Function 10

8/6/2019 Function 10

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lim

xaf(x) The limit of f(x) as

x approaches a

limxa f(x) = k [f(x) approaches k]

f(x) k

[x approaches a]

x a

as


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Example : f(x) = x + 5

2-

x approaches 2from the left

1.9 1.99 1.999x

6.9 6.99 6.999f(x)

1.9 + 5

1.99 + 5

1.999 + 5

7

f(x) approaches 7

limx2-

f(x) = 7 Left-hand limit

2+

x approaches 2from the right

2.012.0012.0001 2.1

7.017.0017.0001 7.1

2.1 + 5

2.01 + 5

2.01 + 5

7

f(x) approaches 7

limx2+

f(x) = 7Right-hand limit

2.0001 + 5

equal


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equal

limx2

f(x) = 7


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limxa

f(x) = k

If and only if

limxa- f(x) = limxa+ f(x) = k


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limxa

c = c

limxa

x = a

limxa [f(x) s g(x)] =

limxa f(x)

slim

xa g(x)

limxa

[f(x).g(x)] = limxa

f(x). limxa

g(x)

limxa

f(x)

g(x)= ,

limxa

f(x)

lim

xa

g(x)

limxa

g(x) { 0

limxa cf(x) = c

limxa f(x)

c = constant


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Example :

a) limx -1 (x3

x + 4) = ( )3

( ) + 4-1 -1= 4

b) limx 1

(x2 + 3x 1)

x2 + x + 2= ( )

2 + 3( ) 1

( )2 + ( ) + 2

1 1

1 1

= 3

4


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Example :

c) limx 3(x2 9)

x 3 =(x 3)(x + 3)

x 3

Cannot substitute

directly x = 3

Denominator = 0

The functionis not defined

The

function

need

to be

simplified

first= x + 3

= ( ) + 33

= 6

limx 3


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Example :

d) limx 1(x2 3x + 2)

x2 + x 2 = limx 3( )( )

( )( )

x x1 2 x x1 2 +

= limx 1

x 2

x + 2

=( ) 2

( ) + 2

1

1

=_ 1

3


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Example :

Determine whether the limit exists for each of thefollowing.

a) limx -1

|x + 1|

x + 1

b) limx 0

f(x) where f(x) =x 1 , x < 0

x + 1 , x > 0


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Solution :

For the limit to exist : limx-1-

f(x) = limx-1+

f(x)

a) limx -1|x + 1|

x + 1

limx -1-

|x + 1|

x + 1=

-1-1.0001

limx -1-

-(x + 1)

x + 1

|-1.0001 + 1|-1.0001 + 1

= |-0.0001|-0.0001

= -1

0.0001

-0.0001

= -1


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Solution :


f(x) = limx-1+

f(x)

a) limx -1|x + 1|

x + 1

limx -1+

|x + 1|

x + 1=

-1 -0.999

limx -1+

(x + 1)

x + 1

|-0.999 + 1|-0.999 + 1

= |0.001|0.001

= 1

0.001

0.001

= 1


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Solution :


f(x) = limx-1+

f(x)

a) limx -1|x + 1|

x + 1

limx -1-

|x + 1|x + 1

limx -1+

|x + 1|x + 1{

-1 { 1

limx -1

|x + 1|

x + 1Therefore , does not exist


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b) limx 0

f(x) where f(x) =x 1 , x < 0

x + 1 , x > 0

For the limit to exist : limx0-

f(x) = limx0+

f(x)

limx 0- (x 1) =

0-0.0001

x < 0

1

-0.0001 1

= -1.0001


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b) limx 0

f(x) where f(x) =x 1 , x < 0

x + 1 , x > 0


f(x) = limx0+

f(x)

limx 0+ (x + 1) =

0 0.0001

x > 0

1

0.0001 + 1

= 1.0001


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Solution :


f(x) = limx0+

f(x)

limx 0-

(x 1) limx 0+

(x + 1){

-1 { 1

limx 0

f(x)Therefore , does not exist

b) lim

x 0f(x) where f(x) =

x 1 , x < 0

x + 1 , x > 0


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Limits involving infinity

Example :

a) limx

5x + 3

2x 5b) lim

x

x2 + 1

2x3 1

= limx

The

functionneed

to be

simplified

first1

x

1

x(5x + 3)

(2x 5)

=lim

x

(5 + )3

x

(2 )5x

=5

2


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= limx

The functionneed to be

simplified first

1

x3

1

x3(x2 + 1)

(2x3 1)

=lim

x

1

x3

1x3

= 0

1

x+

2


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f is continuous

at x = a if :

limxa-

f(x) = limxa+

f(x)

= f(a)

The graph of f

is unbroken

Example 1 :

y

0 x

f(x) = ln x , xR+


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f is continuous

at x = a if :

limxa-

f(x) = limxa+

f(x)

= f(a)

The graph of f

is unbroken

Example 2 :

y

0 xf(x) = ex , xR


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f is continuous

at x = a if :

limxa-

f(x) = limxa+

f(x)

= f(a)

The graph of f

is unbroken

Example 3 :

y

0 xf(x) = |x| , xR


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f is continuous

at x = a if :

limxa-

f(x) = limxa+

f(x)

= f(a)

The graph of f

is unbroken

Example 4 :

y

0 x

f(x) = x(x-1)(x-3) , xR

1 3


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example :

Sketch the graphs of each of the following functionsand state whether the function is a continuous

function.

a) f(x) = x3

, xe

210 x , x > 2

b) g(x) = x2

+ 2 , x < 0x , x u 0


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Solution :

a) f(x) = x3 , x e 210 x , x > 2

y

0 x2

8

10

The graph is

unbroken

f is a continuousfunction


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Solution :

b) g(x) = x2 + 2 , x < 0x , x u 0

y

0 x

2

The graph isbroken

f is not a

continuous

function


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Example :

The function f is defined by

f(x) =x2 , x e 2

x + 2 , x > 2

a) Find an d .limx2-

f(x) limx2+

f(x)

Hence, determine whether f is continuous at x = 2.

b) Sketch the graph of f.


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Solution :

limx2-

f(x) = limx2-

f(x) =

x2 , x e 2

x + 2 , x > 2

(x2)

21.999

x < 2

= 4

limx2+

f(x) = limx2+

(x + 2)

2 2.001

x > 2

= 4


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limx2-

f(x) = limx2-

(x2)

= 4

limx2+

f(x) = limx2+

(x + 2)

= 4

Solution :

f(x) =

x2 , x e 2

x + 2 , x > 2

Not equal

limx2-

f(x) limx2+

f(x){

limx2

f(x) does not existf is not

continuous

at x = 2


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Solution :

f(x) =

x2 , x e 2

x + 2 , x > 2

Sketch the graph

0 x

y

2

-4 f(x)=-(-2)2

= -4

4 f(x)=2+2

= -4


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Example :

The function f is defined by

f(x) =(x 2)2 , x e 2

1 + , x > 2a

2

If f is continuous at x = 2, determine the value of

a and sketch the graph of f.


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Solution :

f is continuous

at x = a :

limx2-

f(x) = limx2+

f(x)

f(x) =(x 2)2 , x e 2

1 + , x > 2a

x

limx2-

f(x) = limx2-

(x 2)2

21.999

x < 2

= 0

limx2+

f(x) = limx2+

(1+ )

2 2.0001

x > 2

= 1 +

a

xa

2


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Solution :

f is continuous

at x = a :

limx2-

f(x) = limx2+

f(x)

f(x) =(x 2)2 , x e 2

1 + , x > 2a

2

limx2-

f(x) = limx2-

(x 2)2

= 0

limx2+

f(x) = limx2+

(1+ )

= 1 +

a

2a

2

equal

0 = 1 + a2a

2 1 =a = 2


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x

Solution :

f(x) =(x 2)2 , x e 2

1 + , x > 2-2x1 2/x

Sketch the graph

0

y

4

2

1

1

xy=

1

xy= -

x-intercept?0 = 1 2/x2/x = 1

x = 2


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Function 10

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