Mathematics

Session

Functions, Limits and Continuity-1

Function

Domain and Range

Some Standard Real Functions

Algebra of Real Functions

Even and Odd Functions

Limit of a Function; Left Hand and Right Hand Limit

Algebraic Limits : Substitution Method, Factorisation Method, Rationalization Method

Standard Result

Session Objectives

Function

If f is a function from a set A to a set B, we represent it by ƒ : A B→

If A and B are two non-empty sets, then a rule which associateseach element of A with a unique element of B is called a functionfrom a set A to a set B.

( )y = ƒ x .x A to y B,∈ ∈If f associates then we say that y is the image of the

element x under the function or mapping and we write

Real Functions: Functions whose co-domain, is a subset of R are called real functions.

Domain and Range

The set of the images of all the elements under the mapping or function f is called the range of the function f and represented by f(A).

( ) ( ){ }The range of f or ƒ A = ƒ x :x A∈ ( )and ƒ A B⊆

The set A is called the domain of the function and the set B is called co-domain.

ƒ : A B→()

Valueofafunction:

IfaA,thenfaiscalledthevalueoffata. ∈

Domain and Range (Cont.)

For example: Consider a function f from the set of natural

numbers N to the set of natural numbers N

i.e. f : N →N given by f(x) = x2

Domain is the set N itself as the function is defined for all values of N.

Range is the set of squares of all natural numbers.

Range = {1, 4, 9, 16 . . . }

Example– 1

Find the domain of the following functions:

( ) ( ) 2i f x = 9-x ( ) 2

xii f(x)=

x -3x+2

( ) 2Solution: We have f x = 9-x

( )The function f x is defined for

[ ]-3 x 3 x -3, 3⇒ ≤ ≤ ⇒ ∈

( ) ( )2 29-x 0 x -9 0 x-3 x+3 0≥ ⇒ ≤ ⇒ ≤

Domain of f = -3, 3∴

( ) 2

xSolution: ii We have f(x)=

x -3x+2

The function f(x) is not defined for the values of x for which the

denominator becomes zero

Hence, domain of f = R – {1, 2}

Example– 1 (ii)

( ) ( )2i.e. x -3x+2=0 x-1 x-2 =0 x=1, 2⇒ ⇒

Example- 2

[ )Hence, range of f = 0 , ∞

Find the range of the following functions:

( ) ( )i f x = x-3 ( ) ( )ii f x = 1+ 3cos2x

( ) ( )Solution: i We have f x = x-3

( )f x is defined for all x R.

Domain of f = R

∈

∴

| x - 3 | 0 for all x R≥ ∈

| x - 3 | for all x R0⇒ ≤ < ∞ ∈

( )f x for all x R0⇒ ≤ < ∞ ∈

-1 ≤ cos2x ≤ 1 for all x∈R

⇒-3 ≤ 3cos2x ≤ 3 for all x∈R

⇒-2 ≤ 1 + 3cos2x ≤ 4 for all x∈R

⇒ -2 ≤ f(x) ≤ 4

Hence , range of f = [-2, 4]

Example – 2(ii)

( ) ( )Solution : ii We have f x = 1+ 3cos2x

( )Domain of cosx is R. f x is defined for all x R

Domain of f = R

∴ ∈

∴

Q

Some Standard Real Functions (Constant Function)

( )A function f : R R is defined by

f x = c for all x R, where c is a real number.fixed

→∈

O

Y

X

(0, c) f(x) = c

Domain = R

Range = {c}

Domain = R

Range = R

Identity Function

( )A function I : R R is defined by

I x = x for all x R

→∈

X

Y

O450

I(x) = x

Modulus Function

( )

A function f : R R is defined by

x, x 0f x = x =

-x, x < 0

→≥

f(x) = xf(x) = - x

OX

Y

Domain = RRange = Non-negative real numbers

y = sinx

– π O

y

2 π

1

x– 2 π π

– πO

y

– 1

2 π

1

x– 2 π π

y = |sinx|

Example

Greatest Integer Function

= greatest integer less than or equal to x.

( )A function f : R R is defined by

f x = x for all x R

→∈

For example : 2.4 = 2, -3.2 = -4 etc.

Algebra of Real Functions

1 2Let ƒ :D R and g:D R be two functions. Then,→ →

1 2Addition: ƒ +g:D D R such that∩ →

( ) ( ) ( ) ( ) 1 2ƒ+g x = ƒ x +g x for all x D D∈ ∩

1 2Subtraction: ƒ - g:D D R such that∩ →

( ) ( ) ( ) ( ) 1 2ƒ - g x = ƒ x - g x for all x D D∈ ∩

Multiplication by a scalar: For any real number k, the function kf isdefined by

( ) ( ) ( ) 1kƒ x =kƒ x such that x D∈

Algebra of Real Functions (Cont.)

1 2Product : ƒg:D D R such that∩ →

( ) ( ) ( ) ( ) 1 2ƒg x = ƒ x g x for all x D D∈ ∩

( ){ }1 2ƒ

Quotient : D D - x :g x = 0 R such thatg: ∩ →

( ) ( )( ) ( ){ }1 2

ƒ xƒx = for all x D D - x :g x = 0

g g x

∈ ∩ ÷

Composition of Two Functions

1 2Let ƒ :D R and g:D R be two functions. Then,→ →

( ) ( )( ) ( ) ( )2fog:D R such that

fog x = ƒ g x , Range of g Domain of ƒ

→

⊆

( ) ( )( ) ( ) ( )1gof :D R such that

gof x =g f x , Range of f Domain of g

→

⊆

Let f : R → R+ such that f(x) = ex and g(x) : R+ → R such that g(x) = log x, then find

(i) (f+g)(1) (ii) (fg)(1) (iii) (3f)(1) (iv) (fog)(1) (v) (gof)(1)

(i) (f+g)(1) (ii) (fg)(1) (iii) (3f)(1) = f(1) + g(1) =f(1)g(1) =3 f(1) = e1 + log(1) =e1log(1) =3 e1

= e + 0 = e x 0 =3 e = e = 0

Example - 3

Solution :

(iv) (fog)(1) (v) (gof)(1) = f(g(1)) = g(f(1)) = f(log1) = g(e1) = f(0) = g(e) = e0 = log(e) =1 = 1

Find fog and gof if f : R → R such that f(x) = [x] and g : R → [-1, 1] such that g(x) = sinx.

Solution: We have f(x)= [x] and g(x) = sinx

fog(x) = f(g(x)) = f(sinx) = [sin x]

gof(x) = g(f(x)) = g ([x]) = sin [x]

Example – 4

Even and Odd Functions

Even Function : If f(-x) = f(x) for all x, then f(x) is called an even function.

Example: f(x)= cosx

Odd Function : If f(-x)= - f(x) for all x, then f(x) is called an odd function.

Example: f(x)= sinx

Example – 5

( ) 2Solution : We have f x = x - | x |

( ) ( )2f -x = -x - | -x |∴

( ) 2f -x = x - | x |⇒

( ) ( )f -x = f x⇒

( )f x is an even function.∴

Prove that is an even function.2x - | x |

Example - 6

Let the function f be f(x) = x3 - kx2 + 2x, x∈R, then

find k such that f is an odd function.

Solution: The function f would be an odd function if f(-x) = - f(x)

⇒ (- x)3 - k(- x)2 + 2(- x) = - (x3 - kx2 + 2x) for all x∈R

⇒ 2kx2 = 0 for all x∈R

⇒ k = 0

⇒ -x3 - kx2 - 2x = - x3 + kx2 - 2x for all x∈R

Limit of a Function

2(x - 9) (x -3)(x+3)If x 3, f(x)= = =(x+3)

x -3 (x -3)≠

x 2.5 2.6 2.7 2.8 2.9 2.99 3.01 3.1 3.2 3.3 3.4 3.5

f(x) 5.5 5.6 5.7 5.8 5.9 5.99 6.01 6.1 6.2 6.3 6.4 6.5

2x -9f(x)= is defined for all x except at x = 3.

x -3

As x approaches 3 from left hand side of the number line, f(x) increases and becomes close to 6

-x 3lim f(x)= 6i.e.→

Limit of a Function (Cont.)

Similarly, as x approaches 3 from right hand side of the number line, f(x) decreases and becomes close to 6

+x 3i.e. lim f(x)= 6

→

x takes the values2.912.952.9991..2.9999 ……. 9221 etc.

x 3≠

Left Hand Limit

x

3

Y

OX

-x 3lim→

x takes the values 3.13.0023.000005……..3.00000000000257 etc.

x 3≠

Right Hand Limit

3X

Y

Ox

+x 3lim→

Existence Theorem on Limits

( ) ( ) ( )- +x a x a x a

lim ƒ x exists iff lim ƒ x and lim ƒ x exist and are equal.→ → →

( ) ( ) ( )- +x a x a x a

lim ƒ x exists lim ƒ x = lim ƒ xi.e.→ → →

⇔

Example – 7

Which of the following limits exist:

( )x 0

xi lim

x→[ ]

5x

2

(ii) lim x→

( ) ( ) xSolution : i Let f x =

x

( ) ( ) ( )- h 0 h 0 h 0x 0

0 - h -hLHL at x = 0 = lim f x = limf 0 - h =lim =lim = -1

0 - h h→ → →→

( ) ( ) ( )+ h 0 h 0 h 0x 0

0 + h hRHL at x = 0 = lim f x = limf 0 + h =lim =lim = 1

0 + h h→ → →→

( ) ( )- +x 0 x 0

lim f x lim f x→ →

≠Qx 0

xlim does not exist.

x→∴

Example - 7 (ii)

( ) [ ]Solution:(ii) Let f x = x

( )h 0 h 05

x2

5 5 5LHL at x= = lim f x =limf -h =lim -h =2

2 2 2− → →→

÷ ÷

( )h 0 h 05

x2

5 5 5RHL at x= = lim f x =limf +h =lim +h =2

2 2 2+ → →→

÷ ÷

( ) ( )5 5

x x2 2

lim f x lim f x− +

→ →

=Q [ ]5

x2

lim x exists.→

∴

Properties of Limits

( )x a x a x a

i lim [f(x) g(x)]= lim f(x) lim g(x)= m n→ → →

± ± ±

( )x a x a

ii lim [cf(x)]= c. lim f(x)= c.m→ →

( ) ( )x a x a x a

iii lim f(x). g(x) = lim f(x) . lim g(x)= m.n→ → →

( ) x a

x ax a

lim f(x)f(x) m

iv lim = = , provided n 0g(x) lim g(x) n

→

→→

≠

If and

where ‘m’ and ‘n’ are real and finite then

x alim g(x)=n

→x alim f(x)= m

→

The limit can be found directly by substituting the value of x.

Algebraic Limits (Substitution Method)

( )2

x 2For example : lim 2x +3x+4

→

( ) ( )2= 2 2 +3 2 +4 = 8+6+4 =18

2 2

x 2

x +6 2 +6 10 5lim = = =

x+2 2+2 4 2→

Algebraic Limits (Factorization Method)

When we substitute the value of x in the rational expression it

takes the form 0

.0

2

2x 3

x -3x+2x-6=lim

x (x-3)+1(x-3)→

2x 3

(x-3)(x+2)=lim

(x +1)(x-3)→

2 2x 3

x-2 3-2 1=lim = =

10x +1 3 +1→

2

3 2x 3

x -x-6 0For example: lim form

0x -3x +x-3→

Algebraic Limits (Rationalization Method)

When we substitute the value of x in the rational expression it

takes the form0

, etc.0

∞∞

[ ]2 2

2 2x 4

x -16 ( x +9+5)=lim × Rationalizing the denominator

( x +9 -5) ( x +9+5)→

22

2x 4

x -16=lim ×( x +9+5)

(x +9-25)→

22

2x 4

x -16=lim ×( x +9+5)

x -16→

2 2

x 4=lim( x +9+5)= 4 +9+5 =5+5=10

→

2

2x 4

x -16 0For example: lim form

0x +9 -5→

Standard Result

n nn-1

x a

x -alim =n a

x-a→

If n is any rational number, then

0form

0

3

2x 5

x -125Evaluate: lim

x -7x+10→

( ) 333

2 2x 5 x 5

x - 5x -125Solution: lim =lim

x -7x+10 x -5x-2x-10→ →

Example – 8 (i)

2

x 5

(x-5)(x +5x+25)=lim

(x-2)(x-5)→

2

x 5

(x +5x+25)=lim

x-2→

25 +5×5+25 25+25+25= = =25

5-2 3

2

x 3

1 1Evaluate: lim (x -9) +

x+3 x-3→

2

x 3

1 1Solution: lim (x -9) +

x+3 x-3→

x 3

x-3+x+3=lim(x+3)(x-3)

(x+3)(x-3)→

Example – 8 (ii)

=2×3=6

x 3=lim 2x

→

x a

a+2x - 3xEvaluate:lim

3a+x -2 x→

x a

a+2x - 3xSolution: lim

3a+x -2 x→

[ ]x a

a+2x - 3x 3a+x +2 x=lim × Rationalizing the denominator

3a+x -2 x 3a+x +2 x→

Example – 8 (iii)

x a

a+2x - 3x=lim × 3a+x +2 x

3a+x-4x→

[ ]x a

3a+x +2 x a+2x + 3x=lim × a+2x - 3x× Rationalizing thenumerator

3(a-x) a+2x + 3x→

x a

3a+x +2 x a+2x-3x=lim ×

3(a-x)a+2x + 3x→

Solution Cont.

x a

3a+x +2 x a-x=lim ×

3(a-x)a+2x + 3x→

x a

3a+x +2 x 1=lim ×

3a+2x + 3x→

3a+a+2 a 1 2 a+2 a 1= × = ×

3 3a+2a+ 3a 3a+ 3a

4 a 1 2= × =

32 3a 3 3

2x 1

3+x - 5-xEvaluate: lim

x -1→

2x 1

3+x - 5-xSolution: lim

x -1→

[ ]2x 1

3+x - 5-x 3+x + 5-x=lim × Rationalizing the numerator

x -1 3+x + 5-x→

Example – 8 (iv)

2x 1

3+x-5+x 1=lim ×

x -1 3+x + 5-x→ x 1

2(x-1) 1=lim ×

(x-1)(x+1) 3+x + 5-x→

( ) ( )x 1

2=lim

x+1 3+x + 5-x→

2 1= =

42( 4+ 4)

( ) ( )2

=1+1 3+1+ 5-1

5 5

x a

x -aIf lim = 405, find all possible values of a .

x -a→

5 5

x a

x -aSolution: We have lim = 405

x-a→

Example – 8 (v)

n n5-1 n-1

x a

x -a5 a = 405 lim =na

x-a→

⇒ ÷

Q

4a =81⇒

a=± 3⇒

Thank you