Page 1
CONTINUITY
Continuous functions are a key concept in mathematics, particularly in calculus and
analysis. They describe functions where small changes in input lead to small changes in
output, without sudden jumps. This document explores the definition and properties of
continuous functions, types of discontinuities, and important theorems like the
Intermediate Value Theorem.
1. CONTINUOUS FUNCTIONS:
A function for which a small change in the independent variable causes only a small
change and not a sudden jump in the dependent variable are called continuous
functions. Naively, we may say that a function is continuous at a fixed point if we can
draw the graph of the function around that point without lifting the pen from the plane
of the paper.
A function ?? ( ?? ) is said to be continuous at ?? = ?? , if ?? ?? ?? ?? ? ?? ? ?? ( ?? ) exists and is equal to
?? ( ?? ) . Symbolically ?? ( ?? ) is continuous at ?? = a if ?? ?? ?? h ? 0
? ?? ( ??  h ) = ?? ?? ?? h ? 0
? ?? ( ?? + h ) =
?? ( ?? ) = finite quantity.
i.e. ?????? 
?? = ?? = ?? ?? ?? 
?? = ?? = value of ?? ( ?? ) 
?? = ?? = finite quantity. ( h > 0 )
figure (1)
Page 2
CONTINUITY
Continuous functions are a key concept in mathematics, particularly in calculus and
analysis. They describe functions where small changes in input lead to small changes in
output, without sudden jumps. This document explores the definition and properties of
continuous functions, types of discontinuities, and important theorems like the
Intermediate Value Theorem.
1. CONTINUOUS FUNCTIONS:
A function for which a small change in the independent variable causes only a small
change and not a sudden jump in the dependent variable are called continuous
functions. Naively, we may say that a function is continuous at a fixed point if we can
draw the graph of the function around that point without lifting the pen from the plane
of the paper.
A function ?? ( ?? ) is said to be continuous at ?? = ?? , if ?? ?? ?? ?? ? ?? ? ?? ( ?? ) exists and is equal to
?? ( ?? ) . Symbolically ?? ( ?? ) is continuous at ?? = a if ?? ?? ?? h ? 0
? ?? ( ??  h ) = ?? ?? ?? h ? 0
? ?? ( ?? + h ) =
?? ( ?? ) = finite quantity.
i.e. ?????? 
?? = ?? = ?? ?? ?? 
?? = ?? = value of ?? ( ?? ) 
?? = ?? = finite quantity. ( h > 0 )
figure (1)
figure (2)
figure (3)
figure (4)
Page 3
CONTINUITY
Continuous functions are a key concept in mathematics, particularly in calculus and
analysis. They describe functions where small changes in input lead to small changes in
output, without sudden jumps. This document explores the definition and properties of
continuous functions, types of discontinuities, and important theorems like the
Intermediate Value Theorem.
1. CONTINUOUS FUNCTIONS:
A function for which a small change in the independent variable causes only a small
change and not a sudden jump in the dependent variable are called continuous
functions. Naively, we may say that a function is continuous at a fixed point if we can
draw the graph of the function around that point without lifting the pen from the plane
of the paper.
A function ?? ( ?? ) is said to be continuous at ?? = ?? , if ?? ?? ?? ?? ? ?? ? ?? ( ?? ) exists and is equal to
?? ( ?? ) . Symbolically ?? ( ?? ) is continuous at ?? = a if ?? ?? ?? h ? 0
? ?? ( ??  h ) = ?? ?? ?? h ? 0
? ?? ( ?? + h ) =
?? ( ?? ) = finite quantity.
i.e. ?????? 
?? = ?? = ?? ?? ?? 
?? = ?? = value of ?? ( ?? ) 
?? = ?? = finite quantity. ( h > 0 )
figure (1)
figure (2)
figure (3)
figure (4)
figure (5)
figure (6)
In figure (1) and (2), ?? ( ?? ) is continuous at ?? = ?? and ?? = 0 respectively and in figure (3)
to (6) ?? ( ?? ) is discontinuous at ?? = ?? .
Note 1: Continuity of a function must be discussed only at points which are in the
domain of the function.
Note 2: If ?? = ?? is an isolated point of domain then ?? ( ?? ) is always considered to be
continuous at ?? = ?? .
Problem 1: If ?? ( ?? ) = { ?? ?? ?? ?
????
2
, ?? < 1 ? [ ?? ] ? ?? = 1 ? then find whether ?? ( ?? ) is continuous or not
at ?? = 1, where [ ] denotes greatest integer function.
Solution:
?? ( ?? ) = { ?? ?? ?? ?
????
2
, ?? < 1 ? [ ?? ] ? ?? = 1 ?
For continuity at ?? = 1, we determine, ?? ( 1 ) , ?? ?? ?? ?? ? 1
 ? ?? ( ?? ) and ?? ?? ?? ?? ? 1
+ ? ?? ( ?? ) .
Now, ?? ( 1 ) = [ 1 ] = 1
Page 4
CONTINUITY
Continuous functions are a key concept in mathematics, particularly in calculus and
analysis. They describe functions where small changes in input lead to small changes in
output, without sudden jumps. This document explores the definition and properties of
continuous functions, types of discontinuities, and important theorems like the
Intermediate Value Theorem.
1. CONTINUOUS FUNCTIONS:
A function for which a small change in the independent variable causes only a small
change and not a sudden jump in the dependent variable are called continuous
functions. Naively, we may say that a function is continuous at a fixed point if we can
draw the graph of the function around that point without lifting the pen from the plane
of the paper.
A function ?? ( ?? ) is said to be continuous at ?? = ?? , if ?? ?? ?? ?? ? ?? ? ?? ( ?? ) exists and is equal to
?? ( ?? ) . Symbolically ?? ( ?? ) is continuous at ?? = a if ?? ?? ?? h ? 0
? ?? ( ??  h ) = ?? ?? ?? h ? 0
? ?? ( ?? + h ) =
?? ( ?? ) = finite quantity.
i.e. ?????? 
?? = ?? = ?? ?? ?? 
?? = ?? = value of ?? ( ?? ) 
?? = ?? = finite quantity. ( h > 0 )
figure (1)
figure (2)
figure (3)
figure (4)
figure (5)
figure (6)
In figure (1) and (2), ?? ( ?? ) is continuous at ?? = ?? and ?? = 0 respectively and in figure (3)
to (6) ?? ( ?? ) is discontinuous at ?? = ?? .
Note 1: Continuity of a function must be discussed only at points which are in the
domain of the function.
Note 2: If ?? = ?? is an isolated point of domain then ?? ( ?? ) is always considered to be
continuous at ?? = ?? .
Problem 1: If ?? ( ?? ) = { ?? ?? ?? ?
????
2
, ?? < 1 ? [ ?? ] ? ?? = 1 ? then find whether ?? ( ?? ) is continuous or not
at ?? = 1, where [ ] denotes greatest integer function.
Solution:
?? ( ?? ) = { ?? ?? ?? ?
????
2
, ?? < 1 ? [ ?? ] ? ?? = 1 ?
For continuity at ?? = 1, we determine, ?? ( 1 ) , ?? ?? ?? ?? ? 1
 ? ?? ( ?? ) and ?? ?? ?? ?? ? 1
+ ? ?? ( ?? ) .
Now, ?? ( 1 ) = [ 1 ] = 1
?? ?? ?? ?? ? 1

? ?? ( ?? ) = ?? ?? ?? ?? ? 1

? ?? ?? ?? ?
????
2
= ?? ?? ?? ?
?? 2
= 1 ?? ?? ?? ?? ?? ?? ?? ? 1
+
? ?? ( ?? ) = ?? ?? ?? ?? ? 1
+
? [ ?? ] = 1
so
?? ( 1 ) = ?? ?? ?? ?? ? 1

? ?? ( ?? ) = ?? ?? ?? ?? ? 1
+
? ?? ( ?? )
? ? ?? ( ?? ) is continuous at ?? = 1
Problem 2: Let ?? ( ?? ) = {
?? 2
3
? ?? < 0 ? 3 ? ?? = 0 ? ( 1 + (
???? + ????
?? 2
) )
1
?? ? ?? > 0 ?
If ?? is continuous at ?? = 0, then find out the values of ?? , ?? , ?? and ?? .
Solution: ? Since ?? ( ?? ) is continuous at ?? = 0, so at ?? = 0, both left and right limits must
exist and both must be equal to 3 .
Now ?? ?? ?? ?? ? 0
 ?
?? ( 1  ?? ???? ?? ? ?? ) + ?? ?? ?? ?? ? ?? + 5
?? 2
= ?? ?? ?? ?? ? 0
 ?
( ?? + ?? + 5 ) + (  ?? 
?? 2
) ?? 2
+ ?
?? 2
= 3
(By the expansions of ?? ?? ?? ? ?? and ?? ?? ?? ? ?? )
If ?? ?? ?? ?? ? 0
 ? ?? ( ?? ) exists then ?? + ?? + 5 = 0 and  ?? 
?? 2
= 3 ? ?? =  1 and ?? =  4
since ?? ?? ?? ?? ? 0
+ ? ( 1 + (
???? + ?? ?? 3
?? 2
) )
1
?? exists ? ?? ?? ?? ?? ? 0
+ ?
???? + ?? ?? 3
?? 2
= 0 ? ?? = 0
Now ?? ?? ?? ?? ? 0
+ ? ( 1 + ???? )
1
?? = ?? ?? ?? ?? ? 0
+ ? [ ( 1 + ???? )
1
????
]
?? = ?? ??
So ?? ?? = 3 ? ?? = ?? n 3 ,
Hence ?? =  1 , ?? =  4 , ?? = 0 and ?? = ???? ? 3.
2. CONTINUITY OF THE FUNCTION IN
AN INTERVAL:
(a) A function is said to be continuous in (a,b) if ?? is continuous at each & every point
belonging to (a, b).
(b) A function is said to be continuous in a closed interval [ ?? , ?? ] if:
(i) ?? is continuous in the open interval ( ?? , ?? )
(ii) ?? is right continuous at 'a' i.e. ?? ?? ?? ?? ? ?? + ? ?? ( ?? ) = ?? ( ?? ) = ?? finite quantity
(iii) ?? is left continuous at ' ?? ' i.e. ?? ?? ?? ?? ? ??  ?? ( ?? ) = ?? ( ?? ) = ?? finite quantity
Page 5
CONTINUITY
Continuous functions are a key concept in mathematics, particularly in calculus and
analysis. They describe functions where small changes in input lead to small changes in
output, without sudden jumps. This document explores the definition and properties of
continuous functions, types of discontinuities, and important theorems like the
Intermediate Value Theorem.
1. CONTINUOUS FUNCTIONS:
A function for which a small change in the independent variable causes only a small
change and not a sudden jump in the dependent variable are called continuous
functions. Naively, we may say that a function is continuous at a fixed point if we can
draw the graph of the function around that point without lifting the pen from the plane
of the paper.
A function ?? ( ?? ) is said to be continuous at ?? = ?? , if ?? ?? ?? ?? ? ?? ? ?? ( ?? ) exists and is equal to
?? ( ?? ) . Symbolically ?? ( ?? ) is continuous at ?? = a if ?? ?? ?? h ? 0
? ?? ( ??  h ) = ?? ?? ?? h ? 0
? ?? ( ?? + h ) =
?? ( ?? ) = finite quantity.
i.e. ?????? 
?? = ?? = ?? ?? ?? 
?? = ?? = value of ?? ( ?? ) 
?? = ?? = finite quantity. ( h > 0 )
figure (1)
figure (2)
figure (3)
figure (4)
figure (5)
figure (6)
In figure (1) and (2), ?? ( ?? ) is continuous at ?? = ?? and ?? = 0 respectively and in figure (3)
to (6) ?? ( ?? ) is discontinuous at ?? = ?? .
Note 1: Continuity of a function must be discussed only at points which are in the
domain of the function.
Note 2: If ?? = ?? is an isolated point of domain then ?? ( ?? ) is always considered to be
continuous at ?? = ?? .
Problem 1: If ?? ( ?? ) = { ?? ?? ?? ?
????
2
, ?? < 1 ? [ ?? ] ? ?? = 1 ? then find whether ?? ( ?? ) is continuous or not
at ?? = 1, where [ ] denotes greatest integer function.
Solution:
?? ( ?? ) = { ?? ?? ?? ?
????
2
, ?? < 1 ? [ ?? ] ? ?? = 1 ?
For continuity at ?? = 1, we determine, ?? ( 1 ) , ?? ?? ?? ?? ? 1
 ? ?? ( ?? ) and ?? ?? ?? ?? ? 1
+ ? ?? ( ?? ) .
Now, ?? ( 1 ) = [ 1 ] = 1
?? ?? ?? ?? ? 1

? ?? ( ?? ) = ?? ?? ?? ?? ? 1

? ?? ?? ?? ?
????
2
= ?? ?? ?? ?
?? 2
= 1 ?? ?? ?? ?? ?? ?? ?? ? 1
+
? ?? ( ?? ) = ?? ?? ?? ?? ? 1
+
? [ ?? ] = 1
so
?? ( 1 ) = ?? ?? ?? ?? ? 1

? ?? ( ?? ) = ?? ?? ?? ?? ? 1
+
? ?? ( ?? )
? ? ?? ( ?? ) is continuous at ?? = 1
Problem 2: Let ?? ( ?? ) = {
?? 2
3
? ?? < 0 ? 3 ? ?? = 0 ? ( 1 + (
???? + ????
?? 2
) )
1
?? ? ?? > 0 ?
If ?? is continuous at ?? = 0, then find out the values of ?? , ?? , ?? and ?? .
Solution: ? Since ?? ( ?? ) is continuous at ?? = 0, so at ?? = 0, both left and right limits must
exist and both must be equal to 3 .
Now ?? ?? ?? ?? ? 0
 ?
?? ( 1  ?? ???? ?? ? ?? ) + ?? ?? ?? ?? ? ?? + 5
?? 2
= ?? ?? ?? ?? ? 0
 ?
( ?? + ?? + 5 ) + (  ?? 
?? 2
) ?? 2
+ ?
?? 2
= 3
(By the expansions of ?? ?? ?? ? ?? and ?? ?? ?? ? ?? )
If ?? ?? ?? ?? ? 0
 ? ?? ( ?? ) exists then ?? + ?? + 5 = 0 and  ?? 
?? 2
= 3 ? ?? =  1 and ?? =  4
since ?? ?? ?? ?? ? 0
+ ? ( 1 + (
???? + ?? ?? 3
?? 2
) )
1
?? exists ? ?? ?? ?? ?? ? 0
+ ?
???? + ?? ?? 3
?? 2
= 0 ? ?? = 0
Now ?? ?? ?? ?? ? 0
+ ? ( 1 + ???? )
1
?? = ?? ?? ?? ?? ? 0
+ ? [ ( 1 + ???? )
1
????
]
?? = ?? ??
So ?? ?? = 3 ? ?? = ?? n 3 ,
Hence ?? =  1 , ?? =  4 , ?? = 0 and ?? = ???? ? 3.
2. CONTINUITY OF THE FUNCTION IN
AN INTERVAL:
(a) A function is said to be continuous in (a,b) if ?? is continuous at each & every point
belonging to (a, b).
(b) A function is said to be continuous in a closed interval [ ?? , ?? ] if:
(i) ?? is continuous in the open interval ( ?? , ?? )
(ii) ?? is right continuous at 'a' i.e. ?? ?? ?? ?? ? ?? + ? ?? ( ?? ) = ?? ( ?? ) = ?? finite quantity
(iii) ?? is left continuous at ' ?? ' i.e. ?? ?? ?? ?? ? ??  ?? ( ?? ) = ?? ( ?? ) = ?? finite quantity
Note:
(i) All polynomials, trigonometrical functions, exponential & logarithmic functions are
continuous in their domains.
(ii) If ?? ( ?? ) & ?? ( ?? ) are two functions that are continuous at ?? = ?? then the function defined
by: ?? 1
( ?? ) = ?? ( ?? ) ± ?? ( ?? ) ; ?? 2
( ?? ) = ???? ( ?? ) , where ?? is any real number ; ?? 3
( ?? ) = ?? ( ?? ) · ?? ( ?? )
are also continuous at ?? = ?? .
Further, if ?? ( ?? ) is not zero, then ?? 4
( ?? ) =
?? ( ?? )
?? ( ?? )
is also continuous at ?? = ?? .
Problem 3: Discuss the continuity of ?? ( ?? ) = {  ?? + 1  ? ? , ?? <  2 ? 2 ?? + 3 ,  2 = ?? < 0 ? ?? 2
+ 3 ,
0 = ?? < 3 ? ?? 3
 15 ? , ?? = 3 ?
Solution: We write ?? ( ?? ) as ?? ( ?? ) = {  ??  1 ? , ?? <  2 ? 2 ?? + 3 ? ,  2 = ?? < 0 ? ?? 2
+ 3 ? , 0 = ?? <
3 ? ?? 3
 15 ? , ?? = 3 ?
As we can see, ?? ( ?? ) is defined as a polynomial function in each of intervals (  8 ,  2 ) ,
(  2 , 0 ) , ( 0 , 3 ) and ( 3 , 8 ) . Therefore, it is continuous in each of these four open intervals.
Thus we check the continuity at ?? =  2 , 0 , 3.
At the point ?? =  2
?? ?? ?? ?? ?  2
 ? ?? ( ?? ) = ?? ?? ?? ?? ?  2
 ? (  ??  1 ) = + 2  1 = 1
?? ?? ?? ?? ?  2
+ ? ?? ( ?? ) = ?? ?? ?? ?? ?  2
+ ? ( 2 ?? + 3 ) = 2 . (  2 ) + 3 =  1
Therefore, ?? ?? ?? ?? ?  2
? ?? ( ?? ) does not exist and hence ?? ( ?? ) is discontinuous at ?? =  2.
At the point ?? = 0
?? ?? ?? ?? ? 0
 ? ?? ( ?? ) = ?? ?? ?? ?? ? 0
 ? ( 2 ?? + 3 ) = 3
?? ?? ?? ?? ? 0
+ ? ?? ( ?? ) = ?? ?? ?? ?? ? 0
+ ? ( ?? 2
+ 3 ) = 3
?? ( 0 ) = 0
2
+ 3 = 3
3. TYPES OF DISCONTINUITIES:
Type1: (Removable type of discontinuities): In case ?? ?? ?? ?? ? ?? ?? ( ?? ) exists but is not equal
to ?? ( ?? ) ?? ( ?? ) is defined) then the function is said to have a removable discontinuity or
discontinuity of the first kind. In this case we can redefine the function such that
?? ?? ?? ?? ? ?? ?? ( ?? ) = ?? ( ?? ) & make it continuous at ?? = ??
Problem 4: Examine the function, ?? ( ?? ) = { ??  1 ? ? , ?? < 0 ? 1 / 4 ? ? , ?? = 0 ? ?? 2
 1 ? , ?? > 0 ?.
Discuss the continuity, and if discontinuous remove the discontinuity by redefining the
function (if possible).
Solution: ? Graph of ?? ( ?? ) is shown, from graph it is seen that
?? ?? ?? ?? ? 0
 ? ?? ( ?? ) = ?? ?? ?? ?? ? 0
+ ? ?? ( ?? ) =  1, but ?? ( 0 ) = 1 / 4
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