Teacher: Liubiyu

Chapter 1-2

Contents

§1.2 Elementary functions and graph

§2.1 Limits of Sequence of number

§2.2 Limits of functions

§1.1 Sets and the real number

§2.3 The operation of limits

§2.4 The principle for existence of limits

§2.5 Two important limits

§2.6 Continuity of functions

§2.7 Infinitesimal and infinity quantity, the order for infinitesimals

§2.5 Two important limits §2.5 Two important limits

The following two important limits should be memorized. They will often be used to find limits of other more complicated functions.

0

sin1 lim 1

x

x

x

We now prove it precisely.

Proof Proof First, we suppose that 0 .

2Draw an unit circle. We may see that

x

o A C

D

B

1x

Area of the triangle OBC < Area of the sector OBC < Area of the triangle OCD (1)

Now, by the formula for these area.

sinArea of the triangle ;

2

xOBC

Area of the sector ;2

xOBC

tanArea of the triangle .

2

xOCD

So inequalities (1) take the form

2

tan

22

sin xxx

o A C

D

B

1x

Next we manipulate these inequalities to make the

sinmiddle term and yield the inequalities

x

x

1sin

cos x

xx

Thus, as 0,both cos and 1 approach 1. Hence

sinsqueezed between cos and 1 , must also

approach 1. Thus

x x

xx

x

1sin

lim0

x

xx

Example 1Example 1x

xx 2

3tanlim Find

0

SolutionSolutionlet 3

0 0 0

0

tan 3 tan 3 tanlim lim lim

22 23

3 sin 1 3lim

2 cos 2

u x

x u u

u

x u u

x uu

u

u u

Example 2Example 2 20 3

cos1lim Find

x

xx

SolutionSolution 22

2 20 0 0

2sin sin1 cos 1 12 2lim lim lim .3 3 6 6

2

x x x

x xx

xx x

Example 3Example 3x

xx

1arcsinlim Find

SolutionSolution1 1

Let arcsin , so that sin . There is no harm

in restricting since . Hence the variable2

can only approach 0 as . Thus,

t tx x

t x

t x

0 0

1 1lim arcsin lim lim 1.

sinsinx t t

tx

tx tt

Example 4Example 4xx

xxxx sin

cossin1lim Find

0

SolutionSolution

2

0 0

1 sin cos 1 sin coslim lim

sin sin ( 1 sin cos )x x

x x x x x x

x x x x x x x

1)sin

1(lim2

1

)cossin1

1

sin

sinsin(lim

0

2

0

x

xxxxxx

xxx

x

x

Example 5Example 5 xxx 2

tan)1(lim Find1

SolutionSolution

)1(2

cot)1(lim2

tan)1(lim11

xxxxxx

)1(2

sin

)1(2

cos)1(lim

1x

xx

x

1

2cos (1 ) (1 )2 2lim( )

sin (1 )2

2

x

x x

x

12 lim 1

x

xe

x

Proof Proof 1

We first prove that lim 1 .n

ne

n

1Firstly, we will prove the sequence 1 is

monotone increasing.

n

n

1Let 1 . By the binomial formula we know

n

nxn

2 3

1 ( 1) 1 ( 1)( 2) 11

2! 3!( 1)( 2) ( 1) 1

!

n

n

n n n n nx n

n n nn n n n n

n n

)1

1()2

1)(1

1(!

1

)2

1)(1

1(!3

1)

11(

!2

111

n

n

nnn

nnn

1

1 1 1 1 2so, 1 1 (1 ) (1 )(1 )

2! 1 3! 1 11 1 2 1

(1 )(1 ) (1 )! 1 1 1

nxn n n

n

n n n n

)1

1()1

21)(

1

11(

)!1(

1

n

n

nnn

1

1

1

Comparing the expressions of and , we can see

that has one more term than , and also that,

starting from the third term, each term of is bigger

than the corresponding term of . Henc

n n

n n

n

n

x x

x x

x

x

e, the sequence

is monotone increasing.nx

Second, we will prove boundedness above.

In the expression of changing all the factors in the

brackets of each term into 1 will make the expression

larger, and thus we have

nx

2 1

1 1 11 1 ...

2! 3! !1

11 1 1 21 1 ... 1 312 2 2 12

n

n

n

xn

so that { } is bounded above. Thus lim exists.n nn

x x

1The limit of the sequence 1 is denoted by ,

1thus, lim 1 .

n

n

n

en

en

It can be proved that is an irrational number, and

2.718281828459

e

e 1

We next prove that lim 1 .x

xe

x

1

Suppose that , then 1, so that

1 1 1 1 1 1

1

n x n

n x n x n

n x n

1

Since as and

11

1 1 lim 1 lim

11 11

n

n

n n

n x

ne

nn

11 1 1

lim 1 lim 1 1n n

n ne

n n n

An application of the squeeze theorem gives

1 lim 1 .

x

xe

x

1The proof of lim 1 lefts to you as exercise.

x

xe

x

NotationsNotations

11 The function 1 is of the form , which

is called a power exponent function. The charater of this

xg x

f xx

limit is that when , 1 and . This

kind of limit is called an indeterminate form of type 1 .

Hence, this limit may often be used for finding the limit

of a given powe-exponent function.

x f x g x

Example 7Example 7 x

x x3)

21(lim Find

SolutionSolution

thus, as , then,2

Let xtx

t

6663 e])1

1(lim[)1

1(lim)2

1(lim

t

t

t

t

x

x ttx

Example 8Example 8 12)1

2(lim Find

x

x x

x

SolutionSolution

2 1 2 1 2( 1) 12 1 1lim( ) lim(1 ) lim(1 )

1 1 1x x x

x x x

x

x x x

( 1) 2 1 21 1[lim(1 ) ] [lim(1 )] e

1 1x

x xx x

Example 9Example 9 aax

ax x

x find ,9)(limLet

SolutionSolution2

22 1lim( ) lim(1 ) [lim(1 ) ]

2

x a axx x a x a

x x x

x a ax ax a x a

a

2lim 2e e .

ax

x ax a 3ln,9e Thus, 2 aa