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Advanced Mathematics. Teacher: Liubiyu. Chapter 1 Function and its graph. Chapter 2 Limits and Continuity. Chapter 1-2. Contents. §1.1 Sets and the real number. §1.2 Elementary functions and graph. §2.1 Limits of Sequence of number. §2.2 Limits of functions. - PowerPoint PPT Presentation
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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