Functions limits and continuity

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 associates
each element of A with a unique element of B is called a function
from 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→

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:
( ) ( ) 2
i f x = 9- x ( ) 2
x
ii f(x)=
x -3x+2
( ) 2
Solution: We have f x = 9- x
( )The function f x is defined for
[ ]-3 x 3 x -3, 3⇒ ≤ ≤ ⇒ ∈
( ) ( )2 2
9- x 0 x -9 0 x-3 x+3 0≥ ⇒ ≤ ⇒ ≤
Domain of f = -3, 3∴   

( ) 2
x
Solution: 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)
( ) ( )2
i.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
O
450
I(x) = x

Modulus Function
( )
x, x 0
f x = x =
-x, x < 0
→
≥


f(x) = xf(x) = - x
O
X
Y
Domain = R
Range = Non-negative real numbers

y = sinx
– π O
y
2 π
1
x
– 2 π π
– π
O
y
– 1
2 π
1
x
– 2 π π
y = |sinx|
Example

Greatest Integer Function
= greatest integer less than or equal to x.
( )
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 is
defined 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 2
ƒ
Quotient : D D - x : g x = 0 R such that
g
: ∩ →
( )
( )
( )
( ){ }1 2
ƒ xƒ
x = 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) =e1
log(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
( ) 2
Solution : We have f x = x - | x |
( ) ( )2
f -x = -x - | -x |∴
( ) 2
f -x = x - | x |⇒
( ) ( )f -x = f x⇒
( )f x is an even function.∴
Prove that is an even function.
2
x - | 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
2
x - 9
f(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 3
lim 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 3
i.e. lim f(x) = 6
→

x takes the values
2.91
2.95
2.9991
..
2.9999 ……. 9221 etc.
x 3≠
Left Hand Limit
x
3
Y
O
X
-x 3
lim
→

x takes the values 3.1
3.002
3.000005
……..
3.00000000000257 etc.
x 3≠
Right Hand Limit
3
X
Y
O
x
+x 3
lim
→

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
x
i lim
x→
[ ]5
x
2
(ii) lim x
→
( ) ( )
x
Solution : i Let f x =
x
( ) ( ) ( )- h 0 h 0 h 0x 0
0 - h -h
LHL at x = 0 = lim f x = limf 0 - h =lim =lim = -1
0 - h h→ → →→
( ) ( ) ( )+ h 0 h 0 h 0x 0
0 + h h
RHL 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
→ →
≠Q x 0
x
lim does not exist.
x→
∴

Example - 7 (ii)
( ) [ ]Solution:(ii) Let f x = x
( ) h 0 h 05
x
2
5 5 5
LHL at x = = lim f x =limf -h =lim -h =2
2 2 2− → →
→
     
 ÷  ÷       
( ) h 0 h 05
x
2
5 5 5
RHL at x = = lim f x =limf +h =lim +h =2
2 2 2+ → →
→
     
 ÷  ÷       
( ) ( )5 5
x x
2 2
lim f x lim f x− +
→ →
=Q [ ]5
x
2
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 a
x a
lim f(x)
f(x) m
iv lim = = , provided n 0
g(x) lim g(x) n
→
→
→
≠
If and
where ‘m’ and ‘n’ are real and finite then
x a
lim g(x)= n
→x a
lim f(x)= m
→

The limit can be found directly by substituting the value of x.
Algebraic Limits (Substitution Method)
( )2
x 2
For example : lim 2x +3x + 4
→
( ) ( )2
= 2 2 +3 2 + 4 = 8+6+ 4 =18
2 2
x 2
x +6 2 +6 10 5
lim = = =
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 0
For 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 form
0
, etc.
0
∞
∞
[ ]
2 2
2 2x 4
x -16 ( x +9 +5)
=lim × Rationalizing the denominator
( x +9 -5) ( x +9 +5)→
2
2
2x 4
x -16
=lim ×( x +9 +5)
(x +9-25)→
2
2
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 0
For example: lim form
0x +9 -5→
 
  

Standard Result
n n
n-1
x a
x - a
lim = n a
x - a→
If n is any rational number, then
0
form
0
 
 
 

3
2
x 5
x -125
Evaluate: lim
x -7x+10→
( )
333
2 2x 5 x 5
x - 5x -125
Solution: 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→
2
5 +5×5+25 25+25+25
= = =25
5-2 3

2
x 3
1 1
Evaluate: lim (x -9) +
x+3 x-3→
 
  
2
x 3
1 1
Solution: 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 - 3x
Evaluate:lim
3a+x -2 x→
x a
a+2x - 3x
Solution: 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- x
Evaluate: lim
x -1→
2x 1
3+x - 5- x
Solution: 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 -a
If lim = 405, find all possible values of a.
x-a→
5 5
x a
x -a
Solution: We have lim = 405
x-a→
Example – 8 (v)
n n
5-1 n-1
x a
x -a
5 a = 405 lim = na
x-a→
 
⇒  ÷
 
Q
4
a =81⇒
a=± 3⇒

Functions limits and continuity

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