Demo-limits ,Continuity and Differentiability

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Relations,Functions,limits,continuityanddifferentiability

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REAL FUNCTIONS A function which has the mapping :f R R is called a real function i.e The domain of a real function is a sub set of real numbers R Note: if the range of a real function is also a sub set of real numbers R , then the function is said to be real valued function. Examples of real functions : 1.Polynomial functions 2. Trigonometric functions 3.Exponential functions 4. Logarithmic function 5.Signum function 6.Greatest integer function 7.Fractional function 9.Modulus function 10.Invaerse trigonometric functions and many. (The detailed explanations of those functions are given under the topic REAL FUNCTIONS ) Functional equations of real functions

(i) Functional equation: 1( ) 1f x fx

= Real function: ( ) nf x x=

(ii) Functional equation: ( ) ( ) ( )f x f y f xy= Real function: ( ) nf x x=

(iii) Functional equation : ( ) ( ) ( )f x y f x f y+ = + Real function : ( )f x ax=

(iv) Functional equation : ( ) ( ) ( )f x y f x f y+ = Real function : ( ) xf x a=

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(v) Functional equation : ( ) ( ) ( )f xy f x f y= + Real function : ( ) logef x a x=

(vi) Functional equation : ( ) ( )1x yf f x f y

xy + = +

Real function : 1( ) ( )f x aTan x=

LIMIT OF A REAL FUNCTION Let ( )y f x= be a real function and , are two smallest positive numbers such that ( )f x l

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Basic results of limits

1. 1limn n

n

x a

x a nax a

=

2.0

sinlim 1x

xx

=

3.0

tanlim 1x

xx

=

4.0

log (1 )lim 1ex

xx+ =

5.0

1lim 1x

x

ex =

6.0

1lim logx

ex

a ax =

7. ( ) lim ( ) ( )( )lim 1 ( ) ,when f(a)=0 and g(a)=x a f x g xg xx a

f x e + =

8. ( )10

lim 1 xx

x e

+ =

9.1lim 1

x

xe

x + =

10. lim ( ( )) lim ( )x a x a

f g x f g x =

11. [ ]lim ( ) lim ( ) nnx a x a

f x f x =

12.lim ( )( )lim ( ) lim ( ) ,x a

g xg x

x a x af x f x = when both lim ( ) and lim ( ) existsx a x af x g x .

LHospital rule Let both ( )f x and ( )g x are differential functions ,then

( ) ( )lim lim , ( ) 0( ) ( )x a x af x f x g ag x g x

= and ( )( )f ag a

should be either 00

or ,otherwise apply the same until we get a real number(limit). Continuity of a real function Let ( )y f x= be a real function and it is continuous at x = a if lim ( ) ( )

x af x f a =

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i.e lim ( ) lim ( ) ( )x a x a

f x f x f a + = = ( )f x is continuous at x = a. Points of discontiuity 1. lim ( ) lim ( ) ( )

x a x af x f x f a + = ,Removable type of discontinuity.

2. lim ( ) lim ( )x a x a

f x f x + , Second kind of discontinuity. 3. both lim ( ) and lim ( ) does not exist

x a x af x f x + .

Continuous functions 1. Constant function is always continuous. 2. Identity function is always continuous. 3. Polynomial function is always continuous in its domain. 4. Trigonometric function is continuous in its domain. 5. Sum of two continuous functions is continuous. 6. Difference of two continuous functions is continuous. 7. Product of two continuous functions is continuous. 8.Quetiont of two functions is continuous in the domain of second function. 9. Composition of two continuous functions is always continuous. First principle of differentiation Let ( )y f x= be a real function and its derivative for any value of x is defined

by 0

0

lim

( ) ( ) = lim

x

x

dy ydx x

f x x f xx

=+

Left derivative and right derivative

( ) ( ) ( )( ) limx a

f x f aL f xx a = and ( )

( ) ( )( ) limx a

f x f aR f xx a+ =

Theorems of derivatives

1. [ ]( ) ( ) ( ) ( )d f x g x f x g xdx

=

2. [ ]( ) ( ) ( ) ( ) ( ) ( )d f x g x f x g x f x g xdx

= +

3. 2( ) ( ) ( ) ( ) ( )( ) ( )

d f x f x g x f x g xdx g x g x

=

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4. [ ]( ) ( )d kf x kf xdx

=

Derivative of real functions

( )f x ( )f x ( )f x ( )f x xe xe 1

x 2

1x

nx 1nnx 2

1x

32x

sinx Cosx 1nx

1nn

x +

Cosx -Sinx log x

1x

Tanx 2Sec x xx ( )1 logxx x+

secCo x secCo xTanx ( )( )g xf x

[ ]( ) ( ) ( )( ) ( ) log ( )( )

g x g x f xf x g x f xf x

+

Secx tanSecx x ( ( ))f g x

( ( )) ( )f g x g x

Cotx secCo xCotx

1Sin x 2

11 x

1Cos x 2

11 x

1Tan x 2

11 x+

1Cot x 2

11 x+

1Sec x 2

11x x

1secCo x 4 2

1x x

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Derivative of parametric equations Let ( ) and ( )x f t y g t= = ,then

( ) ( )

dy dy dtdx dt dx

g tf t

= =

nth derivative of real function

1. in2n

y S x y Sin x n = = +

2.2n

y Cosx y Cos x n = = + 3. !n nx y n =

4. ( )( ) 11 !1

n n

n n

n ay y

ax b ax b += =+ +

Examples 1.Let ( ) ( ) ( )f x y f x f y+ = ,x y R and (0) 0f , then prove that

( )2( )( )

1 ( )f xF xf x

= + is an even function.

Solution If a real function ( )f x satisfies a functional equation ( ) ( ) ( )f x y f x f y+ =

,x y R , then ( ) xf x a= and (0) 0f Given: ( )2

( )( )1 ( )

f xF xf x

= +

Now , ( ) ( )2 2( ) ( )( ) ( )

1 ( ) 1 ( )f x f xF x F xf x f x

= + +

( ) ( )2 21 1x x

x x

a aa a

= + +

( ) ( )2 2( ) ( ) 01 1x x

x x

a aF x F xa a

= =+ +

F(x)=F(-x) Therefore, ( )F x is an even function.

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2.If ( ) ( ) 2 ( ) ( ) ,f x y f x y f x f y x y+ + = ,then examine whether the function is even or odd. Solution Given : ( ) ( ) 2 ( ) ( ) f x y f x y f x f y+ + = ..(1) Replacing x by y

( ) ( ) 2 ( ) ( ) f y x f y x f y f x+ + = ..(2) From the results (1) & (2) , ( ) ( ) 0f x y f y x =

( ) ( )f x y f y x = i.e ( ) ( );f a f a= where a x y= This proves that the functions isan even function.

3.If ( )2( +1)= ( ) + ( ) x,y Rf x y f x f y+ and (0) 1f = , then find the value of ( 3)f .

Solution

Given; ( )2( +1)= ( ) + ( ) x,y Rf x y f x f y+ and (0) 1f = 0x y= = ( )2(1) 1 1f = +

1, 0 x y= = ( )2 2(2) (1) 1 (2 1)f f= + = + 2, 0 x y= = ( ) ( )2 2(3) (2) 1 3 1f f= + = +

In general ( )2( ) 1f x x= + and ( )2( 3) 3 1 4f = + = 4.If ( ) ( ) ( ) ( ) ( ), ,f x y f x f y f a x f a y x y R = + and (0) 1f = ,then find the value of (2 )f a x Solution Given: ( ) ( ) ( ) ( ) ( ), ,f x y f x f y f a x f a y x y R = + (1) Let 2 ,x a y x= = and (2 ) (2 ) ( ) ( ) (2 )f a x f a f x f a f a = (2) Put 0x = in (1),then ( ) ( )f y f y = ( )f x is an even function and

( ) ( ) 0f a f a = = Put x y a= = in(1)then (0) ( ) ( ) (0) (2 )f f a f a f f a= (2 ) 1f a = Applying these in (2) , (2 ) ( 1) ( ) (0)( 1)f a x f x = (2 ) ( )f a x f x =

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5.If ( ) ( )2 ( ) ( ) ( ) ,y xf xy f x f y x y R= + and (1) 2f = , then find the value of 9

1( )

nf n

= . Solution

Given: ( ) ( )2 ( ) ( ) ( ) ,y xf xy f x f y x y R= + .(1) Put 1y = in(1),then ( )2 ( ) ( ) (1) xf x f x f= + ( ) 2xf x = And

9

1( )

nf n

= (1) (2) (3) ... (9)f f f f= + + + +

=9

2 3 4 9 2(2 1)2 2 2 2 ... 22 1

+ + + + + = 9

1

( )n

f n= = 102 2 1022 = 6.If ( )f x is a polynomial such that ( ) ( ) ( ) ( ) ( ) ,f x f y f xy f x f y x y R= and

(2) 7f = ,then find the value of ( 3)f . Solution Given: ( ) ( ) ( ) ( ) ( ) ,f x f y f xy f x f y x y R= ( ) ( ) ( ) ( ) ( )f x f y f x f y f xy+ + = ( ) ( ) ( ) ( ) 1 ( ) 1f x f y f x f y f xy+ + + = + ( )( )( ) 1 ( ) 1 ( ) 1f x f y f xy+ + = + ( ) ( ) ( )g x g y g xy= ( ) ( ) 1g x f x= + From the law of functional equation , ( ) ng x x=

( ) ( ) 1ng x x f x= = + i.e ( ) 1nf x x= Given: (2) 7 1nf x= = 8nx = 3n = Therefore, 3( 3) ( 3) 1 28f = = 7.If ( )f x is a polynomial such that 1 1( ) ( ) 0,f x f f x f x y R

x x = + and

( 2) 33f = ,then find the value of (1)f .

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Solution

Given: 1 1( ) ( ) 0,f x f f x f x y Rx x

= +

1 1( ) ( ) 0,f x f f x f x y Rx x

( ) 11 ( ) 1 1f x fx

=

1( ) 1g x gx

=

From the functional equation ( ) 1 ( ) ng x f x x= =

( )( ) 1 1n nf x x x= = and ( 2) 33f = 5( ) 1f x x = and therefore , (1) 0f =

Problems based on difference equations

Let ( )f n be a function defined with domain { }0,1,2,3,4,5... is called a sequence (i) Equation : ( 1) ( )f n pf n q+ + =

Solution : ( ), 1

( ) 1 ( )A(-p) + , 1

1

nn

A qn pf n q p

pp

+ = = +,where (0)A f=

(ii) Equation: ( 2) ( 1) ( ) 0f n pf n qf n+ + + + =

Solution: ( ),

( ),

n n

n

A Bf n

A Bn

+ = + =

,where , are the roots of the equation 2 0x px q+ + = and (0)A f= and B also can be found from the value f(1).

8.Solve ( 1) ( ) 2, (0) 1f x f x f+ + = = Solution

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As per difference equation 1, 2p q= = ( )1 ( )

( ) ( )1

nn q pf x A p

p = + + , (0) 1A f= =

= ( )2 1 ( 1)

( 1) 12

nn + =

Therefore , ( ) 1f x = 9.If ( 2) 4 ( 1) 4 ( ) 0, (0) 0, (1) 1f x f x f x f f+ + + = = = ,then find the value of

(10)f Solution Given: ( 2) 4 ( 1) 4 ( ) 0, (0) 0, (1) 1f x f x f x f f+ + + = = = From the law of difference equation , , are the roots of the equation

2 4 4 0y y + = (Replace ( 2)f x + by 2y and ( 1) by f x y+ ) 2 = = and (1) 1(0) 0,

2 2fA f B= = = =

i.e ( )( ) xf x A Bx = + ( )1( ) 2xf x x =

Therefore , ( )9(10) 10 2 5120f = = . 10.If ( 1) ( 1) 3 ( ), (2) 2f x f x f x f+ + = = ,thenfindthevalueof (4)f .Solution

Given: ( 1) ( 1) 3 ( ), (2) 2f x f x f x f+ + = = ..(1)Replacexbyx+1in(1)

( 2) 3 ( 1) ( ) 0f x f x f x+ + + = Let , are the roots of 2 3 1 0y y + = and 3Roots

2i= ,imaginary roots

Therefore 3 3( )2 2

x xi if x A B

+ = +

( )6 2x xf x Acis Bcis = + and (2) 2f =

( ) 46xf x Cos =

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And therefore , 2(4) 4 23

f Cos = =

Limits and continuity Real functions as an infinite series

1.2 3

1 ........1! 2! 3!

x x x xe = + + + +

2.2 3

1 ........1! 2! 3!

x x x xe = + +

3.2 3 4

ln(1 ) ........2 3 4x x xx x+ = + +

4.2 3 4

ln(1 ) ........2 3 4x x xx x =

5.3 5 7

........3! 5! 7!x x xSinx x= + +

6.2 4 6 8

1 ........2! 4! 6! 8!x x x xCosx = + + +

7.3

52 ...3 15xTanx x x= + + +

8.2 4

cot 1 ...3 45x xx x = +

9.2

451 ....2 24xSecx x= + + +

10.2

47cos 1 ........6 360xx ecx x= + + +

11.3 5 7

1 1 1 3 1 3 5sin .....2 3 2 4 5 2 4 6 7x x xx x = + + + +

12.3 5 7

1tan .....3 5 7x x xx x = + +

13. ( ) ( ) ( )2 32 31 ln ln ln ......1! 2! 3!

x x x xa a a a= + + + +

14. ( ) 2111 1 .....2 241 x xxx e + + + =

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Examples

11. 2 2 2 2lim 1 2 3 ....

1 1 1 1n

x n n n n + + + +

Solution

2 2 2 2

lim 1 2 3 ....1 1 1 1

nx n n n n

+ + + + = 2lim 1 2 3 4 5 ...

1n

x n+ + + + + +

= ( )2lim ( 1) 1

22 1n n

x n+ = ,

(Hint coefficient of 2n both in Nr and Dr)

12.lim

x x x xx

+ + Solution

limx x x x

x + + =

lim x x x xx x x xx x x x x

+ + + + + + + +

=lim x x

x x x x x

+ + + +

(Take x as a commonfactor from Nr and Dr.)

=

11lim

1 11 1

xx

x x x

+ + + +

= 12

13. 4 21

lim1

n

r

rn r r=

+ + Solution

4 21

lim1

n

r

rn r r=

+ + = 2 2 21lim 1 1 1

2 1 1

n

rn r r r r= + + +

= 2 2lim 1 1 1

2 1 1 1 1n n n + + +

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= 12

14. ( )1 21

limcot 2

n

rr

n

= Solution

( )1 21

limcot 2

n

rr

n

= = 1 21lim 1tan

2

n

rn r

=

= 1 2

1

lim 2tan1 4 1

n

rn r

=

+

= ( )11

lim 2 1 (2 1)tan

1 (2 1)(2 1)

n

r

r rn r r

=

+ + +

= ( ) ( )1 11

limtan 2 1 tan 2 1

n

rr r

n

=

+ = ( )1 1lim tan (2 1) tan 2(1) 1n

n +

= 1 1tan tan 1

= 2 4 4 =

15. ( )( )( ) 13lim 1 2 5x x x xx

+

Solution

We know that 3 3

2 2

a ba ba ab b

= + + ,therefore

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( )( )( ) 13lim 1 2 5x x x xx

+ =

( )( )( )( )( )( )( ) ( )( )( )( )

3

2 1/3 23

lim 1 2 5

1 2 5 1 2 5

x x x xx x x x x x x x x

+ + + + +

(Taking co efficient of 2x from Nr and Dr)

( )( )( ) 13lim 1 2 5x x x xx

+ =1 1

1 1 1 3=+ +

16. [ ] [ ] [ ] ( )

2

3 5 ..... 2 1lim x x x n xn n

+ + + +

Solution

[ ] [ ] [ ] ( )2

3 5 ..... 2 1lim x x x n xn n

+ + + + =

( )2

1

2 1lim n

r

r xn n=

+

=( ) ( ){ }

21

2 1 2 1lim n

r

r x r xn n=

+ +

=2

2

0n x xn =

17.If ( ) 2, '( ) 1, ( ) 1, '( ) 2f a f a g a g a= = = = , then find the value of lim ( ) ( ) ( ) ( )g x f a g a f xx a x a

Solution

lim ( ) ( ) ( ) '( )g x f a g a f xx a x a

=

( ) ( ) ( ) ( ) 00

g a f a g a f aa a =

Applying Lhospithal rule

lim ( ) ( ) ( ) ( )g x f a g a f xx a x a

=

lim ( ) ' ( ) ( ) '( )1

g x f a g a f xx a

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=lim ( ) ' ( ) ( ) '( )

1g a f a g a f x

x a

= (2)(2) ( 1)(1) 51 =

18.If '(2) 6, '(1) 4f f= = , then find the value of 2

2

lim (2 2 ) (2)0 ( 1) ( 1)f h h f

h f h h f+ +

+

Solution

2

2

lim (2 2 ) (2)0 ( 1) (1)f h h f

h f h h f+ +

+ =(2) (2)(1) (1)

f ff f

= 00

( indeterminant form)

Therefore there must exist a limit , Applyinf LHospital rule

2

2

lim (2 2 ) (2)0 ( 1) (1)f h h f

h f h h f+ +

+ =2

2

lim '(2 2 )(2 2 )0 '( 1)(1 2 )f h h h

h f h h h+ + +

+

= '(2)(2)'(1)(1)

ff

= 6 2 34 =

19.Evaluate 2

2sin

cos

1

lim0

xnec x

rr

x =

Solution

2

2sin

cos

1

lim0

xnec x

rr

x = = ( )

22 2 2 2 sincos cos cos coslim 1 2 3 ...

0x

ec x ec x ec x ec xnx

+ + + +

Multiply and Divide every term by 2cosec xn

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= 2

2 2 2 2 sincos cos cos coslim 1 2 3 1... 10

xec x ec x ec x ec xn n

x n n n n

+ + + + +

= 0(0 0 ... 0 1) n+ + + + 1 2 11, 1,.... 1nn n n

< <

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1

lim0 3

x x x xa b cx

+ + =

1lim

10 3

x x x xa b cxe

+ +

=

( )1 1 1lim 10 3

x x xa b c

x xe + +

=

lim1 1 1 103

x x xa b cx x x xe

+ +

= ( )13 1log 3abce abc=

Continuity and differentiation Leibnitze theorem to find nth derivative If y uv= ,then 1 1 1 2 2 2 .....n n nn n n ny u v C u v C u v uv = + + + + 22.Let

1 ,0 2( )

3-x ,2

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123

1 2 3

It is shown that there are 2 points of discontinuity in ( )1,3 .23.If ( )f x is a continuous function and

0

( )x

f t dt as x ,then find the number of points in which the line y mx= intersects the curve

2

0

( ) 2x

y f t dt+ Solution

Sub y mx= in 20

( ) 2x

y f t dt+ 2 2

0

( ) 2 0x

m x f x dt+ = and Let ( )g x = 2 20

( ) 2x

m x f x dt+ (0) 2 0g = < and ( ) 0g = > ( )g x =0 for atleast one value of [ )0,x Geometrical interpretation

0

24.If f is a differentiable function such that 2( ) ( ) , ,f x f y x y x y R and (0) 0f = ,then find the value of (1)f .

Solution2( ) ( ) , ,f x f y x y x y R

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( ) ( )f x f y x yx y

lim lim( ) ( )f x f y x yx y x yx y

'( ) 0 but '( ) 0f x f x

'( ) 0f x = F(x) is a constant function. And therefore , (1) (0) 0f f= =

25.If

2 3

sin cos

( ) ! sin cos2 2

nx x xn nf x n

a a a

= ,thenfind ny .

Solution

2 3

sin cos

( ) ! sin cos2 2

nx x xn nf x n

a a a

=

( )

2 3

(sin ) (cos )

( ) ! sin cos2 2

n n n n

n

D x D x D x

n nf x n

a a a

=

2 3

! sin cos2 2

( ) ! sin cos2 2

n

n nn x x

n nf x n

a a a

+ + =

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2 3

! sin cos2 2

(0) ! sin cos2 2

n

n nn

n nf n

a a a

= =0

26.Prove that the function ( ) 1 sinf x x= + is continuous but nor differentiable. Solution

1 sin , if sinx > 0

( )1-sinx , if sinx < 0

xf x

+= Graphical representation

2 31y =

1 siny x= +

It is observed fron the graph that the function is continuous but it is not differentiable at 0, , 2 , 3 ,....x = 27. If ( ) ( ) , ,

2 2x y f x f yf x y R+ + = , (0) 1, '(0) 1f f= = , then find the value

of (2).f Solution

If ( ) ( )2 2

x y f x f yf + + = then ( )f x a bx= + From the given data (0) 1f = 1a = and '(0) 1f = 1b = Therefore , ( ) 1f x x=

(2) 1 2 1f = = 28.Prove that the function { }( ) max 1 ,1 ,2f x x x= + is differentiable at all points except 1x =

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Solution Graphical interpretation

1y x= 1y x= +2y =

1x = 1x =

This figure shows that the functions is differentiable at all point except 1x = 29.Let

1, 0( )

x-1 , x 0x x

f x+

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1x = 1x =2x =

This figure shows that the functions is differentiable at all point except 1x = 30.If 1( ) sin (sin )f x x= ,then find '( )f x . Solution

,2 2

( ) , 32 2 2

3 5x-2 , 2 2

x x

f x x x

x

<

Demo-limits ,Continuity and Differentiability

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Transcript of Demo-limits ,Continuity and Differentiability