This document discusses functions, limits, and continuity. It begins by defining functions, domains, ranges, and some standard real functions like constant, identity, modulus, and greatest integer functions. It then covers limits of functions including one-sided limits and properties of limits. Examples are provided to illustrate evaluating limits using substitution and factorization methods. The overall objectives are to understand functions, domains, ranges, limits of functions and methods to evaluate limits.

1. Mathematics

2. Session
Functions, Limits and Continuity-1

3.  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

4. 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.

5. 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→

6. 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 . . . }

7. 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∴   

8. ( ) 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⇒ ⇒

9. 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⇒ ≤ < ∞ ∈

10. -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

11. 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}

12. 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

13. Modulus Function
( )
A function f : R R is defined by
x, x 0
f x = x =
-x, x < 0
→
≥


f(x) = xf(x) = - x
O
X
Y
Domain = R
Range = Non-negative real numbers

14. 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

15. 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.      

16. 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∈

17. 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
 
∈ ∩ ÷
 

18. 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
→
⊆

19. 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

20. 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

21. 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

22. 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 |

23. 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

24. 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.
→

25. 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
→

26. 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
→

27. 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
→

28. 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.
→ → →
⇔

29. 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→
∴

30. 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.
→
∴

31. 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
→

32. 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→

33. 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→
 
  

34. 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→
 
  

35. 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
 
 
 

36. 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

37. 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
→

38. 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→

39. 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

40. 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

41. 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⇒

42. Thank you