.3 Limit of Function(1)

1.3.1 The concept of limit of a function

1.3.2 The properties and operation

rules of functional limits

1.3.3 Two important limits

1.3.1 Limit of function

1. Limit of a function for x tending to infinity

(1) Preface

(2) Limit of functions

lim n n

a A  

 

f ( ) x  A x ,

0, ,such that as long as. n

 N N a A  n N 

      



 



f ( ) x  A x ,

0

x



0

x



0

x

f ( ) x  , x



 



f ( ) x  , x

0

x



0

x



0

x

f ( ) x   , x



 



f ( ) x   , x

0

x



0

x



0

x

   0,

1 1 0 ,

| |

n n x x

   

And we need only

Proof

We require

1 | | ,

n x





1 1 | | ( ).

n or x





By taking we have

1 1 ( ) ,

n X





1 0 for all | |. n

x X

x

   

Hence

1

lim 0. n x   x



(1) Neighbourhood

0

2. Limit of a function for x tending to a finite value x

1) A neighbourhood with centre and radius

0

x^ ^ : 0

U x ( , )

0

x

0

x   0

x  

0 0

( x   , x  ) 0

 | x  x | 

2) A delected neighbourhood with centre and radius

0

x^  :

0

U x ( , )



0

x

0

x   0

x  

0 0 0 0

( x  , x )  ( x , x ) 0

 0 | x  x | 

3) Left neighbourhood with centre and radius

0

x (^)  :

0 0

( x  , x ) 0 0

 x    x  x

4) Right neighbourhood with centre and radius

0

x  :

0 0

( x , x  ) 0 0

 x  x  x  

5

, 0 1,

( ) ( )? as 1;

1,1 2,

x x

f x f x x

x x

^ ^ 

 (^)   

   

2

4

( ) 1? as 1;

x

x

f x x x

x

1

1

2

-

2

4 1 1 1

1

lim ( ) lim lim( 1) 2 x x (^) 1 x

x

f x x   (^) x 



   



1

1

2

3

2

5 1

lim ( )

x

f x



does not exist

( ) as lim ( ) x

f x A x f x A  

    

(2) Exploring the definition

Comparing

and

   0,  X  0,such that f ( ) x  A  , as long as x  X.

0

0

( ) as lim ( ) x x

f x A x x f x A 

   

0

  0,   0,such that f ( ) x  A , as long as 0< | x - x |< .

(3) Definition

0

  0,   0,such that f ( ) x  A , as long as 0< | x - x |< .

0

0

lim ( ) or ( ) as x x

f x A f x A x x 

   

A A  

A  

x 0   x 0 x 0  

(4) Geometric meaning

Prove that when x^ ^0 , the limit of the function

1 f ( ) x sin

x



0 0

0

lim ( ) any sequence { } ( ), lim ( ).

n

n n x x x x

f x A x U x f x A  

   



(8) Aggregation principle of functional limits

Theorem 1.3.1 (Hene’s theorem)

Example 1.3.

does not exist.

Proof By taking we have

(1) 1 , n

x

n 

 (^) (1) 0

lim ( ) lim sin 0; x n n

f x n  ^  

 

By taking we have

(2) 1 ,

2 / 2

x n

n  





( 2) (^0)

lim ( ) lim sin(2 / 2) 1. x n n

f x n   ^  

  

Therefore, does not exist.

0

1 lim sin x  x

(1) ( uniqueness ) The limit of f ( x ) must be unique as x  x 0 ;

 M  0 and  0, such that

0

lim ( ) , x x

f x A 



1.Theorem 1.3.2 ( Some properties of function limits )

Suppose that then

of

1.3.2 The properties and operation

rules of functional limits

(2) ( local boundedness ) f ( x ) is bounded in a deleted neighborhood

0

x. That is,

| f ( ) | x  M for all

x  U x ( 0 , );



(3) ( local preservation of sign ) if A > 0 ( < 0 ), then and a

constant q > 0, such that

   0

f ( ) x  q  0( f ( ) x  q  0) for all x such that

0

0 | x  x | .

f ( ) x   ( ) x  g x ( )  f ( xn )  ( xn )  g x ( n )

0

lim ( )

n

n x x

 x A



^ 

0 0

lim ( ) lim ( ) n

n x x x x

f x A f x A  

0 0

lim ( ) lim ( ) n

n x x x x

g x A g x A  

0

lim ( ) x x

 x A



2.Theorem 1.3.3 ( rational operation rules)

Suppose that then

0 0

lim ( ) , lim ( ) , x x x x

f x A g x B  

0 0 0

lim[ ( ) ( )] lim ( ) lim ( ); x x x x x x

f x g x A B f x g x   

0 0 0

lim[ ( ) ( )] lim ( ) lim ( ); x x x x x x

f x g x AB f x g x   

0

0

0

lim ( ) ( ) lim ,

( ) lim ( )

x x

x x

x x

f x f x A

g x B g x







provided B^ 0,^ g x ( )^^ 0 if^ x^ ^ U(^ x 0 ,^ ).



3.Corollary 1.3.

0 0 0

lim[ ( ) ( )] lim ( ) lim ( ); x x x x x x

af x bg x a f x b g x   

0 0

lim[ ( )] [lim ( )] ,.

n n

x x x x

f x f x n N   

If both exist, a and b are constants, then

0 0

lim ( ) lim ( ) x x x x

f x and g x  

4.Composition of operation rule)

Suppose that is the composition of the

function y = f (u) and u = g (x). And that the composition,

y  ( f  g )( ) x  f g x [ ( )]

f  g ,

defined in a deleted neighborhood of x 0^ ,^ U x (^0 ).



If

0 0

0

lim ( ) , lim ( ) , x x u u

g x u f u A  

And such that for all , then

0

   0, g x ( )^  u^0 0 0

x  U x ( ,  )



0 0

lim [ ( )] lim ( ), x x u u

f g x A f u  

When x > 0, since

Example 1.3.8 Find

It is an indeterminate form of type   .

Solution

2 1

lim( ). x  (^) x 1 x 1

1 2

lim( ) x  (^) x 1 x 1

1 2

lim x 1

x

 x

lim x  x 1

Example 1.3.9 Prove

Taking then

0

(1) lim 1;

x

x

e



0

0

(2) lim ( 0).

x x

x x

a a a



Proof (1)

(^0 0 )

lim 1 lim lim 1

x x x

x (^) x x

e e e

 ^   

x x

e   e   

x

 e     x  ln(  1)

  ln(  1), 0 | 1|.

x

 x    e    Thus

0

lim 1.

x

x

e  

When x < 0, let u = - x , then u > 0,

x u^1

u

e e

e

  

and u^^0 as x^0

    0 0

lim lim

x

u x u

e

e

   

0

lim

u

u

e  

(2) Since

x x 0 (^) x x 0 a a a



Proof^ 

0 0

0 0

( )ln

lim lim

x x^ x^ x^ a

x x x x

a a e



 

0

u ( x  x ) ln a ,

0 0 ln^ x x^ x a a e

  0 0

x ( x x )ln a a e

 

then u^ ^0 as x^  x 0 and

0

0

lim

x u

u

a e



x

 a

0

0

(2) lim ( 0).

x x

x x

a a a 

 

, let