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 off 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 off = -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 off = 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 off = 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 xR

-3 3cos2x 3 for all xR

-2 1 + 3cos2x 4 for all xR

-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 off = R

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

O

450

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 = R

Range = Non-negative real numbers

y = sinx

– O

y

2

1

x–2

–

O

y

–1

2

1

x–2

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 : ƒ g: D D R such that

1 2ƒ g 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, xR, 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 xR

2kx2 = 0 for all xR

k = 0

-x3 - kx2 - 2x = - x3 + kx2 - 2x for all xR

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 iffl im 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

x 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

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)=nx a

lim 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 - aIfl im = 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

4a =81

a=± 3

Thank you