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ECTCHENNAI
Relations,Functions,limits,continuityanddifferentiability
1
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=
ECTCHENNAI
Relations,Functions,limits,continuityanddifferentiability
2
(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
ECTCHENNAI
Relations,Functions,limits,continuityanddifferentiability
3
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 =
ECTCHENNAI
Relations,Functions,limits,continuityanddifferentiability
4
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
=
ECTCHENNAI
Relations,Functions,limits,continuityanddifferentiability
5
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
ECTCHENNAI
Relations,Functions,limits,continuityanddifferentiability
6
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.
ECTCHENNAI
Relations,Functions,limits,continuityanddifferentiability
7
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 =
ECTCHENNAI
Relations,Functions,limits,continuityanddifferentiability
8
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 .
ECTCHENNAI
Relations,Functions,limits,continuityanddifferentiability
9
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
ECTCHENNAI
Relations,Functions,limits,continuityanddifferentiability
10
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 =
ECTCHENNAI
Relations,Functions,limits,continuityanddifferentiability
11
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 + + + =
ECTCHENNAI
Relations,Functions,limits,continuityanddifferentiability
12
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 + + +
ECTCHENNAI
Relations,Functions,limits,continuityanddifferentiability
13
= 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
ECTCHENNAI
Relations,Functions,limits,continuityanddifferentiability
14
( )( )( ) 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
ECTCHENNAI
Relations,Functions,limits,continuityanddifferentiability
15
=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
ECTCHENNAI
Relations,Functions,limits,continuityanddifferentiability
16
= 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
< <
ECTCHENNAI
Relations,Functions,limits,continuityanddifferentiability
17
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
ECTCHENNAI
Relations,Functions,limits,continuityanddifferentiability
18
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
ECTCHENNAI
Relations,Functions,limits,continuityanddifferentiability
19
( ) ( )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
+ + =
ECTCHENNAI
Relations,Functions,limits,continuityanddifferentiability
20
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 =
ECTCHENNAI
Relations,Functions,limits,continuityanddifferentiability
21
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+
ECTCHENNAI
Relations,Functions,limits,continuityanddifferentiability
22
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
<