2. functions, limit, and continuity

7/25/2019 2. Functions, Limit, And Continuity

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FUNCTIONS, LIMIT, ANDCONTINUITY OF COMPLEX

VARIABLESD r. I r . Ar m a n D j o h a n D i p o n eg o r o , M . S c

12th August


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TOPICS

Functions of a Complex Variable

Limits : Theorems on Limits

Continuity : Derivatives, Differentiation Formulas

Cauchy-Riemann Equations

Polar Coordinates;

Analytic Functions

Harmonic Functions


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FUNCTIONS OF COMPLEX VARIABLE

In the last chapter, although we saw a couple of functions with complex a

z, we spent most of our time talking about complex numbers.

Now we will introduce complex functions and begin to introduce conce

the study of calculus like limits and continuity.

Many important points in the first few chapters will be covered several t

dont worry if you dont understand everything right away.


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FUNCTIONS OF COMPLEX VARIABLE

We define a function of a complex variablew =f(z) as a rule that assigns

z C a complex numberw.

If the function is defined only over a restricted setS, thenw =f(z) assign

z S, the complex numberw and we callSthe domain of the function.

The value of a function atz=a is indicated by writingf(a).


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FUNCTIONS OF A COMPLEX VARIABL

Letsbe a set complex numbers. A functionfdefined onSis a rule that a

eachzinSa complex numberw.

S

Complex

numbersf

S

Complex

numbers

z

The domain of definition off The range off

w


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Suppose thatw =u +iv is the value of a functionfatz=x +iy, so

Thus each of real numberu andv depends on the real variablesx

meaning that

Similarly if the polar coordinatesrand , instead ofx andy, ar

we get

( )u iv f x iy

( ) ( , ) ( , )f z u x y iv x y

( ) ( , ) ( , )f z u r iv r


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Example 1 :

Iff(z)=z2, then

case #1:

case #2:

2 2 2( ) ( ) 2f z x iy x y i x 2 2

( , ) ; ( , ) u x y x y v x y

z x iy

iz re

2 2 2 2 ( ) ( ) cos 2 i if z re r e r

2 ( , ) cos 2 ; ( , )u r r v r

When v = 0,f is a real-valued function


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FUNCTIONS OF A COMPLEX VARIABLE

Example 2

A real-valued function is used to illustrate some important concep

chapter is

Polynomial function

where n is zero or a positive integer and a0, a1, an are complex cons

Rational function

the quotientsP(z)/Q(z) of polynomials

2 2 2( ) | | 0f z z x y i

2

0 1 2( ) ... n

nP z a a z a z a z

The domain of definition

The domain of definition


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Multiple-valued functionA generalization of the concept of function is a rule that assi

than one value to a pointzin the domain of definition.

f

S

Complex

numbers

z

S

Complex

numbers

w1

w2

wn


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Example 3 :

Letzdenote any nonzero complex number, thenz1/2 has th

If we just choose only the positive value of

1/2 exp( )2

z r i

Multiple-valued function

r

1/2 exp( ), 02

z r i r

Single-valued function


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LIMITS

For a given positive value , there exists a positive value (depensuch that when

0 < |z-z0| < , we have |f(z)-w0| <

meaning the point w =f(z) can be made arbitrarily chose to w0 if w

the pointzclose enough toz0but distinct from it.

00lim ( )

z zf z w


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LIMITS

The uniqueness of limitIf a limit of a function f(z) exists at a pointz0, it is unique.

Proof: suppose that

then

when

Let , when 0


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LIMITS Example 1

Show that in the open disk |z|


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LIMITS

Example 2If then the limit does not exist.( )

zf z

z

0lim ( )z

f z

0

0lim 1

0x

x i

x i

( ,0)z x

(0, )z y0

0lim 1

0y

iy

iy


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THEOREMS ON LIMITS Theorem 1

Let

and

then

if and only if

00lim ( )

z zf z w

0 00

( , ) ( , )lim ( , )

x y x y

u x y u

0 00

( , ) ( , )lim ( , )

x y x yv x y v

( ) ( , ) ( , )f z u x y iv x y z x iy

0 0 0 0 0 0;z x iy w u iv

and

(a)

(b)


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THEOREMS ON LIMITS

Theorem 2Let and

then

00lim ( )

z zf z w

00lim ( )

z z

F z W

00 0lim[ ( ) ( )]

z zf z F z w W

00 0lim[ ( ) ( )]

z zf z F z w W

0

00

0

( )lim[ ] , 0

( )z z

wf zW

F z W


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THEOREMS ON LIMITS

( ) ( , ) ( , ), ( ) ( , ) ( , )f z u x y iv x y F z U x y iV x y

0 0 0 0 0 0 0 0 0; ;z x iy w u iv W U iV

( ) ( ) ( ) ( )f z F z uU vV i vU uV

Let

00lim ( )

z zf z w

00lim ( )

z z

F z W

When (x,y)(x0,y0);

u(x,y)u0; v(x,y)v0; & U(x,y)U0; V(x,y)V0;

0 0 0 0( )u U v V Re(f(z)F(z)):

0 0 0 0( )v U u V Im(f(z)F(z)):w0W

0

0 0lim[ ( ) ( )]z z

f z F z w W

0 0

lim ( )z z

f z w

0 0

lim ( )z z

F z W

&


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CONTINUITY

ContinuityA function is continuous at a point z0 if

meaning that

1. the function f has a limit at point z0 and2. the limit is equal to the value of f(z0)

00lim ( ) ( )

z zf z f z

For a given positive number , there exists a positive number , s.t.

0| |z z When 0| ( ) ( ) |f z f z

00 | | ?z z


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CONTINUITY

Theorem 1

A composition of continuous functions is itself continuous.

Suppose w=f(z) is a continuous at the point z0;

g=g(f(z)) is continuous at the point f(z0)

Then the composition g(f(z)) is continuous at the point z0


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CONTINUITY Theorem 2

If a function f (z) is continuous and nonzero at a point z0, then

throughout some neighborhood of that point.

Proof0

0lim ( ) ( ) 0z z

f z f z

0| ( ) | 0, 0, . .2

f zs t

0| |z z When

00

| ( ) || ( ) ( ) |

2

f zf z f z

f(z0)

f(z)

If f(z)=0, then0

0

| ( ) || ( ) |

2

f zf z

Contradiction!

0| ( ) |f z

Why?


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CONTINUITY

Theorem 3

If a function f is continuous throughout a region R that is

both closed and bounded, there exists a nonnegative real

number M such that

where equality holds for at least one such z.

| ( ) |f z M for all points z in R

2 2| ( ) | ( , ) ( , )f z u x y v x y Note:

where u(x,y) and v(x,y) are continuous real functions


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DERIVATIVES

DerivativeLet f be a function whose domain of definition contains a ne

|z-z0|


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DERIVATIVES Illustration of Derivative

0 00

0

( ) ( )'( ) lim

z

f z z f zf z

z

0

limz

dw w

dz z

0 0( ) ( )w f z z f z

0

00

0

( ) ( )'( ) lim

z z

f z f zf z

z z

O u

v

f(z0)

f(z0+z)

w

0z z z

Any position


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DERIVATIVES

Example 1

Suppose that f(z)=z2. At any point z

since 2z + z is a polynomial in z. Hence dw/dz=2z or f(z)=2z.

2 2

0 0 0

( )lim lim lim(2 ) 2

z z z

w z z z

z z z

z z


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DERIVATIVES Example 2

If f(z)=z, then w z z z z z z z z z z z

Case #1:x0,y=0

( , ) (0, 0)z x y In any direction

O x

y

Case #1

Case #2

0

0lim 1

0x

z x i

z x i

Case #2:x=0,y0

0

0lim 1

0x

z i y

z i y

Since the limit is unique, this function does not exist anywhere


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DERIVATIVES Example 3

Consider the real-valued function f(z)=|z|2. Here

2 2| | | | ( )( )w z z z z z z z z z z z z z

z z z z

Case #1: x0, y=0

0 0

0lim( ) lim( )0x x

z x iz z z z x z z zz x i

Case #2: x=0, y0

0 0

0lim ( ) lim ( )

0y y

z i yz z z z i y z z z

z i y

0z z z z z dw/dz can not exist when z is not 0


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DERIVATIVES

Continuity & DerivativeContinuity Derivative

Derivative Continuity

For instance,

f(z)=|z|2 is continuous at each point, however, dw/dz does not exists when z is not 0

0 0 0

00 0 0

0

( ) ( )lim[ ( ) ( )] lim lim( ) '( )0 0

z z z z z z

f z f zf z f z z z f z

z z

Note: The existence of the derivative of a function at a point implies the continuity

of the function at that point.


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DIFFERENTIATION FORMULAS Differentiation Formulas

[ ( ) ( )] '( ) '( )d

f z g z f z g zdz

[ ( ) ( )] ( ) '( ) '( ) ( )d

f z g z f z g z f z g zdz

2

( ) '( ) ( ) ( ) '( )[ ]

( ) [ ( )]

d f z f z g z f z g z

dz g z g z

1[ ]n nd

z nzdz

0; 1; [ ( )] '( )d d d

c z cf z cf z dz dz dz

( ) ( ( ))F z g f z

0 0 0'( ) '( ( )) '( )F z g f z f z

dW dW dw

dz dw dz


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DIFFERENTIATION FORMULAS

Example

To find the derivative of (2z2+i)5, write w=2z2+i and W

Then

2 5 4 2 4 2 4

(2 ) (5 ) ' 5(2 ) (4 ) 20 (2 )

dz i w w z i z z z i

dz


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OPERATOR DEL


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COMPLEX DIFFERENTIAL OPERATOR


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GRADIENT


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DIVERGENCE


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CURL


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LAPLACIAN


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CAUCHY-RIEMANN EQUATION

Theorem

Suppose that

and that f(z) exists at a point z0=x0+iy0. Then the first-order

partial derivatives of u and v must exist at (x0,y0), and they

must satisfy the Cauchy-Riemann equations

then we have

( ) ( , ) ( , )f z u x y iv x y

;x y y x

u v u v

0 0 0 0 0'( ) ( , ) ( , )x xf z u x y iv x y


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CAUCHY-RIEMANN EQUATION Proof:

Let 0 0 0;z x iy z x i y

0 0

0 0 0 0 0 0 0 0

( ) ( )

[ ( , ) ( , )] [ ( , ) ( , )]

w f z z f z

u x x y y u x y i v x x y y v x y

0 0

0 0 0 0 0 0 0 0

( , ) (0,0)

'( ) lim

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

z

x y

wf z

zu x x y y u x y i v x x y y v x y

x i y

Note that (x,y) can be tend to (0,0) in any manner .

Consider the horizontally and vertically directions


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CAUCHY-RIEMANN EQUATION Horizontally direction (y=0)

Vertically direction (x=0)

0 0 0 0 0 0 0 00

0

( , ) ( , ) [ ( , ) ( , )]'( ) lim0x

u x x y y u x y i v x x y y v x yf zx i

0 0 0 0 0 0 0 0

0

( , ) ( , ) [ ( , ) ( , )]limx

u x x y u x y i v x x y v x y

x

0 0 0 0( , ) ( , )x xu x y iv x y

0 0 0 0 0 0 0 00

0

( , ) ( , ) [ ( , ) ( , )]'( ) lim

0y

u x y y u x y i v x y y v x yf z

i y

2

0 0 0 0 0 0 0 0

0

{[ ( , ) ( , )] [ ( , ) ( , )]}lim

( )y

i u x y y u x y i v x y y v x y

i i y

0 0 0 0( , ) ( , )y yv x y iu x y

;x y y xu v u v

Cauchy-Riemann equations


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Example 1

is differentiable everywhere and that f(z)=2z. To verify that

the Cauchy-Riemann equations are satisfied everywhere,

write

2 2 2( ) 2f z z x y i xy

2 2

( , )u x y x y ( , ) 2v x y xy

2x y

u x v 2y x

u y v

'( ) 2 2 2( ) 2f z x i y x iy z


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Example 2

2( ) | |f z z

2 2( , )u x y x y ( , ) 0v x y

If the C-R equations are to hold at a point (x,y), then

0x y y x

u u v v

0x y

Therefore, f(z) does not exist at any nonzero point.


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POLAR COORDINATES

Assuming that z0

0

cos , sinx r y r

u u x u y

r x r y r

u u x u y

x y

cos sinr x yu u u

sin cosx yu u r u r

Similarly cos sinr x y

v v v sin cosx y

v v r v r

If the partial derivatives of u and v with respect to x and y satisfy the Cauchy-Riemann equatio

;x y y xu v u v

;r rru v u rv


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POLAR COORDINATES

TheoremLet the function f(z)=u(r,)+iv(r,) be defined througho

neighborhood of a nonzero point z0= r0exp(i0) and suppose tha

(a)the first-order partial derivatives of the functions u and v with resp

exist everywhere in the neighborhood;

(b)those partial derivatives are continuous at (r0, 0) and satisfy the prur= v, u = rvrof the Cauchy-Riemann equations at (r0, 0)

Then f(z0) exists, its value being 0 0 0 0 0'( ) ( ( , ) ( , ))i

r rf z e u r iv r


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POLAR COORDINATES Example 1

Consider the function1 1 1 1

( ) (cos sin ), 0ii

f z e i zz re r r

cos sin( , ) , ( , )u r v r

r r

Then

cos sin&

r rru v u rv

r r

2 2 2 2 2

cos sin 1 1'( ) ( )

( )

ii i

i

ef z e i e

r r r re Z


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ANALYTICITY AT A POINT

A function is analytic at every pointzis said to be an entire function.

Polynomial functions are entire functions.

A complex function w =f(z) is said to be analytic at

a pointz0 if f is differentiable atz0 and at every point

in some neighborhood ofz0.


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CRITERION FOR ANALYTICITY

Suppose the real-valued function u(x,y) and v(x,y) are

continuous and have continuous first-order partial

derivatives in a domainD. If u and v satisfy the

Cauchy-Riemann equations at all points ofD, then thecomplex functionf(z) = u(x,y) + iv(x,y) is analytic inD.


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ANALYTIC FUNCTION Analytic at a point z0

A function f of the complex variable z is analytic at a point zderivative at each point in some neighborhood of z0.

Analytic function

A function f is analytic in an open set if it has a derivative e

that set.

Note that iffis analytic at a point z0, it must be analytic at each point

neighborhood of z0

Note that iff is analytic in a set S which is not open, it is to be under

analytic in an open set containing S.


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ANALYTIC FUNCTION

Analytic vs. Derivative For a point

Analytic Derivative

DerivativeAnalytic

For all points in an open set

Analytic Derivative

DerivativeAnalytic

f is analytic in an open set D if f is derivative in D


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ANALYTIC FUNCTION

Singular point (singularity)

If functionf fails to be analytic at a point z0 but is analytic a

point in every neighborhood of z0, then z0 is called a singular p

For instance, the function f(z)=1/z is analytic at every point in the fini

except for the point of (0,0). Thus (0,0) is the singular point of funct

Entire Function

An entire function is a function that is analytic at each point

entire finite plane.

For instance, the polynomial is entire function.


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ANALYTIC FUNCTION Property 1

If two functions are analytic in a domain D, then their sum and product are both analytic in D

their quotient is analytic in D provided the function in the

denominator does not vanish at any point in D

Property 2From the chain rule for the derivative of a composite function, a

composition of two analytic functions is analytic.

( ( )) '[ ( )] '( )d

g f z g f z f zdz


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ANALYTIC FUNCTION

Theorem

If f (z) = 0 everywhere in a domain D, then f (z) must be

constant throughout D.

( )du

gradu Uds

x=u ygradu i u j

0 & 0x y x y

u u v v

'( ) 0x x y yf z u iv v iu


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EXAMPLES

Example 1The quotient

is analytic throughout the z plane except for the singular

points

3

2 2

4( )

( 3)( 1)

zf z

z z

3&z z i


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EXAMPLES

Example 3

Suppose that a function and its

conjugate are both analytic in a

given domain D. Show that f(z) must be constant

throughout D.

( ) ( , ) ( , )f z u x y iv x y

( ) ( , ) ( , )f z u x y iv x y

( ) ( , ) ( , )f z u x y iv x y is analytic, then

,x y y x

u v u v ( ) ( , ) ( , )f z u x y iv x y is analytic, thenProof:

,x y y x

u v u v

0, 0x xu v '( ) 0x xf z u iv

Based on the Theorem in pp. 74, we have that f is constant through


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EXAMPLES Example 4

Suppose that f is analytic throughout a given region D, andthe modulus |f(z)| is constant throughout D, then the

function f(z) must be constant there too.

Proof:

|f(z)| = c, for all z in D

where c is real constant.

If c=0, then f(z)=0 everywhere in D.

If c 0, we have 2( ) ( )f z f z c2

( ) , ( ) 0( )

cf z f z inD

f z

Both f and it conjugate are analytic, thus fmust be constant in D. (Refer to Ex. 3)


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HARMONIC FUNCTIONS

A Harmonic Function

A real-valuedfunctionHof two real variablesx and y is said to

a given domain of the xy plane if, throughout that domain, it has

partial derivatives of the first and second order and satisfies the p

differential equation

Known as Laplaces equation.

( , ) ( , ) 0xx yyH x y H x y


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HARMONIC FUNCTIONS

Theorem 1

If a function f (z) = u(x, y) + iv(x, y) is analytic in a domain

D, then its component functions u and v are harmonic in D.

Proof: ( ) ( , ) ( , )f z u x y iv x y is analytic in D

&x y y xu v u v

Differentiating both sizes of these equations with respect to x and y respectively, we have

a function is analytic at a point, then its real and imaginary components

have continuous partial derivatives of all order at that point

continuity

0 & 0xx yy xx yy

u u v v

&xx yx yx xx

u v u v

&xy yy yy xy

u v u v

&xx xy xy xxu v u v

&xy yy yy xy

u v u v


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HARMONIC FUNCTIONS

Example 3

The function f(z)=i/z2 is analytic whenever z0 and since

The two functions

2 2 2

2 2 2 22 2

( ) 2 ( )

( )( )

i i z xy i x y

z x yz z

2 2 2

2( , )

( )

xyu x y

x y

2 2

2 2 2( , )

( )

x yv x y

x y

are harmonic throughout any domain in the xy plane that does not contain the origin.


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HARMONIC FUNCTIONS

Harmonic conjugate

If twogiven function u and v are harmonic in a domain D

and their first-order partial derivatives satisfy the Cauchy-

Riemann equation throughout D, then v is said to be a

harmonic conjugate of u.

If u is a harmonic conjugate of v, then

&x y y x

u v u v

Is the definition symmetry for u and v?

Cauchy-Riemann equation &x y y x

u v u v


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HARMONIC FUNCTIONS Theorem 2

A function f (z) = u(x, y) + iv(x, y) is analytic in a domain D ifand only if v is a harmonic conjugate of u.

Example 4

The function is entire function, and its real andimaginary components are

Based on the Theorem 2, v is a harmonic conjugate of u

throughout the plane. However, u is not the harmonic

conjugate of v, since is not an analytic

function.

2 2( , ) & ( , ) 2u x y x y v x y xy

2

( )f z z

2 2( ) 2 ( )g z xy i x y


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HARMONIC FUNCTIONS Example 5

Obtain a harmonic conjugate of a given function.

Suppose that v is the harmonic conjugate of the given function

Then

3 2( , ) 3u x y y x y

&x y y x

u v u v

6x yu xy v

2 23 3

y xu y x v

2

3 ( )v xy x

2 2 23 3 ( 3 '( ))y x y x

2 3'( ) 3 ( )=x x x x C

2 33v xy x C


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HOMEWORK

1.

2.

3.

4.


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HOMEWORK (CONTINUED)

5.

Please do the homework on a paper. This exercise should be submitted on Thur19th 2013 before the class begins.

2. functions, limit, and continuity

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