02ns-limitcontinuity

8/19/2019 02ns-limitContinuity

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SRI / Calculus / Multivariable Functions / Limits & Continuity 1

LIMITS

For a function of one variable – two one-sided limits at , namely0 x

)(lim0

x f x x −→ and )(lim0 x f

x x +→

reflecting the fact that there are only two directions (the right and left)

For function of two variables – the statement means that the point is allowed to

approach from any ‘direction’.

),(),( 00 y x y x → ),( y x

),( 00 y x

y

x

z

DEFINITION

If C is a smooth parametric curve in 3-space that is represented by the equations

)(t x x = , )(t y y =

and if )(),( 0000 t y yt x x == , then the limits are defined by

))(),((lim),(lim0)0,0(),(

t yt x f y x f t t

C along

y x y x →→=

y

x

z

C

))(),(( t yt x

)))(),((),(),(( t yt x f t yt x

L

),( 00 y x

As the point ( x(t ), y(t )) moves along the curve in the xy-plane toward .),( 00 y x


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The point ( x(t ), y(t ), f ( x(t ), y(t ))) moves directly above it along the graph of

with approaching the limiting value L.

),( y x f z =

))(),(( t yt x

If the value of

),(lim)0,0(),(

y x f y x y x →

is not the same for all possible approaches, or paths to , the limit does

not exist.

),( 00 y x

DEFINITION (GENERAL)

Let f be a function of two variables defined, except possibly at on an open disc

centered at and let L be a real number. Then

),( 00 y x

),( 00 y x L y x f

y x y x=

→),(lim

)0,0(),(

If for each 0>ε there corresponds a 0>δ that ε


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ALTERNATIVE NOTATION

L y x f y x y x

=→

),(lim)0,0(),(

also be written as

L y x f →),( as ),(),( 00 y x y x →

PROPERTIES OF LIMITS OF FUNCTIONS OF TWO VARIABLES

The following rules hold if

L y x f y x y x

=→

),(lim)0,0(),(

and M y xg y x y x

=→

),(lim)0,0(),(

1. ScalarMultiple

kL y x f k =),(lim

(k -constant)

2. Sum orDifference

[ ]

M L

y xg y x f

y xg y x f

±=

±=

±

),(lim),(lim

),(),(lim

3. Product [ ][ ] [

LM

y xg y x f

y xg y x f

=

= ),(lim),(lim

),(),(lim

]

4. Quotient M

L

y xg

y x f

y xg

y x f ==

),(lim

),(lim

),(

),(lim ; 0≠ M

The limit of quotient is the quotient of thelimit provided the limit of denominator isnonzero.

5. Power [ ]

[ ]n

n

n

L

y x f

y x f

== ),(lim

),(lim

6. Root

n

n

n

L

y x f

y x f

=

= ),(lim

),(lim

provided if n is even0≥ L


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Limit of multivariable functions is more complicated to be determined than limitfor the functions of single variable.

Step To Find The Limit

1st attempt :

- Determine whether defined or not.),( ba f

- If not, try 2nd attempt.

2nd attempt :

- Simplify f (a, b) (if not applicable, try 3rd attempt), then check againwhether f (a, b) defined or not.

3rd attempt :

- Show that the limit of the function f (a, b) does not exists.- Choose any two curves (or lines) which pass through the point (a, b).- Show the values of the limit are not the same from those directions.- Conclude that the limit for that function does not exits.

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Example 1

Evaluate .)3312(lim25

)2,1(),( +−→ y x x y x Solution

3lim3lim12lim)3312(lim)2,1(),(

25

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

25

)2,1(),( →→→→+−=+−

y x y x y x y x y x x y x x

3

3)2)(1(3)1(12 25

=

+−=

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Example 2

Evaluate 22)4,3(),(

lim y x y x

+−→

.

Solution

22

2

)4,3(),(

2

)4,3(),(

22

)4,3(),(

)4(3

limlimlim

−+=

+=+−→−→−→

y x y x y x y x y x

525

==


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Example 3

The function is a polynomial, so find its limit at

.

5472),( 3224 −+−= y x xy y x y x f

)2,1(−

Solution

5472lim 3224)2,1(),(

−+−−→

y x xy y x y x

49

532148

5)2()1(4)2)(1(7)2()1(2

5lim4lim7lim2lim

3224

)2,1(),(

32

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

24

)2,1(),(

=

−++=

−−+−−−=

−+−=−→−→−→−→ y x y x y x y x

y x xy y x

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Example 4

Find y x

xy x

y x −

−→

2

)0,0(),(lim .

Solution

1st attempt :

0

0lim

2

)0,0(),(=

−

−→ y x

xy x

y x (indeterminate form)

2nd attempt :

y x

xy x

y x −

−→

2

)0,0(),(lim

⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

+

+⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−

−=

→ y x

y x

y x

xy x

y x

2

)0,0(),(lim

0

)00(0

)(lim

))((lim

)0,0(),(

)0,0(),(

=

+=

+=

−+−=

→

→

y x x

y x

y x y x x

y x

y x


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Example 5

Let22

),( y x

xy y x f

+= .

Find the limit of as along),( y x f )0,0(),( → y x

(a) the x-axis (c) the parabola 2 x y =

(b) the y-axis (d) the line x y =

Solution

(a) 1st method

the x-axis has parametric equations ,t x = 0= y with corresponding to)0,0( 0=t

So,)0,(lim),(lim

0)0,0(),(t f y x f

t y x →→=

0

0)0(lim

20

=

+=

→ t

t t

2nd method

along x-axis )0( = y

00

)0(lim),(lim

22)0,0(),()0,0(),(=

+=

→→ x

x y x f

y x y x

(b) along y-axis )0( = x

22)0,0(),(22)0,0(),( 0

)0(limlim

y

y

y x

xy

y x y x +=

+ →→0=

(c) along 2 x y =

222

2

)0,0(),(22)0,0(),( )(limlim

x x

xx

y x

xy

y x y x +=

+ →→

0

1lim 2)0,0(),(

=+= → x

x

y x

(d) along x y =

22)0,0(),(22)0,0(),(limlim

x x

xx

y x

xy

y x y x +=

+ →→

21

2lim

2

2

)0,0(),(

=

=→ x

x

y x


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Example 6

Show that22

22

)0,0(),(lim

y x

y x

y x +

−→

does not exist.

Solution

1st attempt :

0

0lim

22

22

)0,0(),(=

+

−→ y x

y x


3rd attempt :

along x-axis ;)0( = y0

0limlim

2

2

)0,0(),(22

22

)0,0(),( +

−=

+

−→→ x

x

y x

y x

y x y x1=

along y-axis ;)0( = x 2

2

)0,0(),(22

22

)0,0(),( 0

0limlim

y

y

y x

y x

y x y x +

−=

+

−→→

1−=

Since has two different limits along two different lines, therefore the limit does

not exist.

),( y x f

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Example 7

If42

2

),( y x

xy y x f

+= , does exist?),(lim

)0,0(),( y x f

y x →

Solution

1st attempt :

0

0lim

42

2

)0,0(),(=

+→ y x

xy


3

rd

attempt :

along x-axis ;)0( = y 00

)0(limlim

42

2

)0,0(),(42

2

)0,0(),(=

+=

+ →→ x

x

y x

xy

y x y x

along y-axis ;)0( = x42

2

)0,0(),(42

2

)0,0(),( 0

)0(limlim

y

y

y x

xy

y x y x +=

+ →→0=

along parabola ;2 y x =422

22

)0,0(),(42

2

)0,0(),( )(limlim

y y

y y

y x

xy

y x y x +=

+ →→ 2

1=

Then the limit does not exist


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CONTINUITY

The intuitive meaning of continuity is that if the point ( x, y) changes by a smallamount, then the value of f ( x, y) changes by a small amount.

This means that a surface that is the graph of continuous function has no hole or break.

Definition

A function f ( x, y) is continuous at if),( 00 y x

1. exists,),(lim)0,0(),(

y x f y x y x →

2. f is defined at ),( 00 y x

3. ),(),(lim 00

)0,0(),( y x f y x f

y x y x=

→

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Example 1

Where is the function22

22

2

2),(

y x

y x y x f

+

−= discontinuous?

Solution

1st attempt :

0

0

2

2lim

22

22

)0,0(),(=

+

−→ y x

y x


3rd attempt :

along x-axis ;)0( = y02

)0(2lim

2

2lim

2

2

)0,0(),(22

22

)0,0(),( +

−=

+

−→→ x

x

y x

y x

y x y x 2

1=

along y-axis ;)0( = x 22

)0,0(),(22

22

)0,0(),( )0(220lim

22lim

y y

y x y x

y x y x +−=

+−

→→2−=

1. Since f ( x, y) has two different limits along two different lines, therefore

does not exists.),(lim)0,0(),(

y x f y x y x →

2. is undefined.)0,0( f

Then the function is discontinuous at (0, 0) because it is not defined there.),( y x f


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Example 2

Let

⎪⎩

⎪⎨

⎧

=

≠−

−=

)0,4(),(0

)0,4(),(2

4),(

y x

y x x

y xy

y xg

Show that g( x, y) continuous at (4, 0).

Solution

1st attempt :

0

0

2

4lim

)0,4(),(=

−

−→ x

y xy


2nd attempt :

2

2

2

4lim

2

4lim

)0,4(),()0,4(),( +

+⋅

−

−=

−

−→→ x

x

x

y xy

x

y xy

y x y x

0

)2(lim

4

)2)(4(lim

)0,4(),(

)0,4(),(

=

+=

−

+−=

→

→

x y

x

x x y

y x

y x

1. Therefore, exists.),(lim)0,4(),(

y xg y x →

2. 0)0,4( =g

3. )0,4(0),(lim

)0,4(),(g y xg

y x==

→

Then, the function g( x, y) is continuous at (4, 0).

02ns-limitcontinuity

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