Continuity stage (1)

7/29/2019 Continuity stage (1)

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Calculus 3rdsecondary Continuity--

( )ax

f xlim

BUT

Revision on Limits

Determined & unspecified and undefined Quantities

(1) a determined quantity is that which gives a fixed result

ex.-8

5, 3 , 0 ,

(2) an unspecified quantity is that which gives not fixed result

ex.0

0,

, ,

(3) an undefined quantity is that which gives for example0

realnumber

-------------------------------------------------------------------------------------------------------

Steps to solve L imi ts questions

(1) first , substitute in the function :

So if f(a)

If the solution is found to be a real number , so this is a determined value

Guide Examples

1) Then the answer is 2(2) = 4 " determined value "

2)1x

2 3x

3 2xlim

then the answer is

2 3 1 1

3 2 1 5

" determined value "

3)2

23x

x 1

x 5x 6lim

then the answer is

2

2

3 1 10

03 5 3 6

" Limit doesnt exist "

So the conclusion i s , after substitution

number exist number Doesn' t exist

Limitsmaybe

What will happen if we substitute in the function and found that0

( )

0axf xlim

Or

Or Or any unspecified value

22

xxlim


2/28




Calculus 3rdsecondary Continuity-2-

x a x a

x a

x 0 x

x 0 x 0

x 0

1)lim sin x sin a 2)lim cos x cos a

3)lim tan x tan a

sin x x4)lim 1 5)lim 1

x sin x

sin a x sin a x a6)lim a 7)limx sin b x b

sin a x8)lim

b

x 0

x 0 x 0

x x 0

a b x b9) lim

x b sin a x a

tan x tan a x a10)lim 1 11)lim

x b x b

tan ax a b x b12)lim 13)lim

tan b x b tan a x a

To solve this problem , we have to use one of the following

2 2

3 3

2 2

usually use wewill take1 ) common factor

conjugate if it soon2 ) x yyou find

x y x ysquare roots

3 ) x y

x y x xy y

4 ) trinomial

l on g d ivi si on C on j u gat e R ul es f actor izati on

-------------------------------------------------------------------------------------------------------

we can solve limits also by a rule which is

n n

n m

m mx a

x a nlim a

x a m

------------------------------------------------------------------------------------------------------

Limits of trigonometric functions

------------------------------------------------------------------------------------------------------


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Trigonometric function of Double the angle

(1) Sin 2A = 2 Sin A Cos A---------------------------------------------------------------------------------------------------------------------

(2) Cos 2A = (i) Cos2

ASin2

A

(ii) 2 Cos2A1

(iii) 12 Sin2A

---------------------------------------------------------------------------------------------------------------------Very Important note :-

2

2

2 2 2

A A1 ) Sin A 2 Sin Cos

2 2A A A A

2 ) Cos A Cos Sin 2Cos 1 1 2 Sin2 2 2 2

A2Tan

23 ) Tan AA

1 Tan2

---------------------------------------------------------------------------------------------------------------------


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Examples

1)2

23x

7 xx 2

lim

Answer

2

2

223x

7 37 x 7 3 4 2

x 2 3 2 5 53 2lim

------------------------------------------------------------------------------------------------------ -

2) 20x

3xx xlim

Answer

20x

3x 0

x x 0lim

so x is the factor must be eliminated

0 0x x3x 3 3

3x x 1 x 1 0 1

lim lim

------------------------------------------------------------------------------------------------------ -

3)2

2x

x 4

x 2lim

Answer2

2x

x 4 0

x 2 0lim

so ( x + 2 ) is the factor must be eliminated

2 2x xx 2 x 2 x 2 -2 2 -4x 2lim lim

------------------------------------------------------------------------------------------------------ -


5/28





4)2

0x

x 9 3

x 2xlim

Answer

20x

x 9 3 0

x 2x 0lim

so x is the factor must be eliminated , use the conjugate .

20x

x 9 3 x 9 3

x 2x x 9 3lim

20 0

0

x x

x

x 9 9 x

x 2x 9 x 3 x x 2 9 x 3

1 1 1

120 2 9 0 3x 2 9 x 3

lim lim

lim

---------------------------------------------------------------------------------------------------------------------

5)x 0

1 x 1 x

2xlim

Answer

x 0

1 x 1 x 0

2x 0lim

, so x is the factor must be eliminated

by using conjugate :

x 0 x 0

x 0 x 0

1 x 1 x1 x 1 x 1 x 1 x

2x 1 x 1 x 2x 1 x 1 x

2x 1

2x 1 x 1 x 1 x 1 x

1 1

21 0 1 0

lim lim

lim lim

------------------------------------------------------------------------------------------------------ -


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6)x 4

24 5x 2

3 x 5lim

Answer

x 4

24 5x 2 0

03 x 5lim

so ( x4 ) is the factor must be eliminated , use the conjugate .

x 4 x 4

x 4 x 4

x 4

24 5x 4 3 x 524 5x 2 24 5x 2 3 x 5

3 x 5 24 5x 2 3 x 5 9 x 5 24 5x 2

5x 20 3 x 5 5 x 4 3 x 5

9 x 5 24 5x 2 x 4 24 5x 2

5 3 x 5 5 3 4 5 307.5

424 5x 2 24 5 4 2

lim lim

lim lim

lim

------------------------------------------------------------------------------------------------------ -

7)6

3x 1

( x 1) 64

( x 1) 8lim

Answer

6 666 3

33 3x 1x 1

x 1 2( x 1) 64 0 6 lim ( 2 ) 16

( x 1) 8 0 3x 1 2lim

------------------------------------------------------------------------------------------------------ -

8)5

4x 3

x 243lim

x 81

Answer5

4x 3

x 243 0limx 81 0

5 5 5 55 4

4 4 4 4x 3 x 3

x ( 3 ) x ( 3 ) 5 15lim lim ( 3 )

x ( 3 ) x ( 3 ) 4 4

------------------------------------------------------------------------------------------------------ -


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9)

4

x 0

x 5 625lim

x

Answer

4

x 0

x 5 625 0lim

x 0

4 4 4 4 43

x 0 x 0 x 0

( x 5 ) 625 ( x 5 ) ( 5 ) ( x 5 ) ( 5 )lim lim lim = 4(5) = 500

x x x 5 5

------------------------------------------------------------------------------------------------------ -

10)5

2x 0

( x 3 ) 243lim

x 27x

Answer

5

2x 0

( x 3 ) 243 0lim

x 27x 0

5 5 5 5

x 0 x 0 x 0

5 5

x 0 x 0

5 1

( x 3 ) ( 3 ) 1 ( x 3 ) ( 3 )lim lim lim

x( x 27 ) x 27 x

1 ( x 3) ( 3 )lim lim

x 27 x 3 31

5 3 150 27

------------------------------------------------------------------------------------------------------ -

11)5

1x

2

32x 1lim

2x 1

Answer

5

1x

2

32x 1 0lim

2x 1 0

5 5 5 5 55 1

1 1 1x x x

2 2 2

32x 1 2 x 1 ( 2x ) (1) 5lim lim lim (1) 5

2x 1 2x 1 ( 2x ) 1 1

------------------------------------------------------------------------------------------------------ -


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12)3

x 0

3x 8 2lim

x

Answer

3

x 0

3x 8 2 0lim

x 0

1 1 1 1 1 1

3 3 3 3 3 3

x 0 x 0 x 0

11

3

( 3x 8 ) ( 8 ) ( 3x 8 ) ( 8 ) ( 3x 8 ) ( 8 )lim lim 3 lim 3

x 3x 3x 8 8

1 13 8

3 4

------------------------------------------------------------------------------------------------------ -

13)

12

x 1

5x 4 1lim

x 1

Answer

12

x 1

5x 4 1 0lim

x 1 0

12 12 12

x 1 x 1 x 1

12 12

x 1

12 1

5x 4 1 5x 4 1 5x 4 1lim 5 lim 5 lim 5

5 x 1 5x 5 5x 4 1

5x 4 1lim 5

5x 4 1

5 12 1 60

------------------------------------------------------------------------------------------------------ -

14)x 0

Sin3x

4xlim

Answer

x 0

Sin3x 3

4x 4lim

------------------------------------------------------------------------------------------------------ -

15)x 0

5x

Cos xlim

Answer

x 0

5x 00

Cos x 1

lim

------------------------------------------------------------------------------------------------------ -


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1 1

1

x x

x

1 x1 x 1

Sin x Sin x

x 1 1 11

Sin x

lim lim

lim

2 2x x

Sin xCos x 1 1 12

12 2 2

2 x x2 2

lim lim

16)x 0

Sin5x

2Sin3xlim

Answer

Divide both numerator and denominator by x

x 0

Sin5x1 1 5 5x

Sin3x2 2 3 6

x

lim

------------------------------------------------------------------------------------------------------ -

17)2

2x 0

Tan 2x

3xlim

Answer

2 22

2 2

2

x 0 x 0 x 0

Tan2xTan 2x 1 1 Tan2x

3x 3 x 3 x

1 42

3 3

lim lim lim

------------------------------------------------------------------------------------------------------ -

18) 1x

1 x

Sin xlim

Answer

------------------------------------------------------------------------------------------------------ -

19)

2x

Cos x

2x

lim

Answer

------------------------------------------------------------------------------------------------------ -


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1 1

1

x x

x

Tan x Tan x

1 x 1 x

Tan x

x

lim lim

lim

x 0 x 0 x 0

Tan Sin5x Tan Sin5xSin5x Sin5x

Sin5x 3x Sin5x 3x

5 51

3 3

lim lim lim

20)

1x

Tan x

1 xlim

Answer

------------------------------------------------------------------------------------------------------ -

21)

x 0

Tan Sin5x

3xlim

Answer

------------------------------------------------------------------------------------------------------ -


11/28




Calculus 3rdsecondary Continuity--

st1 stage

limit of a function

at a point

nd2 stage

Continuity of a function

at a point

1

2

f ( x ) , when x a

f ( x ) , when x a

x ax

f ( x )

x a " a " x a " a " f ( a ) f ( a )

f ( a )

f ( a )

Continuous :a function is continuous if you can draw it without lifting your pen from the

paper. the graph is connected .

I n the opposite fi gures:

left figure represent a Continous function

while the right figure represent

a Discontinous function

------------------------------------------------------------------------------------------------------ -

In this chapter , we have two stages to explain

------------------------------------------------------------------------------------------------------ -

1st

stage : L imit of a function at a point

This happens when a limit contains two different functions , our job is to connect them and todiscuss whether after connecting , limit exists or not

So , let f(x) = This is the domain of the function .

Steps: 1) Draw the real number line

2) 1f ( x ) when x a is called the leftlimit of x = a , and it is denoted by .

3) 2f ( x ) when x a is called the Rightlimit of x = a , and it is denoted by .4) the function y = f(x) only has a limit at x = a , when three conditions are satisfied :

(a) left limit is defined . Nota 0

, , .....

0 0

(b) Right limit is defined .

(c) left limit = right limit f ( a ) f ( a )

Continuit


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x 1x

f ( x ) f (1 ) 3 f (1 ) 2x 1

x 1

x 1

x 1

left limit : f( 1 ) lim 3 3

Right limit : f( 1 ) lim 2x 1 2(1) 1 3

f( 1 ) f( 1 ) lim f ( x ) 3 , So limits exists

x 1 x

f ( x ) f ( 1 ) x 3 f ( 1 ) 2 3x

x 1

x 1

x -1

left limit : f( -1 ) lim x 3 ( 1) 3 2

Right limit : f( 1 ) lim 2 3x 2 3( -1) 5

f( -1 ) f( -1 ) lim f ( x ) doesn' t exists

Example (1)

The opposite figure represents the graph of

3 , when x 1

f ( x )

2x 1 , when x 1

Discuss the existance of limit of this function at x=1

Answer

------------------------------------------------------------------------------------------------------ -

Example (2)


x 3 , when x -1

f ( x )

2 3x , when x -1

Discuss the existance of limit of this function at x= -1

Answer


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x 2x

f ( x ) f ( 2 ) 2x 1 5

f ( 2 )x 2

x 2

x 2

x 2

left limit : f( 2 ) lim 2x 1 2( 2 ) 1 3

5 5Right limit : f( 2 ) lim undefined

x 2 0

lim f(x) , doesn't exists

x 2 x

f ( x ) 2f ( 2 ) x 2 f ( 2 ) 2x 3

2 2

x 2

x 2

x 2

left limit : f( 2 ) lim x 2 ( 2 ) 2 6

Right limit : f( 2 ) lim 2x 3 2( 2 ) 3 1

f( 2 ) f( 2 ) lim f(x) , doesn't exists

Example (3)


2x 1 , when x 2

f ( x ) 5, when x 2

x 2

Discuss the existance of limit of this function at x = 2

Answer

------------------------------------------------------------------------------------------------------ -

Example (4)2

x -2

x 2 : x -2let f(x) = Then , find lim f(x)

2x + 3 : x > -2

Answer

------------------------------------------------------------------------------------------------------ -


14/28





x 1x

f ( x )x 1

x 1

1

2 x 5

x 6x 4

Example (5)

x 1 x 4 x 6

x 1: 1< x < 4

x 1

if f(x) = ,Then find (1) lim f(x) (2) lim f(x) (3) lim f(x)1: 4 x 6

2x 5

Answer

x 1

x 1

x 1 x 1 x 1

(1) for lim f(x) :

f ( x ) is defined only is the right of x 1

x 1 0Right limit : f( 1 ) lim

x 1 0

x 1x 1 x 1 1 1lim lim lim

x 1 2x 1 x 1x 1 x 1

x 4

x 4

x 4

x 4

(2) for lim f(x) :

x 1 1left limit : f( 4 ) lim

x 1 3

1 1Right limit : f( 4 ) lim

2x 5 3

1f( 4 ) f( 4 ) lim f(x)

3

x 6

x 6

(3) for lim f(x) :

1 1left limit only defined : f( 6 ) lim

2x 5 7

------------------------------------------------------------------------------------------------------ -


15/28





x 1x

f ( x )- x

f (1 )x 2

xf (1 )

x 2

x 1

x 1

x 1

- x 1( 3 ) left limit : f( 1 ) lim

x 2 3x 1

Right limit : f( 1 ) limx 2 3

f( 1 ) f( 1 ) lim f(x) , doesn't exists

Example (6)

2

x -2 x 2 x 1 x 0

x x 1if f(x) ,Then find :

x x 2

(1) lim f(x) (2) lim f(x) (3) lim f(x) (4) lim f(x)

Answer

x x 1f ( x ) The domain is R 2 , 1

x 2 x 1

x 1 : x > 1And x 1 =

- x 1 : x < 1

x x 1 x: x > 1 : x > 1

x 2 x 1 x 2f ( x ) = f ( x ) =

xx x 1: x < 1 & x -2: x < 1

x 2x 2 x 1

x 2 x 2 x 2

x 2(1) lim f(x) lim lim f(x) does not exist " normal limit "

x 2 zero

x 2 x 2

x 1(2) lim f(x) = lim = " So limit exists"

x 2 2

x 0 x 0

x(4) lim f(x) = lim = 0

x 2

" So limit exists"

------------------------------------------------------------------------------------------------------ -


16/28





x

f ( x ) f ( 2 ) 3x 1 f ( 2 ) 4x 3

Example (7)

x 2

x 23x : x < 2

let f(x) = Then , find lim f (x)2 x

4x - 3 : x > 2

Answer

x 2 ,x 2 refusedfor x < 2 x 2

2 x , x 2 agreed

34x 3 , x agreed but we will take x > 2

4for x > 2 4x 3

34x 3 , x refused

4

2 x3x 1 ,x 23x ,x 2

f(x) f(x)2 x4x 3 , x 2

4x 3 , x 2

------------------------------------------------------------------------------------------------------ -

x 2

x 2

x 2

x 2

x 2

for lim f(x) :

left limit : f( 2 ) lim 3x 1 5

right limit : f( 2 ) lim 4x 3 5

f( 2 ) f( 2 ) lim f(x) 5 " So limit exists"


17/28





x 1x

f ( x )

nx 1f (1 )

x 1

22x x 6 f (1 )

x 2

Example (8)n

2 x 1 x 2

x 1: x < 1

x 1

let f(x) , if lim f(x) exists , then find n and lim f(x)2x x 6 : x > 1

x 2

Answer

x 1

nn 1

x 1

2

x 1

x 2

x 2

lim f(x) exists : f( 1 ) f( 1 )

x 1left limit : f ( 1 ) lim n 1 n

x 1

2x x 6 Right limit : f( 1 ) lim 5

x 2

n 5

for lim f(x) : Only the Right limit defined

2f( 2 ) : lim

2

x 2 x 2

2x 3 x 2x x 6 0 lim lim 2x 3 7 x 2 0 x 2

------------------------------------------------------------------------------------------------------ -


18/28





x

f ( x )2f (0 ) 3 x f (0 ) 3a 4x

Example (9)3

x 0

3x x: x < 0

let f(x) = Then , find the value of "a" if lim f (x) existsx

3a 4x : x > 0

Answer

3

3

3

3 3

2

2

x ,x 0 refusedfor x < 0 x

x , x 0 agreed

3x x 3x x,x 0,x 0

f(x) f(x) xx

3a 4x , x 03a 4x , x 0

x 3 x3 x , x 0,x 0

f(x) f(x)x3a 4x , x 0

3a 4x , x 0

------------------------------------------------------------------------------------------------------ -

x 0

x 0

2

x 0

x 0

for lim f(x) :

left limit : f( 0 ) lim 3 x 3

right limit : f( 0 ) lim 3a 4x 3a

f( 0 ) f( 0 ) 3a 3 a 1


19/28





x 2 x

f ( x ) 21 x x 21 x x

x 7x 0 x 2

x 2

x 2 2

Example (10)

x 1 x 0 x 2 x 7

1 x x x : -2 < x < 2

let f (x) x 2 : 2 < x < 7x 2 2

Find : (1) lim f ( x ) ( 2 ) lim f ( x ) ( 3 ) lim f ( x ) ( 4 ) lim f ( x )

Answer

2

2

1 x x , -2 < x 0

f ( x ) 1 x x , 0 < x < 2x 2

, 2 < x < 7x 2 2

2

x 1 x 1

2

x 0 x 0

2

x 0

x 0

x 2

(1) for lim f ( x ) : lim (1 x x ) 1 1 1 1

( 2 ) for lim f ( x ) : left limit : f( 0 ) lim(1 x x ) 1

Right limit : f ( 0 ) lim (1 x x ) 1

f( 0 ) f( 0 ) lim f ( x ) 1

( 3 ) for lim f ( x ) : left limit

2

x 2

x 2 x 2

x 2

x 2

x 7 x 7

: f ( 2 ) lim(1 x x ) 1 2 4 3

x 2 x 2 2x 2 x 2 2Right limit : f ( 2 ) lim lim

x 2 4x 2 2 x 2 2

lim( x 2 2 ) 2 2 4

So f(2 ) f ( 2 ) lim f ( x ) does not exist.

(4) lim f ( x ) lim

x 2 7 2

53 2x 2 2

---------------------------------------------------------------------------------------------------------------------


20/28





x

f ( x )x 1

f (0 )x 2

5x tan 3xf (0 )

x 3sin x

x 1 x

f ( x ) 3x 2 ax b

x 3

6 x

Example (11)

x 1 x 3

3x 2 : x < -1

let f(x) ax b : -1 < x < 3 , if lim f(x) and lim f(x) exist , find the value of a and b .

6 x : x > 3

Answer

x 1

x 1

x 1

x 3

x 3

lim f(x) exists : f( 1 ) f( 1 )

f ( -1 ) lim ( 3x 2 ) 3 2 5

f( -1 ) lim ( a x b ) -a b - a b -5 .......(1)

also lim f(x) exists : f( 3 ) f( 3 )

f( 3 ) lim ( a x b ) 3a b

x 3f( 3 ) lim (6 x ) 6 3 3 3 a b 3 .......(2)

By subtracting (2) (1) 4a 8 a 2 , b -3

----------------------------------------------------------------------------------------- -------------Example (12)

Answer

x 0

x 0

x 0

x 0

x 1 1left limit : f( 0 ) lim =

x 2 2

5x Tan3xRight limit : f( 0 ) lim " divide both numerator and denominator by x "

x 3Sin x

Tan3x55 3 1 1xlim f( 0 ) f( 0 ) , hen

Sinx 1 3 2 21 3

x

x 0

1ce lim f(x)

2

x 0

5x tan 3x: x > 0

x 3sin xif f(x) = Then find lim f(x)

x 1: x < 0

x 2


21/28





-x

2

x

f ( x )2 x

sin x

42cos x

3

x

2

x 0

Example (13)

x 0x - x2 2

2x: - < x < 0

sin x 2let f(x) , Find : (1) lim f(x) (2)lim f(x) (3) lim f(x)

42cos x : 0 < x 0

Answer

x 0 x 0

2 2

x 0

2

Sin4x 4left limit : f (0 ): lim Sin 4x Cot 9x lim

Tan 9x 9

Right limit : f (0 ) : lim 4x a a

4 4 2f (0 ) f (0 ) a a

9 9 3

------------------------------------------------------------------------------------------------------ -

Example (15)

x 02

9 xSin : x < 0

x 2let f(x) Then find the value of "a" if lim f ( x ) exists

1

a x 2a : x > 02

Answer

x 0 x 0

2

x 0

xSin

9 x 92left limit : f (0 ) : lim Sin limxx 2 29

1 1Right limit : f (0 ) : lim a x 2a 2a

2 2

1 9 9 1f (0 ) f (0 ) 2a 2a 4 a 2

2 2 2 2

------------------------------------------------------------------------------------------------------ -

x 0

x 0


23/28





x

f ( x )

Sin x2f (1 )5

8 Tan x4

2

x 3 - 2f (1 )

x - 1

Example (16)

Answer

------------------------------------------------------------------------------------------------------ -

x 1

2 2x 1 x 1

2x 1 x 1

x 3 -2 x 3 2 x 3 2Right limit : f (1 ) = lim lim

x 1 x 1 x 3 2x 3 4 x 1

lim limx 1 x 3 2 x 1 x 1 x 3 2

x 1

x 1

1 1lim

8x 1 x 3 2

1 1f (1 ) f (1 ) lim f(x) =

8 8

2

x 1

x 3 - 2: x >1

x - 1

if f(x) = , Then find lim f(x)Sin x2 : x < 15

8 Tan x4

x 1

sin x1 12 2left limit : f ( 1 ) lim

5 58 88 tan x

4 4


24/28





x

f ( x )1 Cos 2x

f (0 )x Tan x

f (0 ) a 1

Example (17)

x 0

1 Cos 2x: x < 0

x Tan xlet f(x) Then find the value of "a" if lim f ( x ) exists

a 1 : x > 0

Answer

2 2x 0 x 0 x 0

2

22

2

x 0 x 0

1 1 2Sin x1 Cos 2x 2Sin xleft limit : f (0 ) : lim lim lim

x Tan x x Tan x x Tan xSo, divide both numerator and denominator by x

Sin xSin x 22xxlim lim 2

Tan x Tan x

x x

Right limit : f (0 ) : lim

x 0a 1 a 1

f (0 ) f (0 ) a 1 2 a 3

------------------------------------------------------------------------------------------------------ -

Example (18)9

6

x 35

x 81 3: x < 3

x 27let f(x) , find lim f(x) .1x : x 3

2

Answer

99

9 9 6

666x 3 x 3

5

x 3

x 3

x 3x 81 3 9 9 3left limit : f( 3 ) lim lim 3

x 27 6 2x 3

1 9 3Right limit : f( 3 ) lim x

2 2

9 3f( 3 ) f( 3 ) lim f(x)

2

------------------------------------------------------------------------------------------------------ -

x 0


25/28





x

f ( x ) f ( 2 ) 1 f ( 2 ) a x 3

Example (19)

2

x 2

x 4x 4: x < 2

let f(x) = Then , find the value of "a" if lim f (x) existsx 2

a x 3 : x > 2

Answer

------------------------------------------------------------------------------------------------------ -

Example (20)

x

2

x 0

Tan 2x: x < 0

log 8let f(x) , find lim f(x) .

2x : x 0

3

Answer

x 2

x 2

x 2

left limit : f( 2 ) lim 1 1

right limit : f( 2 ) lim a x 3 2a 3

f( 2 ) f( 2 ) 2a 3 1 2a 4 a 2

2x 2x 2

: x < 2: x < 2f(x) f(x) x 2x 2

a x 3 : x > 2a x 3 : x > 2

x 2 , x 2 refusedfor x < 2 : x 2

- x 2 , x 2 agreed

- x 21 :: x < 2

f(x) f(x)x 2

a x 3 : x > 2

x < 2

a x 3 : x > 2

3

2 2

x 0

Tan 2x Tan 2x Tan 2x: x < 0 : x < 0 x < 0

xlog 8 xlog 2 3xf(x) f(x) f(x) ,

22 2x x 0x : x 0 x : x 0

33 3

Tan 2x 2left limit : f( 0 ) lim

3x 3

Righ

x 0

x 0

2 2t limit : f( 0 ) lim x3 3

2f( 0 ) f( 0 ) lim f(x) =

3


26/28





x

f ( x ) f ( 2 ) a x b 2f ( 2 ) a x b

Example (21)

7

4

x 0

1 2x 1: x < 0

7xlet f(x) , find the value of "a" if lim f(x) exists.1a : x 0

2

Answer

7 77 74 44 4

x 0 x 0774 74 1

4

x 0

x 0

x 2

1 2x 1 1 2x 11 2left limit : f( 0 ) lim lim

7 x 7 2x

1 2x 12 2 7 1lim 1

7 1 2x 1 7 4 21 1Right limit : f( 0 ) lim a a

2 21 1

lim f ( x ) exists , Then f( 2 ) f( 2 ) a - a 02 2

----------------------------------------------------------------------------------------- -------------Example (22)

2 x 2

a x + b : x < 2let f(x) , If lim f ( x ) 3 ,Then find a and b

a x b : x > 2

Answer

x 2

x 2

2

x 2

lim f ( x ) 3 f( 2 ) f( 2 ) 3

left limit : f( 2 ) lim ( ax b ) 2a b

Right limit : f(2 ) lim ( ax b ) 4a b

2a b 3 .......(1) And 4a b 3 .......(2)

from (1) , (2) and by ad

dition : a 1 , b 1

x 2


27/28





Example (23)

x

2Sinx: x

Continuity stage (1)

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Transcript of Continuity stage (1)