4. Meaning of x approaches to a :
An independent variable x approaches (tending) to a is denoted by
When we write , we shall assume the following:
x a→
x a→
4
When we write , we shall assume the following:
(i) x is a real variable.
(ii) A is a finite real number whose value is fixed.
(iii) x passes successively through a number of values so that if we
consider any particular value taken by x , we may discriminate
the values that precede and the values that follow and none of
the values can be looked upon as the last.
x a→
5. Definition-1 :
Or lim x = a means:
Given any (no matter how small) the successive values of x will
x a→
0ε>
5
Given any (no matter how small) the successive values of x will
ultimately satisfy the inequality i.e ultimately x lies in
the open interval i.e ultimately x satisfies
and
0ε>
0 x a< − <ε
( )a ,a ;x a−ε + ε ≠
a x a−ε < < a x a< < + ε
6. When we say that x tends to a from the left and we write
Or but when we say that x
tends to a from the right and we write or
a x a :−ε < <
x a or x a−
→ − → 0
x a−
→ a x a ;< < + ε
x a→ +tends to a from the right and we write orx a→ +
x a 0 or x a+
→ + →
6
Remember
0 0
x a x a x a− +
→ ⇒ → ∧ →
7. Meaning of x approaches to :
An independent variable x approaches (tending) to is denoted by
When we write , we shall assume the following:
(i) x is a real variable. which assumes successive values so that
x → +∞
x → +∞
+∞
∞
7
(i) x is a real variable. which assumes successive values so that
one can discriminate values that precede and the values that
follow and that no value is the last value attained by x
(ii) is not a number at all; it is a symbol whose meaning will
be made clear in the following lines
now means : the successive values of x ultimately become and
remain more than any arbitrary positive number G (no matter how
large) given in advance
+∞
x → ∞
8. Meaning of x approaches to :
An independent variable x approaches (tending) to is denoted by
When we write , we shall assume the following:
(i) x is a real variable.
x→−∞
x →−∞
−∞
−∞
8
(i) x is a real variable.
(ii) is same as
now means : the successive values of x ultimately become and
remain less than −G (where G is any arbitrary positive number) given
in advance.
x →−∞ ( x)− → +∞
x →−∞
9. Limit of a function at a point:
A function f(x) is said to tend to a limit l as x tends to a if to each
given , there exists a positive number (depending on ) such that0ε > δ εgiven , there exists a positive number (depending on ) such that
| f (x) l | whenever 0 | x a |ε δ− < < − <
i.e,. for all those values of x (except at x =a) which
belongs to . This is denoted by
( ) ( , )f x l lε ε∈ − +
( , )a aδ δ− + lim ( )
x a
f x l
→
=
9
10. Left-hand and right-hand limits
f(x) is said to tend to l as x tends to a through values less than a, if to
each such that0, 0,ε δ> ∃ >
| ( ) | whenf x l a x aε δ− < − < <
So that whenever
The limit in this case is called the left-hand limit (L.H.L) and is
denoted by f(a-0). Thus
10
| ( ) | whenf x l a x aε δ− < − < <
( ) ( , )f x l lε ε∈ − + ( , )x a aδ∈ −
0
( 0) lim ( )
x a
f a f x
→ −
− =
11. Similarly, if f(x) tends to l as x tends a through values which are
greater than a i.e. if given such that0, 0ε δ> ∃ >
| ( ) | when ,f x l a x aε δ− < < < +
then f(x) is said to tend to l from the right and the limit so obtained is
called the right hand limit (R.H.L) and is denoted by f(a+0).
we write
11
| ( ) | when ,f x l a x aε δ− < < < +
0
( 0) lim ( )
x a
f a f x
→ +
+ =
12. Existence of a limit at a point.
f(x) is said to tend to a limit as x tends to ‘a’ if both the left and right
hand limits exist and are equal, and their common value is called the
limit of the functionlimit of the function
i) To find f(a-0) or we first put x = a-h, h>0 in f(x) and then
take the limit as Thus
ii) To find f(a+0) or we first put x = a+h, h>0 in f(x) and
then take the limit as Thus
0
lim ( )
x a
f x
→ −
0h → + 0 0
lim ( ) lim ( )
x a h
f x f a h
→ − → +
= −
0
lim ( )
x a
f x
→ +
0h → + 0 0
lim ( ) lim ( )
x a h
f x f a h
→ + → +
= +
12
How to Find L.H.L & R.H.L ?
13. Limits at Infinity And Infinite Limits
i)
A function f(x) is said to tend to l as if given however
small, number k (depending on ) s.t
lim ( )
x
f x l
→∞
=
x → ∞ 0,ε >
a ve∃ + ε
| ( ) | . ., 1 ( ) .f x l x k i e f x l x kε ε ε− < ∀ ≥ − < < + ∀ ≥
ii)
A function f(x) is said to tend to l as if given however
small, number k(depending on ) s.t
| ( ) | . ., 1 ( ) .f x l x k i e f x l x kε ε ε− < ∀ ≥ − < < + ∀ ≥
lim ( )
x
f x l
→−∞
=
x → −∞ 0,ε >
a ve∃ + ε
| ( ) | . ., ( )f x l x k i e l f x l x kε ε ε− < ∀ ≤ − − < < + ∀ ≤ −
13
14. iii)
A function f(x) is said to tend to as x tends to a, if given k>0,
however large, number s.t.
∞
a+ve∃ δ
lim ( )
x a
f x
→
= ∞
iv)
A function f(x) is said to tend to as x tend to a, if given k>0,
however large, number s.t
( ) for 0 | |f x k x a δ> < − <
− ∞
a+ve∃ δ
lim ( )
x a
f x
→
= −∞
( ) for 0 | |f x k x a δ< − < − <
14
15. v)
A function f(x) is said to tend to as , if given k>0, however
large, number s.t.
∞
a∃
( )f x k x k′> ∀ ≥
lim ( )
x
f x
→∞
= ∞
x → ∞
k 0′>
vi)
A function f(x) is said to tend to as , if given k>0,
however large, number s.t
( )f x k x k′> ∀ ≥
− ∞
a∃
lim ( )
x
f x
→∞
= −∞
( )f x k x k′< − ∀ ≥
15
x → ∞
0k′ >
16. vii)
A function f(x) is said to tend to as , if given k>0, however
large, number s.t.
∞
a∃
( )f x k x k′> ∀ ≤ −
lim ( )
x
f x
→−∞
= ∞
x → −∞
k 0′>
viii)
A function f(x) is said to tend to as , if given k>0,
however large, number s.t
( )f x k x k′> ∀ ≤ −
− ∞
a∃
lim ( )
x
f x
→−∞
= −∞
( )f x k x k′< − ∀ ≤ −
16
x → −∞
0k′ >
17. So From the above all definitions we conclude that there are four types
in limits
at finite point
17
at finite point
at infinite point
lim
value finite
value infinite
18. The existence of limit l of f(x) as is illustrated in the figure
l and being known, one can draw the horizontal lines
if only a of the point
x =a can be found s.t. all points of the graph of y =f(x) corresponding
to points x of the nbd (excepting perhaps
Limit of the Function
Geometric Point of View
x a→
ε
y l,y l and y l ,y l as x a= = −ε = +ε → → nbdδ−
y = f(x)to points x of the nbd (excepting perhaps
for the point x=a) lie within the
strip bounded by the line
The point P(a,l) may or may not
belong to the graph of y = f(x).
We are simply not concerned with it.
18
andy l x a= ±ε = ±δ
y = f(x)
y = f(x)
y = l +ε
y = l -ε
y = l
a a + δa -δ
p(a,l)
20. Indeterminate forms
If a function f(x) takes any of the following forms at ;x a=
0 00
, , ,0 ,0 , ,1
0
∞∞
∞ − ∞ × ∞ ∞
∞
then f(x) is said to be indeterminate at .x a=
Notes:Notes:
1. All indeterminate forms (only limiting value) can be calculated.
2. Infinity (∞ ) is a symbol and not a number and we can not draw ∞ on the paper.
3. Zero (0) is signless quantity i.e., neither positive nor negative and 0 is even.
20
21. 4. (i) ∞ + ∞ = ∞ (ii) ∞ × ∞ = ∞ (iii)
∞
∞ = ∞
(iv) 0 0∞
= (v)
0
∞
= ∞ (vi)
0
0=
∞
(vii)
0
0=
∞
where ais finite
a
(viii)
0
a
is undefined, if 0a ≠
(ix) 0 0ab a= ⇔ = or 0b = or 0a = , 0b = ,a and b are finite.
(x)
, (so not exists) if a>1
1, if a=1
0, if -1<a<1
not exists if a -1
a∞
∞
=
≤
(xi) a ,a ,a , ,0 ,a+ ∞ = ∞ ×∞ = ∞ − ∞ = −∞ ∞ − = ∞ − ∞ = −∞
0 ,+ ∞ = ∞ where a is a non-zero positive quantity.
21
22. Property 1
If f(x) is an even function, then ( ) ( )0 0
lim lim
x x
f x f x
→ − → +
=
Property 2
If f(x) is an odd function and ( )lim f x exists, then. ( )lim 0f x =If f(x) is an odd function and ( )0
lim
x
f x
→
exists, then. ( )0
lim 0
x
f x
→
=
Notes:
If a function involves square root function, modulus function, greatest
integer function, least integer function, fractional part function, signum
function etc. then limit of this function is likely to be two sided limit.
What is two sided limit?
22
23. 1. If ( ) ( ) ( )lim lim lim
x a x a x a
f x l f x f x l
→ → − → +
= ⇔ = = where lis finite and unique.
2. ( ) 0
1
lim lim
x x
f x f
x→±∞ →
=
3. (i)
0
1
lim
x x→ +
= +∞ and
0
1
lim
x x→ −
= − ∞
0
1
lim
x x→
∴ does not exist.
0x x→ + 0x x→ − 0x x→
X
Y
O0x→ − 0x→ +
23
24. (ii) 2 20 0
1 1
lim lim
x xx x→ + → −
∴ = + ∞ = 20
1
lim
x x→
∴ does not exists.
Since ∞ is a symbol, not a number which is not finite.
Y
X
O
24
25. Difference between ( )lim
x a
f x
→
and ( )f a
Let f(x) be a function and a be a point then following cases arise:
(i) ( )lim
x a
f x
→
exists but f(a) does not exist:
Consider the function ( )
2 2
x a
f x
x a
−
=
−x a−
∴ The value of function f(x) at x a= is of the form
0
0
which is indeterminate.
i.e., ( )
0
0
f a =
∴ f(a) does not exist.
25
26. But ( ) ( )
( )
( )
( )
2 2
0 0 0
lim lim lim lim 2 2
x a h h h
a h a
f x f a h a h a
a h a→ − → → →
− −
= − = = − =
− −
and ( ) ( )
( )
( )
2 2
0 0 0
lim lim lim lim 2 2
x a h h h
a h a
f x f a h a h a
a h a→ + → → →
+ −
= + = = + =
+ −0 0 0x a h h ha h a→ + → → →+ −
( ) ( )lim lim
x a x a
f x f x− +
→ →
∴ =
∴ ( )lim
x a
f x
→
exists.
Hence ( )lim
x a
f x
→
exist but f(a) does not exist.
26
27. (ii) ( )lim
x a
f x
→
does not exist but f(a) exists:
(iii) ( )lim f x and ( )f a both exists and are equal:(iii) ( )lim
x a
f x
→
and ( )f a both exists and are equal:
(iv) ( )lim
x a
f x
→
and ( )f a both exist but are unequal:
27
28. Standard Theorems on Limits
Let f be a function defined in a certain interval and containing ‘a’ as
interior point .
If ( )lim
x a
f x l
→
= and ( )lim ,
x a
g x m
→
= where l and m exist and finite, then
(1) ( ) ( ){ } ( ) ( )lim lim limf x g x f x g x l m+ = + = +(1) ( ) ( ){ } ( ) ( )lim lim lim
x a x a x a
f x g x f x g x l m
→ → →
+ = + = +
(2) ( ) ( ){ } ( ) ( )lim lim lim
x a x a x a
f x g x f x g x l m
→ → →
− = − = −
(3) ( ) ( ){ } ( )( ) ( )( )lim . lim lim .
x a x a x a
f x g x f x g x l m
→ → →
= =
(4)
( )
( )
( ){ }
( ){ }
lim
lim ,
lim
x a
x a
x a
f xf x l
g x mg x
→
→
→
= =
provided 0m ≠
28
29. (5) ( ){ }
( )
( ){ }
( )lim
lim lim
g x m
x a x a
x a
g x
f x f x l
→ →
→
= =
(6) ( ){ } ( )lim lim
x a x a
kf x k f x kl
→ →
= =
(7) If ( ) ( )f x g x≤ for all x then ( ) ( )lim limf x g x l m≤ ⇒ ≤(7) If ( ) ( )f x g x≤ for all x then ( ) ( )lim lim
x a x a
f x g x l m
→ →
≤ ⇒ ≤
(8) ( ) ( )lim lim
x a x a
f x f x l
→ →
= =
(9) ( ) ( )lim
lim
f x l
x a
x a
f x
e e e
→
→
= =
29
30. (10) ( ) ( ){ }lim lim ( ),
x a x a
ln f x ln f x ln l
→ →
= = only if 0l >
(11) ( )( ) ( ){ } ( )lim lim
x a x a
f g x f g x f m
→ →
⇔ = provided f is continuous at x m=
(12) Sandwitch / Squeeze theorem:
y
If ( ) ( ) ( )f x g x h x≤ ≤ ( )δ, δx a a∀ ∈ − +
and ( ) ( )lim lim ,
x a x a
f x l h x
→ →
= =
then lim
x a
g( x ) l
→
=
Corollary:
If ( ) ( )<f x g x in a certain nbd of a and lim ( ) 0
→
=
x a
g x , then lim ( ) 0
→
=
x a
f x .
h(x)
f(x)
y = g(x)
0
x
y
30
31. (13) NEIGHBOURHOOD PROPERTY:
If , lim ( )
→
=
x a
f x l then there exists some neighbourhood of a , at every point of which f(x)
will have the same sign as that of l.
Corollary:
Let lim ( ) =f x l , Suppose f(x) > 0 for every x in a certain nbd of a then l is not negative .Let lim ( )
→
=
x a
f x l , Suppose f(x) > 0 for every x in a certain nbd of a then l is not negative .
Similarly when f(x) < 0 for every x in a certain nbd of a then l is not positive.
Also remember that even when f(x) > 0 for all x in a certain nbd of a, still
lim ( )
→x a
f x may be 0.
Eg: f(x) = x 2
If f(x) < g(x) for a – h < x < a + h and
x a x a
limf (x) l andlimg(x) m
→ →
= = then ≤l m ..
31
32. (14) CAUCHY’S CRITERION FOR THE EXISTANCE OF LIMIT OF A FUNCTION:
The necessary and sufficient condition that a function f(x) may tend to a definite finite limit
l as →x a , is that , to any pre-assigned positive quantity ε , however small , there
corresponds a positive number δ , such that 2 1( ) ( )− <εf x f x for every pair x1, x2 of
values of x satisfying 0 < − < δx a and 0 < − < δx a .values of x satisfying 10 < − < δx a and 20 < − < δx a .
And for limits when → ∞x ,
The necessary and sufficient condition that a function f(x) may tend to a definite finite limit
l as → ∞x , is that , to any pre-assigned positive quantity ε , however small , there
corresponds a positive number δ , such that 2 1( ) ( )− <εf x f x for every pair x1, x2
exceed δ
32
33. (15) LIMIT OF A FUNCTION DEFINED IN TERMS OF SEQUENCE:
Let f be defined in some nbd of a
Definition-1:-
lim ( )
0, 0
( )
→
= ⇒
∈ > ∃ δ >
− < ∈
x a
f x l
G iven a n y su ch th a t
f x l( )
0
− < ∈
∈ < − < δ
f x l
fo r a ll x x a
Definition-2:-
{ }
{ }
lim ( ) lim ( )
→
= ⇒ →
→ ∞
n
x a
n
f x l it of thesequence f x l
as n for all sequence x
(in the nbd of a where f is defined) which converge to a.
Remember that the above two definitions are equivalent.
33
34. Conceptual Problems
EXAMPLE-1:
Let
1,
( )
0, c
if x Q
f x
if x Q
∈
=
∈
Prove that lim ( )
x a
f x
→
does not exist where ‘a’ is any real number.
Solution:Solution:
Let { }nx be a sequence of rational numbers such that nx a→ and nx a≠ , n=1,2,3…..;Let { }ny
be a sequence of irrational numbers such that ny a→ and
ny a≠ , n=1,2,3…..;.Clearly ( ) 1, ( ) 0n nf x f y= = for n=1,2,3……
Therefore ( ) 1nf x → and ( ) 0nf y → and lim ( )
x a
f x
→
does not exist. Since ‘a’ is arbitrary, limit
does not exist for every real number.
34
35. EXAMPLE-2:
Let
,
( )
0, c
x if x Q
f x
if x Q
∈
=
∈
Prove that lim ( )
x a
f x
→
exists only if a=0 .
Solution: suppose a ≠ 0.
{ } { }Let { }nx be a sequence of rational numbers such that nx a→ and nx a≠ , n=1,2,3…..;Let { }ny
be a sequence of irrational numbers such that ny a→ and
ny a≠ , n=1,2,3…..;.Clearly ( ) , ( ) 0n n nf x x f y= = for n=1,2,3……
Therefore ( )n nf x x a= → and ( ) 0nf y → and lim ( )
x a
f x
→
does not exist. Since ‘a’ is arbitrary,
limit does not exist for every non-zero real number.
Now for any sequence { }nx , ( )n nf x x= or 0 and so if 0nx → then ( ) 0nf x → .Therefore , by
above definition
0
lim ( ) 0
x
f x
→
=
35
36. Important Expansions
(i) ( )
( ) ( )( )2 31 1 2
1 1 ...
1.2 1.2.3
n n n n n n
x nx x x
− − −
+ = + + + +
(ii)
2 3
1 ...
1! 2! 3!
x x x x
e = + + + +
1! 2! 3!
(iii) ( ) ( )
2 3
3
1 ...
1! 2! 3!
x xlna x x
a lna lna
2
= + + + +
(iv) ( )
2 3 4
1 ...; 1 1
2 3 4
x x x
ln x x x+ = − + − + − ≤ ≤
36
37. (v) 1 2 2 3 1
. ...
n n
n n n nx a
x ax a x a
x a
− − − −−
= + + + +
−
(vi)
3 5
sin ...
3! 5!
x x
x x= − + −
(vii)
2 4
cos 1 ...
2! 4!
x x
x = − + −
(viii)
3
52
tan ...
3 15
x
x x x= + + +
(ix)
2 3 2 2 5 2 2 2
1 71 . 1 .3 . 1 .3 .5
sin . ...
3! 5! 7!
x x
x x x−
= + + + +
37
38. (x)
3 5
1
tan
3 5
x x
x x−
= − + −
(xi)
2 4 6
1 5 61
sec 1 ...
2! 4! 6!
x x x
x−
= + + + +
2! 4! 6!
(xii) ( )
2 2 2
21 2 4 62 2.2 2.2 .4
sin ...
2! 4! 6!
x x x x−
= + + +
(xiii) ( )
1/ 211
1 1 ...
2 24
x x
x e x
+ = − + +
38
39. Standard Limits & Short cuts in Limits
If ( ) 0f x → when ,x a→ then
(1)
( )
( )
1
lim 1
f x
x a
e
f x→
−
=
(2)
( )
( )
( )
1
lim , 0
f x
c
ln c c
f x→
−
= >(2)
( )
( )lim , 0
x a
ln c c
f x→
= >
(3)
( )( )
( )
1
lim 1
x a
ln f x
f x→
+
=
(4)
( )
( )
sin
lim 1
x a
f x
f x→
=
(5)
( )
( )
tan
lim 1
x a
f x
f x→
=
39
40. (6)
( ){ } ( ){ }
( ) ( )
( ){ }
1
lim ,
n n
n
x a
f x f a
n f a n Q
f x f a
−
→
−
= ∈
−
(7) ( ){ }
( )1/
lim 1
f x
x a
f x e
→
+ =
(8) ( ){ }
( )/
lim 1
c f x bc
bf x e+ =(8) ( ){ }
( )/
lim 1
c f x bc
x a
bf x e
→
+ =
(9) ( )
1/
0
1
lim 1 lim 1
n
h
n h
e h
n→∞ →
+ = = +
(10) lim 1
n
a
n
a
e
n→∞
+ =
40
41. (11) ( )
1/
0
lim 1
h a
h
ah e
→
+ =
(12) 1
lim
n n
n
x a
x a
na
x a
−
→
−
=
−
(13) lim
m m
m nx a m
a −−
=(13) lim m n
n nx a
x a m
a
x a n
−
→
−
=
−
(14)
0
1
lim 1
x
x
e
x→
−
=
(15) ( )0
1
lim , 0
x
x
a
ln a a
x→
−
= >
(16)
( )
0
1
lim 1
x
ln x
x→
+
=
41
42. (17)
( )
( )0
log 1
lim log 0, 1a
a
x
x
e a a
x→
+
= > ≠
(18)
1 1
0 0 0 0
sin tan sin tan
lim 1 lim lim lim
x x x x
x x x x
x x x x
− −
→ → → →
= = = =
0
π
(where x is measured in radians) and ( )
0
0
sin
lim 180
180
c
x
x
x
π
π
→
= ° =
°
(19) ( )lim 0 0mx
ln x
m
x→∞
= >
(20)
( )
0
1 1
lim
m
x
x
m
x→
+ −
=
42
43. (21) ( ){ } ( ) ( ){ } ( )lim 1
lim
x a
x ag x f x g x
f x e
→
→
−
=
This is Very Very Important Formula in Limits
provided ( ) ( ), 1g x f x→ ∞ → for x a→
43
44. (22) If ( )lim 0
x a
f x α
→
= > and ( )lim
x a
g x β
→
= (a finite quantity)
then ( ){ }
( ) β
lim α
g x
x a
f x
→
=
n 2
1-cos ax na
(23).
n 2
2x 0
1-cos ax na
lim
x 2→
=
(24).
( )2 2n n
2x 0
n b acos ax cos bx
lim
x 2→
−−
=
(25).
n n n 2
n 2x 0
tan x sin x na
lim
x 2
+
+→
−
=
44
45. (26).
b/x ab
x 0
lim (1 ax) e→
+ =
(27).
bx
ab
x 0
a
lim 1 e
x→
+ =
(28).
d/x bcd
lim(cosax bsincx) e+ =(28).
d/x bcd
x 0
lim(cosax bsincx) e→
+ =
(29).
d/x bcd
x 0
lim (secax b tancx) e→
+ =
(30).
b/xx x x x
1 2 3 n
x 0
a a a a
lim
n→
+ + +−−−−+
= (a1a2------an)b/n
(31). ( )2
x 0
a
lim x ax b x
2→
+ + − =
45
46. (32). ∑ np
is the (p+1)th
degree polynomial function in the variable n and in the expansion.
∑ np
, the coefficient of np+1
is
1
1
+p
(33).
x
f (x)
lim
g(x)→∞
, when f(x) & g(x) are the polynomial functions in x of degrees m & n
respectively. Thenrespectively. Then
I. if m<n, then
x
f (x)
lim 0
g(x)→∞
=
II. if m=n, then
n
nx
f (x) coeff.of x in f(x)
lim
g(x) coeff.of x in g(x)→∞
=
III. If m>n, then
x
f(x)
lim
g(x)→∞
=∞, if coefficient of xm
in f(x) is +ve
= -∞, if coefficient of xm
in f(x) is –ve
46
48. (39). ( )21
2
2
2
1
2
0
bbc
edxcx
e
baxx
baxx
x
Lt −
++
=
++
++
→
(40). ( )functionintegergreatestthedenotes[.]
][
a
x
bax
x
Lt
=
+
∞→
1 1 1Lt
(41).
1 1 1
terms
2
Lt
n
n a( a d ) ( a d )( a d ) ad
+ +−−−−− =
→ ∞ + + +
(42).
[ ] [ ] [ ] [ ]
1
321
1
+
=
+−−−−−+++
∞→ +
p
x
n
xnxxx
n
Lt
p
pppp
where [.]denoted GIF
Remember
1) ( )0
lim 1 0
n
h
n
h
→
→∞
− = 2) ( )0
lim 1
n
h
n
h
→
→∞
+ = ∞
48
49. Methods For Finding Limit of the Function
Method-1: Direct Substitution Method
Directly substitute the point in the given expression if on substitution the expression
does not take indeterminate form and if on substitution we get a finite number, then the finite
number is the limit of the given expression.
Method-2: Factorization Method
( )f x
Consider
( )
( )
lim
x a
f x
g x→
.
If we put the value x a= the rational function
( )
( )
f x
g x
takes the meaningless forms
0
,
0
∞
∞
etc. then ( )x a− is a factor of both ( )f x and ( )g x which must be cancelled. After cancelling
out the common factor ( )x a− we again put x=a in the given expression then we get a
meaningful number or not. We have to repeat the above process till we get a meaningful
number
49
50. Note:
1. ( )( )1 2 3 2 4 3 2 1
...n n n n n n n n
x a x a x x a x a x a a x a− − − − − −
− = − + + + + + +
Where n is even or odd positive integer.
2. ( ) ( )( )11 2 3 2 4 3 1
... 1
nn n n n n n n
x a x a x x a x a x a a
−− − − − −
+ = + − + − + + −2. ( ) ( )( )11 2 3 2 4 3 1
... 1
nn n n n n n n
x a x a x x a x a x a a
−− − − − −
+ = + − + − + + −
Where n is odd positive integer. Also this formulae is not applicable when n is even.
50
51. 3. Synthetic Division Method:
To find the quotient when a polynomial is divisible by a binomial.
Let ( ) 1 2
0 1 2 1.....n n n
n nf x a x a x a x a x a− −
−= + + + + + be a polynomial of degree n and
let it is divisible by the binomial x-h.
If 1 2 3 4
...n n n n
Q b x b x b x b x b b− − − −
−= + + + + + + be the quotient and hereIf 1 2 3 12
... xo nn
Q b x b x b x b x b b −−
= + + + + + + be the quotient and here
remainder = 0, then the coefficient of Q can be found from the following table:
0a
0
0b
1a
0hb
1b
2a
1hb
2b
3b
2hb
3a 1na −
2nhb−
1nb −
......
......
......
na
1nhb −
0
h
Where 0 0 1 1 0 2 2 1; ; ;b a b a hb b a hb= = + = + 3 3 2 1 1 2;.....; n n nb a hb b a hb− − −= + = +
and 1 0n na hb −+ =
51
52. Method-3: Rationalization or Double rationalization Method
This method is used when either numerator or denominator or both have fractional powers
1 1 1
like , ,
2 3 4
etc
.After rationalization of the terms and cancelling common factors in
numerator and in denominator which gives the result.numerator and in denominator which gives the result.
Method-4: Standard limit Method
( )
1
lim where ‘n’ is a rational number
n n
n
x a
x a
n a
x a
−
→
−
=
−
52
53. Method-5: Limit at ∞
Step I: Write down the given expression in the form
( )
( )
f x
g x
(where ( ) 1g x ≠ ), if ( ) 1g x =
then multiplying above and below by conjugate of f(x).
Step II: Then divide each term of the numerator and denominator by k
x where
k is highest power of x in
numerator and denominator.
Step III: Then use the result lim 0kx
c
x→∞
= where c is a constant and 0k > .
Note: If then then | | and f then | |x x x i x x x→ ∞ = → −∞ = −
53
54. Method-6: Exponential and Logarithmic Limits
To evaluate the exponential and logarithmic limits we use the following results:
(i)
0
1
lim 1
x
x
e
x→
−
=
0
form
0
x 0
(ii)
0
1
lim
x
x
a
ln a
x→
−
=
0
form
0
(iii)
( )
0
1
lim 1
x
ln x
x→
+
=
0
form
0
54
55. Method-7:Exponential limits of the form 1∞
To evaluate the exponential limits of the form 1∞
we use the following results:
(i) ( )
1/
0
1
lim 1 lim 1
n
h
n h
e h
n→∞ →
+ = = +
( )
1/
n
ha
(ii) ( )
1/
0
lim 1 lim 1
ha
n h
a
e ah
n→∞ →
+ = = +
(iii) ( ){ }
( ) ( ){ } ( )lim 1
lim x a
f x g xg x
x a
f x e →
−
→
= provided ( ) ( ), 1g x f x→ ∞ → for x a→
Proof:
Let ( ){ }
( )
lim
g x
x a
P f x
→
=
( )( ){ }
( )
lim 1 1
g x
x a
f x
→
= + − ( )( )( )
( ) ( ) ( )1
1/ 1
lim 1 1
f x g x
f x
x a
f x
− −
→
= + −
( ) ( )lim 1
x a
f x g x
e →
−
=
55
56. Method-8: ’ L’ Hospital’s Rule
If ( )f x and ( )g x be two functions, such that ( ) ( ) 0f a g a= = and f and g are both
differentiable at everywhere in some neighbourhood of point a except possibly ‘a’.differentiable at everywhere in some neighbourhood of point a except possibly ‘a’.
Then
( )
( )
( )
( )
1
1
lim lim
x a x a
f x f x
g x g x→ →
= Provided ( )1
f a and ( )1
g a are not both zero
56
57. Note :
1. L’ Hospital’s rule is applicable only for forms
0
0
or
∞
∞
.
2. For other indeterminate forms we have to convert to
0
0
or
∞
∞
and then apply L’ Hospital’s
rule. Sometimes we have to repeat the process if the form is
0
0
or
∞
∞
again.
0 ∞
3. Method to convert other indeterminate form to either 0/0 or ∞ /∞ form.
(i)
1 1 0 0 0
0 0 0.0 0
−
∞ − ∞ = − = = form, in this case take L.C.M.
(ii)
1 1 1 0
0 or or
0 0 0
∞
×∞ = × ×∞ =
∞ ∞
form
(iii) log 1 log 1 0 / 0/0
1 ore e
e e e e e
∞
∞∞ ∞× ∞ ∞
= = = = form
(iv)
0
log 0 0log 00 0 0/0 /
0 ore e
e e e e e×∞ ∞ ∞
= = = =
(v)
0
log 0log0 0 0/0 /
ore e
e e e e e∞ ∞ ×∞ ∞ ∞
∞ = = = =
57
58. 4.
, 1
log 0
, 0 1
a
a
a
−∞ >
=
∞ < <
and
, 1
log
, 0 1
a
a
a
∞ >
∞ =
−∞ < <
5. Newton-Leibnitz’s formula: If
( )
( )
( ) ,
x
x
y f t dt
ψ
φ
= ∫ then
{ } { }1 1
( ) ( ) ( ) ( )
dy
f x x f x x
dx
ψ ψ φ φ= × − ×
This is Very Very Important Formula in Limits
58
61. CONTINUITY AT A POINT AND AN INTERVAL I
Continuity at a point:
Let f be defined on an interval I R⊆ .
Suppose ‘c’ is an interior point of I. The continuity of f at x=c may be defined in the following
way:
Definition-1:-Definition-1:-
f is said to be continuous at x = c if for any arbitrary positive number ε , no matter how small,
∃a positive numberδ such that for all points x x c I∈ − < δ∩ we have ( ) ( )f x f c− < ε .It
means if f is continuous at ‘c’ thenlim ( )
x c
f x
→
exists and is equal to f(c).
That is ‘f’ is continuous at x = c if the functional values f(x) are close to f(c) when x is close to c
.Thus for continuity of f at x = c, f must be defined at x = c and also in some nbd around c;
otherwise we cannot find lim ( )
x c
f x
→
which must exist and equals to f(c).
61
62. Hence for continuity of ‘f’ at a point x = c , we must have
1) ‘f’ is defined at x = c i.e., f(c) exists.
2) lim ( )
x c
f x
→
exists i.e., lim ( ) lim ( )
x c x c
f x f x
→ + → −
=
3) f(c) and lim ( )
x c
f x
→
must be equal. If any one of the above conditions fail, then f is NOT
continuous or discontinuous at x=c.
Definition-2:-
f is said to be continuous at x=c if every convergent sequence{ }nx in I having limit c, the
sequence { }( )nf x is convergent and lim ( ) ( )n
n
f x f c
→∞
= .
And if f is discontinuous at c iff there exist { }nx in I such that { }nx converges to c but
{ }( )nf x does not converges to f(c).
Remember that definition-1 is equivalent to definition-2.
62
63. Hence for continuity of ‘f’ at a point x = c , we must have
1) ‘f’ is defined at x = c i.e., f(c) exists.
2) lim ( )
x c
f x
→
exists i.e., lim ( ) lim ( )
x c x c
f x f x
→ + → −
=
x c→ x c x c→ + → −
3) f(c) and lim ( )
x c
f x
→
must be equal. If any one of the above conditions fail, then f is NOT
continuous or discontinuous at x=c.
63
64. Continuity in an Open Interval
A function f is said to be continuous in an open interval (a,b) if fi is continuous at every
point of (a,b). Thus f is continuous in the open interval (a, b) iff for every
limx c
c ( a,b ), f ( x ) f (c )→
∈ =
64
65. Continuity in a Closed Interval
A function f is said to be continuous in a closed interval [a, b] if it is
i) right continuous at a i.e., lim ( ) ( )x a
f x f a→ +
=
ii) Continuous in (a, b)
iii) Left continuous at b lim ( ) ( )
x b
f x f b
→ −
=
65
66. Continuity on a Set
A function f is said to be continuous on an arbitrary set ( )S R⊂ if for each 0ε < and for
every a S∈ , ∃ a real number 0δ > such that | ( ) ( ) |f x f a− <ε whenever x S∈ and
| |x a− <δ
Equivalently, a function f is said to be continuous on a set S if it continuous at every point
S, i.e. If for every ,lim ( ) ( )
x a
a S f x f a
→
∈ =
66
67. Remember
It should be noted that continuity of a function is the property of interval and is meaningful at
x = a only if the function has a graph in the immediate neighbourhood of x = a, not necessarily at
x = a. Hence it should not be mislead that continuity of a function is talked only in its domain.
Example:- Continuity of
1
f (x)
x 1
=
−
at x = 1 is meaningful but continuity of f (x) = ln x at x = – 2 in
meaningless.
Similarly if f (x) has a graph as shown then continuity at x = 0 is meaninglessSimilarly if f (x) has a graph as shown then continuity at x = 0 is meaningless
-1
1
Also continuity at x = a ⇒ existence of limit at x = a but existence of limit atx a= ⇒/ continuity at x =a.
Note: It should be remembered that all polynomial functions, trigonometric function, exponential and
logarithmic functions are continuous in their domain.
67
68. Discontinuity of a Function
A function f which is not continuous at a point ‘a’ is said to be discontinuous at the
point ‘a’. ‘a’ is called a point of discontinuous of f or f is said to have a discontinuity
at ‘a’.
A function which is discontinuous even at a single point of an interval is said to be
discontinuous in the interval
A function f can be discontinuous at a point x = a because of any one of the following
reasons:
i) f is not defined at ‘a’
ii) lim ( )
x a
f x
→
does not exist i.e., lim ( ) lim ( )
x a x a
f x f x
→ − → +
≠
iii) lim ( ) ( )x a
f x and f a→
both exist but are not equal
68
70. Type-1
(Removable discontinuity)
Here lim ( )
x a
f x
→
necessarily exists, but is either not equal to f (a) or f (a) is not defined.
In this case, therefore it is possible to
redefine the function in such a manner thatredefine the function in such a manner that
lim ( ) ( )
x a
f x f a
→
= and thus making the function continuous.
These discontinuities can be further classified as
70
71. Missing point discontinuity
a)
2
( 1)(9 )
( )
1
x x
f x
x
− −
=
−
at x =1
9
8
b)
2
4
( )
2
x
f x
x
−
=
−
at x=2
4
0 1
3-3
0 2
c)
sin
( )
x
f x
x
=
1
0π−
2
π
−
2
π π
2
π
71
72. Isolated point discontinuity
a) f (x) = [x]+[–x]
0
1
if x I
if x I
∈
=
− ∉
1
y
-2 2-1
b) f(x)=sgn(cos2x–2sinx+3)
= 2(2 + sin x)(1 – sinx)
0 2
2
1 2
if x n
if x n
π
π
π
π
= +
=
+ ≠ +
1
x
-2
1
1 2
2-1
-2 -1
O
1 2
2
if x n
π
π+ ≠ +
has an isolated point at x = 0
discontinuity as x=2n +
2
π
π
x
y
1
3
2
π O
2
π 5
2
π
72
73. Type-2
(Non-Removable discontinuity)
Here
1
lim ( )
x
f x
→
does not exists and therefore it is not possible to redefine the function
in any manner to make it continuous. Such discontinuities can be further classified into
three types.
(a) Finite type (both limits finite and unequal)
(b) Infinite type (at least one of the two limit are infinity)
(c) Oscillatory (limits oscillate between two finite quantities)
73
74. Finite type
Example: 1
1
0
1
lim tan
x x
−
→
(0 )
2
(0 )
2
f
f
π
π
+
−
=
= −
; jump π=
Example: 2 ; jump 2=
| sin |
lim
x
(0 ) 1f +
=
Example: 2 ; jump 2=
0
lim
x x→
(0 ) 1f −
= −
1
; jump
2
=
2
[ ]
lim
x
x
x→
(2 ) 1
1
(2 )
2
f
f
+
−
=
=
Example: 3
Note : In this case non negative difference between the two limits is called the Jump of
discontinuity. A function having a finite number of jumps in a given interval I is called a PIECE
WISE CONTINUOUS or SECTIONALLY CONTINUOUS function in this interval.
74
75. Infinite type
Example: 1 ( ) 1
1
x
f x at x
x
= =
−
(1 )
(1 )
f
f
+
−
= −∞
= +∞
0
2
f
π +
=
Example: 2
Example: 3 2
1
( ) 0f x at x
x
= =
(0 )
(0 )
f
f
+
−
=∞
=∞
tan
( ) 2
2
x
f x at x
π
= =
0
2
2
f
f
π −
=
=∞
75
76. Oscillatory Type
Example: 1
1
( ) sinf x
x
= at x = 0 Oscillates between – 1 & 1
Example: 2
1
( ) cosf x
x
= at x = 0 Oscillates between – 1 & 1
Example: 3 oscillates between 0 & 1 at x = 0
1
( ) 1 sin(ln| |)
3
f x x
= +
76
77. In Brief
Removable Discontinuity
If lim ( )
x a
f x
→
exists but is not equal f(a) , the f is said to have a removable
discontinuity at ‘a’.
This type of discontinuity can be removed by defining a new function g as
( )
( )
lim ( )
x a
f x if x a
g x
f x if x a
→
≠= =
The g is continuous at ‘a’
Note: If lim ( )
x a
f x
→
does not exist, then the function cannot be made continuous, no
matter how define f(a).
77
78. Discontinuity of First Kind (or Jump Discontinuity)
If lim ( )
x a
f x
→ −
and lim ( )
x a
f x
→ +
both exist but are unequal, then f is said to have a
discontinuity of first kind at ‘a’ or jump discontinuity at ‘a’discontinuity of first kind at ‘a’ or jump discontinuity at ‘a’
f is said to have a discontinuity of the first kind from the right at ‘a’ if lim ( )
x a
f x
→ +
exists but is not equal to f(a)
78
79. Discontinuity of Second kind
If neither lim ( )
x a
f x
→ −
nor lim ( )
x a
f x
→ +
exist, then f is said to have a discontinuity of second
kind at ‘a’.
f is said to have a discontinuity of the second kind from the left at ‘a’ if lim ( )
x a
f x
→ −
does not
x a→ −
exist.
f is said to have a discontinuity of the second kind from the right at ‘a’ if lim ( )
x a
f x
→ +
does not
exist
79
80. Mixed Discontinuity
If a function f has a discontinuity of the second kind on one side of ‘a’ and on the other
side, a discontinuity of the first kind or may be continuous, then f is said to have a
mixed discontinuity at ‘a’
Thus f has a mixed discontinuity at ‘a’ if eitherThus f has a mixed discontinuity at ‘a’ if either
(i) lim ( )
x a
f x
→ −
does not exist and lim ( )
x a
f x
→ +
exists, however lim ( )
x a
f x
→ +
may or
may not equal f(a)
(Or)
(ii) lim ( )
x a
f x
→ +
does not exist and lim ( )
x a
f x
→ −
exists, however lim ( )
x a
f x
→ −
may or
may not equal f(a).
80
81. Piecewise continuous Function
A function :f A R→ is said to be piecewise continuous on A if A can be
divided into a finite number of parts so that f is continuous on each part.
Clearly, in such a case, f has a finite number of discontinuities and the set A isClearly, in such a case, f has a finite number of discontinuities and the set A is
divided at the points of discontinuities .
For example, consider : (0,5)f R→ defined by f(x)=[x], then f is discontinuous
at 1, 2, 3 and 4. If the interval (0,5) is divided at 1, 2, 3 and 4, then f is continuous
in (0,1), (1,2), (2, 3), (3, 4) and (4, 5)
∴ f is piecewise continuous
81
82. Property-1:
Let f and g be two functions of x defined in some nbd of c .If f and g are both
continuous at c . Then
Elementary Properties of Continuous Function
(i) f + g, f-g and f.g are all continuous at c
(ii) f/g is also continuous at c provided that ( ) 0g c ≠
(iii) f or g is continuous at c.
82
83. Property-2:-
Let f be defined on a set D with range R .Let g(u) be defined on a set D′ which
contains R . If x D∈ , then ( )f x R∈ and consequently f(x) = u∈ D′ so that g{f(x)}is
defined. The resulting function F, defined by F(x)=g{f(x)}, is called composition of g
with f. we say that the composite function F is defined on D.with f. we say that the composite function F is defined on D.
Now let f and g satisfy the conditions of the above paragraph. If f and g are both
continuous at a point c D∈ , then the composite function F=g(f) is also continuous at c.
Note: Continuous function of a continuous function is continuous.
Property-3:-
If a function f has its values 0≥ ,and if f is continuous at x=c , then f is also
continuous at c.
83
84. Neighbourhood Property:
If f is continuous at a certain point c and if f(c) ≠ 0, then there exists a certainIf f is continuous at a certain point c and if f(c) ≠ 0, then there exists a certain
nbd of c such that ∀ x in that nbd of f(x) has the same sign as that of f(c).
84
85. Deeper Properties Of Continuous Functions
Let f be a real-valued continuous function in a closed interaval [a,b].
Suppose that f(a)and f(b) are of opposite signs i.e f(a).f(b) < 0.
Then there exist at least one point c where a < c < b such that f(c) = 0.
BOLZANO’S THEOREM ON CONTINUITY:
a b
f(a)>0
f(b)<0
α
85
86. Let f be a real-valued continuous function in a closed interaval [a,b].Suppose that
f(a) ≠ f(b).Then f assumes every value between f(a) and f(b) at least once.
INTERMEDIATEVALUE THEOREM:-
86
87. FIXED-POINT THEOREM:-
If f is continuous on[a,b] and f(x) ∈[a,b] for every x∈[a,b] (That is :[ , ] [ , ]f a b a b→ ) then f
has a fixed point i.e ∃ c ∈[a,b] whose f(c)=c
Theorem:-Theorem:-
Let f be a real –valued continuous function in [a,b] .Suppose that f(a) ≠ f(b).Let µbe a number
between f(a) and f(b) .Moreover,assume that f is strictly monotone in[a,b].Then ∃ exactly one
value of x = c where a < c < b such that f(c)= µ.
87
88. CONVESE OF INTERMEDIATE VALUE THEOREM:
If in [a,b], f assumes at least once every value between f(a) and f(b) with f(a) ≠ f(b),
then’f’ need not necessarily be continuous in [a,b].
Example:
Let f be function defined by
0, 0
1 1
0
2 2
1 1
( )
2 2
3 1
1
2 2
1, 1
when x
x when x
f x when x
x when x
when x
=
− < <
= =
− < <
=
88
89. 1 C (1,1)
Clearly , there are discontinuous at x=0,1/2,1.But every value between 0 and 1 is
assumed by f.Take a line parallel to X-axis between y=0 and y=1[e.g.,a dotted line
shown in the figure , y = d(0 1d≤ ≤ ). ∃a corresponding x=c for f(c)=d].Thus f
assumes every value between f(0) and f(1) but f is not continuous in0 1x≤ ≤ .
1
d
C
D
CB
1
(1/2,1/2)1/2
(1,1)
(0,0)
1/2
A
O
89
90. BOUNDEDNESS PROPERTY OF CONTINUOUS FUNCTION:-
If a real valued function f is continuous on a closed interval I=[a,b]then it is bounded there.
NOTE:-A function which is continuos only in (a,b) or [a,b) or (a,b] may not be bounded
there.
Example:- Consider
1
( ) (0,1)f x x
x
= ∀ ∈
Which is continuous in (0,1) also as 0 ( )x f x+
→ ⇒ → +∞Which is continuous in (0,1) also as 0 ( )x f x+
→ ⇒ → +∞
I,e it is not bounded
NOTE:- CONVERSE OF THE ABOVE THEOREM NEED NOT BE TRUE.
i.e a function which is bounded in an interval need not be continuous throughout the
interval .
Example:-Consider
;0 1
( ) 1
; 0,1
2
x x
f x
when x
< <
=
=
Clearly f is bounded in between 0 and 1 but not continuous in 0 1x≤ ≤ .
90
91. ATTAINMENT OF BOUNDS OF CONTINUOUS FUNCTION:
Property-1:
If f be continuous function on a closed interval, then it assumes its least upper bound(l.u.b) and
greatest lower bound (g.l.b) in that interval.
Remember:Remember:
In a bounded set of real numbers if the lub and glb are actually members of the set , then they
are respectively called greatest and least value of the set .So
A function f is continuous in [a,b] then it has greatest and least values in [a,b].
If f is continuous in (a,b) then the above statement need not be true.
91
92. Example:-
Consider
1 ;0 1
( ) 1
; 0
2
x x
f x
when x
− < ≤
=
= 2
At x=0 , there is a point of discontinuity i.e., f is not continuous in [0,1] and f has no greatest
value in [0,1].
Property-2:
If a function f is continuous in [a,b] then f assumes all values between its exact upper and
lower bounds.
92
93. CONTINUITY OF INVERSE FUNCTION
Property-1:
Let f be a strictly increasing function on [a,b].Then
(i) The inverse function f-1
exists and is strictly monotone increasing in its domain of
definition
(ii) If , further , f is continuous in [a,b] then f-1
is also continuous on [ ],α β where(ii) If , further , f is continuous in [a,b] then f is also continuous on [ ],α β where
( )f aα = and ( )f bβ =
Let f be a strictly decreasing function on [a,b].Then
(iii) The inverse function f-1
exists and is strictly monotone decreasing in its domain of
definition
(iv) If , further , f is continuous in [a,b] then f-1
is also continuous on [ ],β α where
( )f aα = and ( )f bβ =
Remember:-
If f is strictly increasing then –f is strictly decreasing and vice – versa
93
94. CONTINUITYAND MONOTONICITY
If f be a continuous in [a,b] and assumes each value between f(a) and f(b) just once then
it is strictly monotonic in the same interval
Remember:
Let f be monotonic increasing in [a,b] .Then if c be s.t a<c<b then f(c-0) and f(c+0)
both exist and sup ( 0)
a x c
f f c
< <
= − and inf ( 0)
c x b
f f c
< <
= + also ( 0) ( ) ( 0)f c f c f c− ≤ ≤ +
94
95. Conceptual Examples
(1) State the number of point of discontinuities and discuss the nature of
discontinuity for the function
1
( )
ln| |
f x
x
= and also sketch its graph.
1
0, 1
ln
( )
1
0, 1
ln( )
if x x
x
f x
if x x
x
> ≠
=
< ≠ − −
function is obviously discontinuous at
x = 0, 1, –1. as it is not defined.
95
96. 0
0
lim ( ) 0
lim ( ) 0
x
x
f x
f x
+
−
→
→
=
=
Limit exists at x = 0. Hence removable discontinuity at x = 0.
(Missing point)
1
lim ( )
x
f x+
→
= ∞
Limit DNE. Hence non removable discontinuity (infinite type) at1
1
lim ( )
x
x
f x−
→
→
= −∞
Limit DNE. Hence non removable discontinuity (infinite type) at
x = 1
1
1
lim ( )
lim ( )
x
x
f x
f x
+
−
→−
→−
= −∞
= ∞
Limit DNE. Hence non removable discontinuity (infinite type)
at x = 0
96
97. Note that f (x) is even ⇒ symmetric about y axis. The graph of f (x) is as follows.
97
98. (2) If f(x+y) = f(x) . f(y) for all x & y & f(x) = 1 + g(x). G(x) where
0
lim ( ) 0
x
g x
→
= &
0
lim ( )
x
G x
→
exist. Prove that f(x) is continuous for all x.
Sol. Let x a R= ∈
0 0 0
lim ( ) lim ( )· ( ) lim ( )[1 ( )· ( )]
h h h
f a h f a f h f a g h G h
→ → →
+ = = +
0 0
( ) 1 lim ( )·lim ( ) ( )
h h
f a g h G h f a
→ →
= + =
98
99. (3) Show that the function f (x) = (x – a)2
(x – b)2
+ x takes the value
2
a b+
for
some value of x ∈ [a, b]
Sol. f (a) = a ; f (b) = b ; Also find f is continuous in [a, b] and
2
a b+
∈ [a, b]
Hence using intermediate value theroem
+
∃ some c ∈ [a, b] such that ( )
2
a b
f c
+
=
99
100. (4) Prove that there exist a number x such that 2005
2
1
x 2005
1 sin x
+ =
+
Sol Let 2005 2 2
f (x) = x + (1 + sin x)−
∴ f is continuous and f (0) = 1 < 2005 and f (2) > 22005
∴ f is continuous and f (0) = 1 < 2005 and f (2) > 2
100
101. (5) Let f be a continuous function defined onto on [0, 1] with range [0, 1].
Show that is some 'c' in [0, 1] such that f (c) = 1 – c.
Sol. Consider g (x) = f (x) – 1 + x
g (0) = f (0) – 1 ≤ 0 (as f (0) ≤ 1)g (0) = f (0) – 1 ≤ 0 (as f (0) ≤ 1)
g (1) = f (1) ≥ 0 (as f (1) ≥ 1)
Hence g (0) and g (1) and of opposite signs
hence ∃ at least one c ∈ (0, 1) such that g (c) = 0
∴ g (c) = f (c) – 1 + c = 0 ; f (c) = 1 – c
101
102. (6) Let f be continuous on the interval [0, 1] to R such that f (0) = f (1). Prove
that there exists a point c in
1
0,
2
such that
1
( )
2
f c f c
= +
Sol. Consider a continuous function
1
g (x) = f f (x)
2
x
+
( g is continuous
1
0,
2
x
∀ ∈
)g (x) = f f (x)
2
x +
( g is continuous 0,
2
x∀ ∈
)
now
1 1
g (0) = f f (0) = f f (1)
2 2
[as f (0) = f (1)]
1 1 1
(1) (1)
2 2 2
g f f f f
= − = −
since g is continuous and g (0) and g(1/2) are of opposite signs hence the
equation g (x) = 0 must have at least one root in
1
0,
2
∴for some
1 1
0, , ( ) 0 ( )
2 2
c g c f c f c
∈ = ⇒ + =
102
103. (7) Let f : [0, 2] → R be continuous and f (0) = f (2). Prove that there exists x1 and
x2 in [0, 2] such that x2 – x1 = 1 and f (x2) = f (x1)
Sol. Consider continuous function g as
g (x) = f (x + 1) – f (x)
now, g (0) = f (1) – f (0) = f (1) – f (2) ....(1)now, g (0) = f (1) – f (0) = f (1) – f (2) ....(1)
g (1) = f (2) – f (1) = f (2) – f (1) ....(2)
hence g (0) and g (1) are of opposite signs, hence ∃ some c ∈ (0, 1) where
g (c) = 0
i.e. f (c + 1) = f (c) [c + 1 ∈ (1, 2) as c ∈ (0, 1) ]
put c = x1 ; c + 1 = x2
∴ f (x2) = f (x1) where x2 – x1 = 1
obviously x1, x2 ∈ [0, 2]
103
104. (8) 2
sin
( )
2
x x
f x
x
=
+
& g (x) = | x | are continuous at x = 0, hence the composite2
2x +
2
sin
(gof) (x)
2
x x
x
=
+
will also be continuous at x = 0.
104
105. (9) Dirichlet’s Function
Let
1,
( )
0, c
if x Q
f x
if x Q
∈
=
∈
Prove that f is NOT continuous at any point of R .
Sol: Let { }nx be a sequence of rational numbers such that nx a→ and nx a≠ , n=1,2,3…..;Let
{ }ny be a sequence of irrational numbers such that ny a→ and{ }n n
ny a≠ , n=1,2,3…..;.Clearly ( ) 1, ( ) 0n nf x f y= = for n=1,2,3……
Therefore ( ) 1nf x → and ( ) 0nf y → and lim ( )
x a
f x
→
does not exist. Since ‘a’ is arbitrary,
limit does not exist for every real number. Hence it is not continuous every where.
(10) Let
0,
( )
1, c
if x Q
f x
if x Q
∈
=
∈
is discontinuous every where.
Sol: Similar to above Solution
105
106. (11) Thomac Function
Let f be defined on R+
by
0,
1
,( )
int , 1
if xisirrational
p
if x where p and q aref x
q q
are ve egers and hcf of p qis
==
+ int , 1are ve egers and hcf of p qis+
Prove that f is continuous at every irrational number in R+
and discontinuous at every
rational number in R+
Sol: Case 1: Suppose ‘a’ is any positive rational number.
Let { }nx be a sequence of irrational numbers in R+
that converges to a. Then
lim ( ) 0n
n
f x
→∞
= while f(a) = a positive number of the form
1
q
where q is a positive integer.
Hence f is discontinuous at ‘a’
106
107. Case 2: Suppose ‘b’ is an irrational number in R+
.
Let ε >0 .Then by the Archimedean property , ∃ a positive integer n0 such that 0 1n ε > .There
are only a finite number of rational numbers with denominator less than n0 in the open
interval (b-1,b+1).interval (b-1,b+1).
Hence δ > 0 can be chosen so small that the nbd ( ),b b− δ + δ contains no rational numbers
with denominators less than n0.
It then follows that for x b− < δ and x ∈R+
we have
0
1
( ) ( ) ( )f x f b f x
n
− = ≤ < ε .
Thus f is continuous at the irrational number b.
107
108. (12). Let
,
( )
1 , c
x if x Q
f x
x if x Q
∈
=
− ∈
Prove that f is NOT continuous at any point of R except at x =
1
2
Sol: At x =
1
2
,
1 1
2 2
f
=
1 1
( ) ,
2 2
f x f x if x Q
− = − ∈
1
(1 ) 1 ,
2
1
,
2
C
C
x if x Q
x evenif x Q
= − − − ∈
= − ∈
Therefore given anyε >0, we can make
1 1
( ) ( )
2 2
C
f x f x Q Q x
− < ε∀ ∈ ∪ − < δ
Where ε =δ i.e f is continuous at x=
1
2
.
108
111. Derivative of a Function at a Point
Let :[ , ]f a b R→ be a function and ( , )c a b∈ , then f is said to be derivable (or
differentiable) at c, if
( ) ( ) ( ) ( )
lim lim
f x f c f c h c c
or
x c h
− + −
−
exists. The limit, indifferentiable) at c, if
0
lim lim
x c h
or
x c h→ →−
exists. The limit, in
case it exists is called the derivative or the differential co-efficient of the function at x =c
and is denoted by ( )f c′
111
112. Left-Hand Derivative(LHD)
Let :[ , ]f a b R→ be a function and ( , )c a b∈
If
( ) ( )
lim
x c
f x f c
x c→ −
−
−
or
0
( ) ( )
lim
h
f c h f c
h→ +
− −
−
exists, the this limit is called by the left-
hand derivative of f at c and is denoted by ( 0)f c′ − or ( )f c′ − or ( )Lf c′hand derivative of f at c and is denoted by ( 0)f c′ − or ( )f c′ − or ( )Lf c′
Right-Hand Derivative(RHD)
Let :[ , ]f a b R→ be a function and ( , )c a b∈
If
( ) ( )
lim
x c
f x f c
x c→ +
−
−
or
0
( ) ( )
lim
h
f c h f c
h→
+ −
exists, then this limit is called the right-
hand derivative of f at c and is denoted by ( 0)f c′ + or ( )f c′ + or ( )Rf c′ .
112
113. Existence of Derivative at a point
A function f is said to be differentiable at x =c iff
i) ( )Lf c′ exists (which is fixed finite quantity)
ii) ( )Rf c′ exists (which is fixed finite quantity)
iii) ( ) ( )Lf c Rf c′ ′=
If any one of the above condition fails to satisfy by the function f(x) then f is said to be not
derivable at x =c.
The derivative ( )f c′ exists iff ( ) ( )Lf c Rf c′ ′= .
Note: When we say that
( ) ( )
lim
x c
f x f c
x c→
−
−
exists, we mean that it is a real number i.e, finite
113
114. Derivability in an interval
A function :[ , ]f a b R→ is said to be derivable in the open interval (a, b) if ( )f c′
exists for each c such that a c b< <
A function :[ , ]f a b R→ is said to be derivable in the closed interval [a,b] ifA function :[ , ]f a b R→ is said to be derivable in the closed interval [a,b] if
(i) ( )f c′ exists for each ( , )c a b∈
(ii) ( )Rf a′ exists, and
(iii) ( )Lf b′ exists
A function :f I R→ is said to be derivable on I if f is derivable at every point of I.
114
115. Derivative of a Function (w.r.t x)
Let f be a function whose domain is an interval I. Let I1 be the set of all those points x of I
at which ( )f x′ exists. Clearly 1I I⊂ . If 1I ≠ φ , then the function f ′ with domain I1 is
called the derivative of f
i.e.,
0
( ) ( )
lim
h
f x h f x
h→
+ −
exists fx D∀ ∈ then f is said to be differentiable w.r.t x and
is denoted by ( ) ( )
d dy
f x or or f x
dx dx
′ . This definition is called ab-intio method or
First principles of Differentiation. And hear
d
dx
differential co-efficient of y=f(x)
w.r.t x.
115
116. Interpretation of Derivative
Geometrical Interpretation :
Slope of the tangent drawn to the curve at x = a if it exists
Physical Interpretation:
Instantaneous rate of change of functionInstantaneous rate of change of function
Remember :
If graph of the function contains any sharp edges, corners, gaps
or peaks then at that points the function is not differentiable.
116
117. Properties, Theorems In Differentiation
Theorem-1:-
If f has a finite derivative at x = c, then f is continuous at x=c.
i.e., derivability ensures continuity
Remember:-
1.Contrapositive of the above statement is “ If a function f is not
continuous at a point x=c , then the function f can not be
derivable there.”
2. Every continuous function need not be derivable
117
118. DARBOUX THEOREM:
Assume that f is defined on the closed interval [a,b] and that f has finite derivative at
each interior point .Assume also that f has a RHD at x=a and LHD at x=b where
( ) ( ) 0f a f b′ ′ < .Then there exist at least one point c in [a,b] such that ( ) 0f c′ = .( ) ( ) 0f a f b′ ′ < .Then there exist at least one point c in [a,b] such that ( ) 0f c′ = .
Note:- If ( ) 0f a′ < and ( ) 0f b′ > then with similar arguments it can be shown that there
exist an interior point [ , ]d a b∈ where ( ) 0f d′ = .In this case d is the point where f
attains the Infimum.
118
119. INTERMEDIATE VALUE PROPERTY FOR DERIVATIVES
Let f be a deriavable function in [a,b] and ( ) ( )f a f b′ ′≠ .Let γ be any number
between ( )f a′ and ( )f b′ .Then ∃atleast one point [ , ]c a b∈ ∋ ( )f c′ = γ
Observation:
Let f be a real-valued continuous function in a closed interaval [a,b].Suppose
that f(a) ≠ f(b).Then f assumes every value between f(a) and f(b) at least once.
This is what is know as the intermediate value property of continuous function
of [a,b].
This property does not characterize continuous function, because we have just
proved that the property is also shared by the class of derivatives ( )f x′ , whether
continuous or not.
119
120. Formulae & Shortcuts of Differentiation
If U = U(x), V = V(x) then
1. ( )
d dU dV
U V
dx dx dx
± = ±
2. ( )
d dU dV
aU bV a b
dx dx dx
+ = ⋅ + ⋅ (where a, b are any constants)2. ( )aU bV a b
dx dx dx
+ = ⋅ + ⋅ (where a, b are any constants)
3. ( )
d dV dU
UV U V
dx dx dx
= ⋅ + ⋅ (Product rule)
4. 2
dU dV
V U
d U dx dx
dx V V
⋅ − ⋅ =
(Quotient Rule)
120
121. 5.
1dy
dxdx
dy
=
6. ( ) ( )
1
( ) ( ) ( )
n nd
f x n f x f x
dx
−
′= ⋅ ⋅
7.
( ) ( )
( )f x f xd
e e f x′= ⋅7.
( ) ( )
( )f x f x
e e f x
dx
′= ⋅
8.
( ) ( )
log ( )f x f xd
a a a f x
dx
′= ⋅ ⋅
9.
1
log ( ) ( )
| ( ) |
d
f x f x
dx f x
′=
10.
1
log
| |
d
x
dx x
=
11. | | , ( 0)
| |
d x
x x
dx x
= ≠
121
122. 12. sin ( ) cos ( )
d
f x f x
dx
=
13. cos ( ) sin ( )
d
f x f x
dx
=−
14. tan ( ) sec² ( )
d
f x f x=14. tan ( ) sec² ( )
d
f x f x
dx
=
15. cot ( ) cosec² ( )
d
f x f x
dx
=−
16. sec ( ) sec ( ) tan ( )
d
f x f x f x
dx
= ⋅
17. cosec ( ) cosec ( ) cot ( )
d
f x f x f x
dx
=− ⋅
122
123. 18.
( )
1
2
1
sin ( ) ( )
1 ( )
d
f x f x
dx f x
−
′= ⋅
−
19.
( )
1
2
1
cos ( ) ( )
1 ( )
d
f x f x
dx f x
−
′=− ⋅
−
20.
1 1
tan ( ) ( )
d
f x f x−
′= ⋅20.
( )
1
2
1
tan ( ) ( )
1 ( )
d
f x f x
dx f x
−
′= ⋅
+
21.
( )
1
2
1
cot ( ) ( )
1 ( )
d
f x f x
dx f x
−
′=− ⋅
+
22.
( )
1
2
1
sec ( ) ( )
| ( ) | ( ) 1
d
f x f x
dx f x f x
−
′= ⋅
−
23.
( )
1
2
1
cosec ( ) ( )
| ( ) | ( ) 1
d
f x f x
dx f x f x
−
′=− ⋅
−
123
124. 24. If y=(f(x))g(x)
, then ( )
1
g(x) g(x) 1dy d f (x)
f (x) (f (x)) g(x). g (x)log(f(x))
dx dx f(x)
= = +
25. If y=(f(x))f(x)
, then
dx
dy
=(f(x))f(x)
(1+log(f(x))f1
(x)25. If y=(f(x)) , then
dx
=(f(x)) (1+log(f(x))f (x)
26. If y=(f(x))y
, then
)log1)((
)(
))(log(1)((
)( 1212
yxf
xfy
xfyxf
xfy
dx
dy
−
=
−
=
27. If yxfxfxfxfy +=−∞−−−−+++= )()()()( , then
12
)(1
−
=
y
xf
dx
dy
124
125. 28. =∞−−−−−=
dx
dy
then,)()()( xfxfxfy f1
(x)
29. If f(x,y)=kg(x,y), where f(x,y) and g(x,y) are the homogenous functions of same order, then
ydy
=
x
y
dx
dy
=
30. If f(x,y)= C, then
f
dy x
dx f
y
∂
∂ = −
∂
∂
31. If f(x+y)=f(x)f(y) ∀ x,y ∈ R and f(x)≠0, f(a)=k, f1
(0) exists, then f1
(a)=kf1
(0)
125
126. 32. If y=Tan-1 ( ) ( )
( ) ( )
=
−
+
dx
dy
then,
)()(
)()(
xfaSinxfbCos
xfbSinxfaCos
f1
(x)
33. If y=Tan-1 ( ) ( )
( ) ( )
=
+
−
dx
dy
then,
)()(
)()(
xfaSinxfbCos
xfbSinxfaCos
-f1
(x)
R
34. If y=
3
2
1
R
R
R
, where R1,R2,R3 are the rows of the determinant, whose the elements are the
polynomials in ‘x’, then
1
3
2
1
3
1
2
1
3
2
1
1
R
R
R
R
R
R
R
R
R
dx
dy
++= and det(A)=det(AT
)
⇒ we can apply the same method in column wise also
126
127. 35. If ( )nnxx
yxkyx −=−+− 22
11 , then n
nn
x
y
y
x
dx
dy
2
21
1
1
−
−
=
−
36. If ( )nnnn
yxkyx +=−−− 22
11 , then n
nn
x
y
y
x
dx
dy
2
21
1
1
−
−
=
−
∂
−
f
37. If aSin( ) ( )
∂
∂
∂
∂
−
==+
y
f
x
f
cyxfbCosyxf
dx
dy
then,),(),(
38. If y=Tan-1
)(2dx
dy
then,
2
22
bCosxa
bax
Tan
ba
ba
+
−
=
+
−
39. If
bCosxa
ba
ba
bay
Tan
+
−
=
+
−
=
22
dx
dy
then,
2
x
Tan
2
127
128. 40. If y=Cos-1
bCosxa
ba
ba
bCosxa
baCosx
+
−
=>
+
+ 22
dx
dy
then),(
41. If y=Cos(mCos-1
x) or y=Cos(mSin-1
x) or y=Sin(mSin-1
x)
or y=Sin(mCos-1
x), then (1-x2
)y2-xy1+m2
y=0
42. If y=Cos(mCos-1
x) (or) y=Cos(mSin-1
x) (or) y=Sin(mCos-1
x) (or) y=Sin(mSin-1
x),42. If y=Cos(mCos-1
x) (or) y=Cos(mSin-1
x) (or) y=Sin(mCos-1
x) (or) y=Sin(mSin-1
x),
then (1-x2
)y2-xy1+m2
y=0
43. If y=k1eax
, then (D-a)y=0, where Dy=
dx
dy
44. If y=(k1+k2x)eax
, then (D-a)2
y=0, where D2
y= 2
2
dx
yd
45. If y=k1eax
+k2ebx
, then (D-a)(D-b)y=0
128
129. 46. If y=eax
Cos(bx+c) (or) y=eax
Sin(bx+c), then ((D-a)2
+b2
)y=a
47. If ax2
+2hxy+by2
=c, then y2= 3
2
)(
)(
byhx
abhc
+
−
48. If y=k1Sin(ax+b)+k2Cos(ax+b), then (D2
-a2
)y=0
49. If y=k1eax
Sin(bx+c)+k2eax
Cos(bx+c), then ((D-a)2
+b2
)y=0
50. If y=xn
logx, then yn=
1 1
n! logx 1
2 n
+ + +−−−−−+
51. If y=xn-1
logx, then yn=
n! 1
x
−
129
