Chapter
stringclasses 18
values | sentence_range
stringlengths 3
9
| Text
stringlengths 7
7.34k
|
|---|---|---|
1 | 1718-1721 | In the process, we
will learn some fundamental theorems in this area 5 2 Continuity
We start the section with two informal examples to get a feel of continuity Consider
the function
1, if
0
( )
2, if
0
x
f x
x
β€
= ο£²ο£±
>
ο£³
This function is of course defined at every
point of the real line |
1 | 1719-1722 | 5 2 Continuity
We start the section with two informal examples to get a feel of continuity Consider
the function
1, if
0
( )
2, if
0
x
f x
x
β€
= ο£²ο£±
>
ο£³
This function is of course defined at every
point of the real line Graph of this function is
given in the Fig 5 |
1 | 1720-1723 | 2 Continuity
We start the section with two informal examples to get a feel of continuity Consider
the function
1, if
0
( )
2, if
0
x
f x
x
β€
= ο£²ο£±
>
ο£³
This function is of course defined at every
point of the real line Graph of this function is
given in the Fig 5 1 |
1 | 1721-1724 | Consider
the function
1, if
0
( )
2, if
0
x
f x
x
β€
= ο£²ο£±
>
ο£³
This function is of course defined at every
point of the real line Graph of this function is
given in the Fig 5 1 One can deduce from the
graph that the value of the function at nearby
points on x-axis remain close to each other
except at x = 0 |
1 | 1722-1725 | Graph of this function is
given in the Fig 5 1 One can deduce from the
graph that the value of the function at nearby
points on x-axis remain close to each other
except at x = 0 At the points near and to the
left of 0, i |
1 | 1723-1726 | 1 One can deduce from the
graph that the value of the function at nearby
points on x-axis remain close to each other
except at x = 0 At the points near and to the
left of 0, i e |
1 | 1724-1727 | One can deduce from the
graph that the value of the function at nearby
points on x-axis remain close to each other
except at x = 0 At the points near and to the
left of 0, i e , at points like β 0 |
1 | 1725-1728 | At the points near and to the
left of 0, i e , at points like β 0 1, β 0 |
1 | 1726-1729 | e , at points like β 0 1, β 0 01, β 0 |
1 | 1727-1730 | , at points like β 0 1, β 0 01, β 0 001,
the value of the function is 1 |
1 | 1728-1731 | 1, β 0 01, β 0 001,
the value of the function is 1 At the points near
and to the right of 0, i |
1 | 1729-1732 | 01, β 0 001,
the value of the function is 1 At the points near
and to the right of 0, i e |
1 | 1730-1733 | 001,
the value of the function is 1 At the points near
and to the right of 0, i e , at points like 0 |
1 | 1731-1734 | At the points near
and to the right of 0, i e , at points like 0 1, 0 |
1 | 1732-1735 | e , at points like 0 1, 0 01,
Chapter 5
CONTINUITY AND
DIFFERENTIABILITY
Sir Issac Newton
(1642-1727)
Fig 5 |
1 | 1733-1736 | , at points like 0 1, 0 01,
Chapter 5
CONTINUITY AND
DIFFERENTIABILITY
Sir Issac Newton
(1642-1727)
Fig 5 1
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
105
0 |
1 | 1734-1737 | 1, 0 01,
Chapter 5
CONTINUITY AND
DIFFERENTIABILITY
Sir Issac Newton
(1642-1727)
Fig 5 1
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
105
0 001, the value of the function is 2 |
1 | 1735-1738 | 01,
Chapter 5
CONTINUITY AND
DIFFERENTIABILITY
Sir Issac Newton
(1642-1727)
Fig 5 1
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
105
0 001, the value of the function is 2 Using the language of left and right hand limits, we
may say that the left (respectively right) hand limit of f at 0 is 1 (respectively 2) |
1 | 1736-1739 | 1
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
105
0 001, the value of the function is 2 Using the language of left and right hand limits, we
may say that the left (respectively right) hand limit of f at 0 is 1 (respectively 2) In
particular the left and right hand limits do not coincide |
1 | 1737-1740 | 001, the value of the function is 2 Using the language of left and right hand limits, we
may say that the left (respectively right) hand limit of f at 0 is 1 (respectively 2) In
particular the left and right hand limits do not coincide We also observe that the value
of the function at x = 0 concides with the left hand limit |
1 | 1738-1741 | Using the language of left and right hand limits, we
may say that the left (respectively right) hand limit of f at 0 is 1 (respectively 2) In
particular the left and right hand limits do not coincide We also observe that the value
of the function at x = 0 concides with the left hand limit Note that when we try to draw
the graph, we cannot draw it in one stroke, i |
1 | 1739-1742 | In
particular the left and right hand limits do not coincide We also observe that the value
of the function at x = 0 concides with the left hand limit Note that when we try to draw
the graph, we cannot draw it in one stroke, i e |
1 | 1740-1743 | We also observe that the value
of the function at x = 0 concides with the left hand limit Note that when we try to draw
the graph, we cannot draw it in one stroke, i e , without lifting pen from the plane of the
paper, we can not draw the graph of this function |
1 | 1741-1744 | Note that when we try to draw
the graph, we cannot draw it in one stroke, i e , without lifting pen from the plane of the
paper, we can not draw the graph of this function In fact, we need to lift the pen when
we come to 0 from left |
1 | 1742-1745 | e , without lifting pen from the plane of the
paper, we can not draw the graph of this function In fact, we need to lift the pen when
we come to 0 from left This is one instance of function being not continuous at x = 0 |
1 | 1743-1746 | , without lifting pen from the plane of the
paper, we can not draw the graph of this function In fact, we need to lift the pen when
we come to 0 from left This is one instance of function being not continuous at x = 0 Now, consider the function defined as
f x
xx
( )
,,
=
=β
ο£²ο£±
ο£³
1
0
2
0
if
if
This function is also defined at every point |
1 | 1744-1747 | In fact, we need to lift the pen when
we come to 0 from left This is one instance of function being not continuous at x = 0 Now, consider the function defined as
f x
xx
( )
,,
=
=β
ο£²ο£±
ο£³
1
0
2
0
if
if
This function is also defined at every point Left and the right hand limits at x = 0
are both equal to 1 |
1 | 1745-1748 | This is one instance of function being not continuous at x = 0 Now, consider the function defined as
f x
xx
( )
,,
=
=β
ο£²ο£±
ο£³
1
0
2
0
if
if
This function is also defined at every point Left and the right hand limits at x = 0
are both equal to 1 But the value of the
function at x = 0 equals 2 which does not
coincide with the common value of the left
and right hand limits |
1 | 1746-1749 | Now, consider the function defined as
f x
xx
( )
,,
=
=β
ο£²ο£±
ο£³
1
0
2
0
if
if
This function is also defined at every point Left and the right hand limits at x = 0
are both equal to 1 But the value of the
function at x = 0 equals 2 which does not
coincide with the common value of the left
and right hand limits Again, we note that we
cannot draw the graph of the function without
lifting the pen |
1 | 1747-1750 | Left and the right hand limits at x = 0
are both equal to 1 But the value of the
function at x = 0 equals 2 which does not
coincide with the common value of the left
and right hand limits Again, we note that we
cannot draw the graph of the function without
lifting the pen This is yet another instance of
a function being not continuous at x = 0 |
1 | 1748-1751 | But the value of the
function at x = 0 equals 2 which does not
coincide with the common value of the left
and right hand limits Again, we note that we
cannot draw the graph of the function without
lifting the pen This is yet another instance of
a function being not continuous at x = 0 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 |
1 | 1749-1752 | Again, we note that we
cannot draw the graph of the function without
lifting the pen This is yet another instance of
a function being not continuous at x = 0 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 Mathematically, it may be phrased precisely as follows:
Definition 1 Suppose f is a real function on a subset of the real numbers and let c be
a point in the domain of f |
1 | 1750-1753 | This is yet another instance of
a function being not continuous at x = 0 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 Mathematically, it may be phrased precisely as follows:
Definition 1 Suppose f is a real function on a subset of the real numbers and let c be
a point in the domain of f Then f is continuous at c if
lim
( )
( )
x
c f x
f c
β
=
More elaborately, if the left hand limit, right hand limit and the value of the function
at x = c exist and equal to each other, then f is said to be continuous at x = c |
1 | 1751-1754 | 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 Mathematically, it may be phrased precisely as follows:
Definition 1 Suppose f is a real function on a subset of the real numbers and let c be
a point in the domain of f Then f is continuous at c if
lim
( )
( )
x
c f x
f c
β
=
More elaborately, if the left hand limit, right hand limit and the value of the function
at x = c exist and equal to each other, then f is said to be continuous at x = c Recall that
if the right hand and left hand limits at x = c coincide, then we say that the common
value is the limit of the function at x = c |
1 | 1752-1755 | Mathematically, it may be phrased precisely as follows:
Definition 1 Suppose f is a real function on a subset of the real numbers and let c be
a point in the domain of f Then f is continuous at c if
lim
( )
( )
x
c f x
f c
β
=
More elaborately, if the left hand limit, right hand limit and the value of the function
at x = c exist and equal to each other, then f is said to be continuous at x = c Recall that
if the right hand and left hand limits at x = c coincide, then we say that the common
value is the limit of the function at x = c Hence we may also rephrase the definition of
continuity as follows: a function is continuous at x = c if the function is defined at
x = c and if the value of the function at x = c equals the limit of the function at
x = c |
1 | 1753-1756 | Then f is continuous at c if
lim
( )
( )
x
c f x
f c
β
=
More elaborately, if the left hand limit, right hand limit and the value of the function
at x = c exist and equal to each other, then f is said to be continuous at x = c Recall that
if the right hand and left hand limits at x = c coincide, then we say that the common
value is the limit of the function at x = c Hence we may also rephrase the definition of
continuity as follows: a function is continuous at x = c if the function is defined at
x = c and if the value of the function at x = c equals the limit of the function at
x = c If f is not continuous at c, we say f is discontinuous at c and c is called a point
of discontinuity of f |
1 | 1754-1757 | Recall that
if the right hand and left hand limits at x = c coincide, then we say that the common
value is the limit of the function at x = c Hence we may also rephrase the definition of
continuity as follows: a function is continuous at x = c if the function is defined at
x = c and if the value of the function at x = c equals the limit of the function at
x = c If f is not continuous at c, we say f is discontinuous at c and c is called a point
of discontinuity of f Fig 5 |
1 | 1755-1758 | Hence we may also rephrase the definition of
continuity as follows: a function is continuous at x = c if the function is defined at
x = c and if the value of the function at x = c equals the limit of the function at
x = c If f is not continuous at c, we say f is discontinuous at c and c is called a point
of discontinuity of f Fig 5 2
Rationalised 2023-24
MATHEMATICS
106
Example 1 Check the continuity of the function f given by f(x) = 2x + 3 at x = 1 |
1 | 1756-1759 | If f is not continuous at c, we say f is discontinuous at c and c is called a point
of discontinuity of f Fig 5 2
Rationalised 2023-24
MATHEMATICS
106
Example 1 Check the continuity of the function f given by f(x) = 2x + 3 at x = 1 Solution First note that the function is defined at the given point x = 1 and its value is 5 |
1 | 1757-1760 | Fig 5 2
Rationalised 2023-24
MATHEMATICS
106
Example 1 Check the continuity of the function f given by f(x) = 2x + 3 at x = 1 Solution First note that the function is defined at the given point x = 1 and its value is 5 Then find the limit of the function at x = 1 |
1 | 1758-1761 | 2
Rationalised 2023-24
MATHEMATICS
106
Example 1 Check the continuity of the function f given by f(x) = 2x + 3 at x = 1 Solution First note that the function is defined at the given point x = 1 and its value is 5 Then find the limit of the function at x = 1 Clearly
1
1
lim
( )
lim(2
3)
2(1)
3
5
x
x
f x
x
β
=β
+
=
+
=
Thus
lim1
( )
5
(1)
x
f x
f
β
=
=
Hence, f is continuous at x = 1 |
1 | 1759-1762 | Solution First note that the function is defined at the given point x = 1 and its value is 5 Then find the limit of the function at x = 1 Clearly
1
1
lim
( )
lim(2
3)
2(1)
3
5
x
x
f x
x
β
=β
+
=
+
=
Thus
lim1
( )
5
(1)
x
f x
f
β
=
=
Hence, f is continuous at x = 1 Example 2 Examine whether the function f given by f(x) = x2 is continuous at x = 0 |
1 | 1760-1763 | Then find the limit of the function at x = 1 Clearly
1
1
lim
( )
lim(2
3)
2(1)
3
5
x
x
f x
x
β
=β
+
=
+
=
Thus
lim1
( )
5
(1)
x
f x
f
β
=
=
Hence, f is continuous at x = 1 Example 2 Examine whether the function f given by f(x) = x2 is continuous at x = 0 Solution First note that the function is defined at the given point x = 0 and its value is 0 |
1 | 1761-1764 | Clearly
1
1
lim
( )
lim(2
3)
2(1)
3
5
x
x
f x
x
β
=β
+
=
+
=
Thus
lim1
( )
5
(1)
x
f x
f
β
=
=
Hence, f is continuous at x = 1 Example 2 Examine whether the function f given by f(x) = x2 is continuous at x = 0 Solution First note that the function is defined at the given point x = 0 and its value is 0 Then find the limit of the function at x = 0 |
1 | 1762-1765 | Example 2 Examine whether the function f given by f(x) = x2 is continuous at x = 0 Solution First note that the function is defined at the given point x = 0 and its value is 0 Then find the limit of the function at x = 0 Clearly
2
2
0
0
lim
( )
lim
0
0
x
x
f x
x
β
=β
=
=
Thus
lim0
( )
0
(0)
x
f x
f
β
=
=
Hence, f is continuous at x = 0 |
1 | 1763-1766 | Solution First note that the function is defined at the given point x = 0 and its value is 0 Then find the limit of the function at x = 0 Clearly
2
2
0
0
lim
( )
lim
0
0
x
x
f x
x
β
=β
=
=
Thus
lim0
( )
0
(0)
x
f x
f
β
=
=
Hence, f is continuous at x = 0 Example 3 Discuss the continuity of the function f given by f(x) = | x | at x = 0 |
1 | 1764-1767 | Then find the limit of the function at x = 0 Clearly
2
2
0
0
lim
( )
lim
0
0
x
x
f x
x
β
=β
=
=
Thus
lim0
( )
0
(0)
x
f x
f
β
=
=
Hence, f is continuous at x = 0 Example 3 Discuss the continuity of the function f given by f(x) = | x | at x = 0 Solution By definition
f(x) =
, if
0
, if
0
x
x
x
x
β
<
ο£²ο£±
β₯
ο£³
Clearly the function is defined at 0 and f(0) = 0 |
1 | 1765-1768 | Clearly
2
2
0
0
lim
( )
lim
0
0
x
x
f x
x
β
=β
=
=
Thus
lim0
( )
0
(0)
x
f x
f
β
=
=
Hence, f is continuous at x = 0 Example 3 Discuss the continuity of the function f given by f(x) = | x | at x = 0 Solution By definition
f(x) =
, if
0
, if
0
x
x
x
x
β
<
ο£²ο£±
β₯
ο£³
Clearly the function is defined at 0 and f(0) = 0 Left hand limit of f at 0 is
0
0
lim
( )
lim (β )
0
x
x
f x
x
β
β
β
=β
=
Similarly, the right hand limit of f at 0 is
0
0
lim
( )
lim
0
x
x
f x
x
+
+
β
=β
=
Thus, the left hand limit, right hand limit and the value of the function coincide at
x = 0 |
1 | 1766-1769 | Example 3 Discuss the continuity of the function f given by f(x) = | x | at x = 0 Solution By definition
f(x) =
, if
0
, if
0
x
x
x
x
β
<
ο£²ο£±
β₯
ο£³
Clearly the function is defined at 0 and f(0) = 0 Left hand limit of f at 0 is
0
0
lim
( )
lim (β )
0
x
x
f x
x
β
β
β
=β
=
Similarly, the right hand limit of f at 0 is
0
0
lim
( )
lim
0
x
x
f x
x
+
+
β
=β
=
Thus, the left hand limit, right hand limit and the value of the function coincide at
x = 0 Hence, f is continuous at x = 0 |
1 | 1767-1770 | Solution By definition
f(x) =
, if
0
, if
0
x
x
x
x
β
<
ο£²ο£±
β₯
ο£³
Clearly the function is defined at 0 and f(0) = 0 Left hand limit of f at 0 is
0
0
lim
( )
lim (β )
0
x
x
f x
x
β
β
β
=β
=
Similarly, the right hand limit of f at 0 is
0
0
lim
( )
lim
0
x
x
f x
x
+
+
β
=β
=
Thus, the left hand limit, right hand limit and the value of the function coincide at
x = 0 Hence, f is continuous at x = 0 Example 4 Show that the function f given by
f(x) =
3
3, if
0
1,
if
0
x
x
x
ο£±
+
β

=

is not continuous at x = 0 |
1 | 1768-1771 | Left hand limit of f at 0 is
0
0
lim
( )
lim (β )
0
x
x
f x
x
β
β
β
=β
=
Similarly, the right hand limit of f at 0 is
0
0
lim
( )
lim
0
x
x
f x
x
+
+
β
=β
=
Thus, the left hand limit, right hand limit and the value of the function coincide at
x = 0 Hence, f is continuous at x = 0 Example 4 Show that the function f given by
f(x) =
3
3, if
0
1,
if
0
x
x
x
ο£±
+
β

=

is not continuous at x = 0 Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
107
Solution The function is defined at x = 0 and its value at x = 0 is 1 |
1 | 1769-1772 | Hence, f is continuous at x = 0 Example 4 Show that the function f given by
f(x) =
3
3, if
0
1,
if
0
x
x
x
ο£±
+
β

=

is not continuous at x = 0 Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
107
Solution The function is defined at x = 0 and its value at x = 0 is 1 When x β 0, the
function is given by a polynomial |
1 | 1770-1773 | Example 4 Show that the function f given by
f(x) =
3
3, if
0
1,
if
0
x
x
x
ο£±
+
β

=

is not continuous at x = 0 Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
107
Solution The function is defined at x = 0 and its value at x = 0 is 1 When x β 0, the
function is given by a polynomial Hence,
lim0
( )
x
f x
β
=
3
3
0
lim (
3)
0
3
3
x
x
β
+
=
+
=
Since the limit of f at x = 0 does not coincide with f(0), the function is not continuous
at x = 0 |
1 | 1771-1774 | Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
107
Solution The function is defined at x = 0 and its value at x = 0 is 1 When x β 0, the
function is given by a polynomial Hence,
lim0
( )
x
f x
β
=
3
3
0
lim (
3)
0
3
3
x
x
β
+
=
+
=
Since the limit of f at x = 0 does not coincide with f(0), the function is not continuous
at x = 0 It may be noted that x = 0 is the only point of discontinuity for this function |
1 | 1772-1775 | When x β 0, the
function is given by a polynomial Hence,
lim0
( )
x
f x
β
=
3
3
0
lim (
3)
0
3
3
x
x
β
+
=
+
=
Since the limit of f at x = 0 does not coincide with f(0), the function is not continuous
at x = 0 It may be noted that x = 0 is the only point of discontinuity for this function Example 5 Check the points where the constant function f(x) = k is continuous |
1 | 1773-1776 | Hence,
lim0
( )
x
f x
β
=
3
3
0
lim (
3)
0
3
3
x
x
β
+
=
+
=
Since the limit of f at x = 0 does not coincide with f(0), the function is not continuous
at x = 0 It may be noted that x = 0 is the only point of discontinuity for this function Example 5 Check the points where the constant function f(x) = k is continuous Solution The function is defined at all real numbers and by definition, its value at any
real number equals k |
1 | 1774-1777 | It may be noted that x = 0 is the only point of discontinuity for this function Example 5 Check the points where the constant function f(x) = k is continuous Solution The function is defined at all real numbers and by definition, its value at any
real number equals k Let c be any real number |
1 | 1775-1778 | Example 5 Check the points where the constant function f(x) = k is continuous Solution The function is defined at all real numbers and by definition, its value at any
real number equals k Let c be any real number Then
lim
( )
x
c f x
β
= lim
x
c k
k
β
=
Since f(c) = k = lim
x
c
β f(x) for any real number c, the function f is continuous at
every real number |
1 | 1776-1779 | Solution The function is defined at all real numbers and by definition, its value at any
real number equals k Let c be any real number Then
lim
( )
x
c f x
β
= lim
x
c k
k
β
=
Since f(c) = k = lim
x
c
β f(x) for any real number c, the function f is continuous at
every real number Example 6 Prove that the identity function on real numbers given by f(x) = x is
continuous at every real number |
1 | 1777-1780 | Let c be any real number Then
lim
( )
x
c f x
β
= lim
x
c k
k
β
=
Since f(c) = k = lim
x
c
β f(x) for any real number c, the function f is continuous at
every real number Example 6 Prove that the identity function on real numbers given by f(x) = x is
continuous at every real number Solution The function is clearly defined at every point and f (c) = c for every real
number c |
1 | 1778-1781 | Then
lim
( )
x
c f x
β
= lim
x
c k
k
β
=
Since f(c) = k = lim
x
c
β f(x) for any real number c, the function f is continuous at
every real number Example 6 Prove that the identity function on real numbers given by f(x) = x is
continuous at every real number Solution The function is clearly defined at every point and f (c) = c for every real
number c Also,
lim
( )
x
c f x
β
= lim
x
c x
c
β
=
Thus, lim
x
c
β f(x) = c = f(c) and hence the function is continuous at every real number |
1 | 1779-1782 | Example 6 Prove that the identity function on real numbers given by f(x) = x is
continuous at every real number Solution The function is clearly defined at every point and f (c) = c for every real
number c Also,
lim
( )
x
c f x
β
= lim
x
c x
c
β
=
Thus, lim
x
c
β f(x) = c = f(c) and hence the function is continuous at every real number Having defined continuity of a function at a given point, now we make a natural
extension of this definition to discuss continuity of a function |
1 | 1780-1783 | Solution The function is clearly defined at every point and f (c) = c for every real
number c Also,
lim
( )
x
c f x
β
= lim
x
c x
c
β
=
Thus, lim
x
c
β f(x) = c = f(c) and hence the function is continuous at every real number Having defined continuity of a function at a given point, now we make a natural
extension of this definition to discuss continuity of a function Definition 2 A real function f is said to be continuous if it is continuous at every point
in the domain of f |
1 | 1781-1784 | Also,
lim
( )
x
c f x
β
= lim
x
c x
c
β
=
Thus, lim
x
c
β f(x) = c = f(c) and hence the function is continuous at every real number Having defined continuity of a function at a given point, now we make a natural
extension of this definition to discuss continuity of a function Definition 2 A real function f is said to be continuous if it is continuous at every point
in the domain of f This definition requires a bit of elaboration |
1 | 1782-1785 | Having defined continuity of a function at a given point, now we make a natural
extension of this definition to discuss continuity of a function Definition 2 A real function f is said to be continuous if it is continuous at every point
in the domain of f This definition requires a bit of elaboration Suppose f is a function defined on a
closed interval [a, b], then for f to be continuous, it needs to be continuous at every
point in [a, b] including the end points a and b |
1 | 1783-1786 | Definition 2 A real function f is said to be continuous if it is continuous at every point
in the domain of f This definition requires a bit of elaboration Suppose f is a function defined on a
closed interval [a, b], then for f to be continuous, it needs to be continuous at every
point in [a, b] including the end points a and b Continuity of f at a means
lim
( )
x
a f x
β+
= f (a)
and continuity of f at b means
limβ
( )
x
b f x
β
= f(b)
Observe that lim
( )
x
a f x
ββ
and lim
( )
x
b f x
β+
do not make sense |
1 | 1784-1787 | This definition requires a bit of elaboration Suppose f is a function defined on a
closed interval [a, b], then for f to be continuous, it needs to be continuous at every
point in [a, b] including the end points a and b Continuity of f at a means
lim
( )
x
a f x
β+
= f (a)
and continuity of f at b means
limβ
( )
x
b f x
β
= f(b)
Observe that lim
( )
x
a f x
ββ
and lim
( )
x
b f x
β+
do not make sense As a consequence
of this definition, if f is defined only at one point, it is continuous there, i |
1 | 1785-1788 | Suppose f is a function defined on a
closed interval [a, b], then for f to be continuous, it needs to be continuous at every
point in [a, b] including the end points a and b Continuity of f at a means
lim
( )
x
a f x
β+
= f (a)
and continuity of f at b means
limβ
( )
x
b f x
β
= f(b)
Observe that lim
( )
x
a f x
ββ
and lim
( )
x
b f x
β+
do not make sense As a consequence
of this definition, if f is defined only at one point, it is continuous there, i e |
1 | 1786-1789 | Continuity of f at a means
lim
( )
x
a f x
β+
= f (a)
and continuity of f at b means
limβ
( )
x
b f x
β
= f(b)
Observe that lim
( )
x
a f x
ββ
and lim
( )
x
b f x
β+
do not make sense As a consequence
of this definition, if f is defined only at one point, it is continuous there, i e , if the
domain of f is a singleton, f is a continuous function |
1 | 1787-1790 | As a consequence
of this definition, if f is defined only at one point, it is continuous there, i e , if the
domain of f is a singleton, f is a continuous function Rationalised 2023-24
MATHEMATICS
108
Example 7 Is the function defined by f(x) = | x |, a continuous function |
1 | 1788-1791 | e , if the
domain of f is a singleton, f is a continuous function Rationalised 2023-24
MATHEMATICS
108
Example 7 Is the function defined by f(x) = | x |, a continuous function Solution We may rewrite f as
f (x) =
, if
0
, if
0
x
x
x
x
β
<
ο£²ο£±
β₯
ο£³
By Example 3, we know that f is continuous at x = 0 |
1 | 1789-1792 | , if the
domain of f is a singleton, f is a continuous function Rationalised 2023-24
MATHEMATICS
108
Example 7 Is the function defined by f(x) = | x |, a continuous function Solution We may rewrite f as
f (x) =
, if
0
, if
0
x
x
x
x
β
<
ο£²ο£±
β₯
ο£³
By Example 3, we know that f is continuous at x = 0 Let c be a real number such that c < 0 |
1 | 1790-1793 | Rationalised 2023-24
MATHEMATICS
108
Example 7 Is the function defined by f(x) = | x |, a continuous function Solution We may rewrite f as
f (x) =
, if
0
, if
0
x
x
x
x
β
<
ο£²ο£±
β₯
ο£³
By Example 3, we know that f is continuous at x = 0 Let c be a real number such that c < 0 Then f(c) = β c |
1 | 1791-1794 | Solution We may rewrite f as
f (x) =
, if
0
, if
0
x
x
x
x
β
<
ο£²ο£±
β₯
ο£³
By Example 3, we know that f is continuous at x = 0 Let c be a real number such that c < 0 Then f(c) = β c Also
lim
( )
x
c f x
β
= lim (
)
β
x
c
x
c
β
β
=
(Why |
1 | 1792-1795 | Let c be a real number such that c < 0 Then f(c) = β c Also
lim
( )
x
c f x
β
= lim (
)
β
x
c
x
c
β
β
=
(Why )
Since lim
( )
( )
x
c f x
f c
β
=
, f is continuous at all negative real numbers |
1 | 1793-1796 | Then f(c) = β c Also
lim
( )
x
c f x
β
= lim (
)
β
x
c
x
c
β
β
=
(Why )
Since lim
( )
( )
x
c f x
f c
β
=
, f is continuous at all negative real numbers Now, let c be a real number such that c > 0 |
1 | 1794-1797 | Also
lim
( )
x
c f x
β
= lim (
)
β
x
c
x
c
β
β
=
(Why )
Since lim
( )
( )
x
c f x
f c
β
=
, f is continuous at all negative real numbers Now, let c be a real number such that c > 0 Then f (c) = c |
1 | 1795-1798 | )
Since lim
( )
( )
x
c f x
f c
β
=
, f is continuous at all negative real numbers Now, let c be a real number such that c > 0 Then f (c) = c Also
lim
( )
x
c f x
β
= lim
x
c x
c
β
= (Why |
1 | 1796-1799 | Now, let c be a real number such that c > 0 Then f (c) = c Also
lim
( )
x
c f x
β
= lim
x
c x
c
β
= (Why )
Since lim
( )
( )
x
c f x
f c
β
=
, f is continuous at all positive real numbers |
1 | 1797-1800 | Then f (c) = c Also
lim
( )
x
c f x
β
= lim
x
c x
c
β
= (Why )
Since lim
( )
( )
x
c f x
f c
β
=
, f is continuous at all positive real numbers Hence, f
is continuous at all points |
1 | 1798-1801 | Also
lim
( )
x
c f x
β
= lim
x
c x
c
β
= (Why )
Since lim
( )
( )
x
c f x
f c
β
=
, f is continuous at all positive real numbers Hence, f
is continuous at all points Example 8 Discuss the continuity of the function f given by f (x) = x3 + x2 β 1 |
1 | 1799-1802 | )
Since lim
( )
( )
x
c f x
f c
β
=
, f is continuous at all positive real numbers Hence, f
is continuous at all points Example 8 Discuss the continuity of the function f given by f (x) = x3 + x2 β 1 Solution Clearly f is defined at every real number c and its value at c is c3 + c2 β 1 |
1 | 1800-1803 | Hence, f
is continuous at all points Example 8 Discuss the continuity of the function f given by f (x) = x3 + x2 β 1 Solution Clearly f is defined at every real number c and its value at c is c3 + c2 β 1 We
also know that
lim
( )
x
c f x
β
=
3
2
3
2
lim (
1)
1
x
c x
x
c
c
β
+
β
=
+
β
Thus lim
( )
( )
x
c f x
f c
β
=
, and hence f is continuous at every real number |
1 | 1801-1804 | Example 8 Discuss the continuity of the function f given by f (x) = x3 + x2 β 1 Solution Clearly f is defined at every real number c and its value at c is c3 + c2 β 1 We
also know that
lim
( )
x
c f x
β
=
3
2
3
2
lim (
1)
1
x
c x
x
c
c
β
+
β
=
+
β
Thus lim
( )
( )
x
c f x
f c
β
=
, and hence f is continuous at every real number This means
f is a continuous function |
1 | 1802-1805 | Solution Clearly f is defined at every real number c and its value at c is c3 + c2 β 1 We
also know that
lim
( )
x
c f x
β
=
3
2
3
2
lim (
1)
1
x
c x
x
c
c
β
+
β
=
+
β
Thus lim
( )
( )
x
c f x
f c
β
=
, and hence f is continuous at every real number This means
f is a continuous function Example 9 Discuss the continuity of the function f defined by f (x) = 1
x , x β 0 |
1 | 1803-1806 | We
also know that
lim
( )
x
c f x
β
=
3
2
3
2
lim (
1)
1
x
c x
x
c
c
β
+
β
=
+
β
Thus lim
( )
( )
x
c f x
f c
β
=
, and hence f is continuous at every real number This means
f is a continuous function Example 9 Discuss the continuity of the function f defined by f (x) = 1
x , x β 0 Solution Fix any non zero real number c, we have
1
1
lim
( )
lim
x
c
x
c
f x
x
c
β
=β
=
Also, since for c β 0,
1
f c( )
=c
, we have lim
( )
( )
x
c f x
f c
β
=
and hence, f is continuous
at every point in the domain of f |
1 | 1804-1807 | This means
f is a continuous function Example 9 Discuss the continuity of the function f defined by f (x) = 1
x , x β 0 Solution Fix any non zero real number c, we have
1
1
lim
( )
lim
x
c
x
c
f x
x
c
β
=β
=
Also, since for c β 0,
1
f c( )
=c
, we have lim
( )
( )
x
c f x
f c
β
=
and hence, f is continuous
at every point in the domain of f Thus f is a continuous function |
1 | 1805-1808 | Example 9 Discuss the continuity of the function f defined by f (x) = 1
x , x β 0 Solution Fix any non zero real number c, we have
1
1
lim
( )
lim
x
c
x
c
f x
x
c
β
=β
=
Also, since for c β 0,
1
f c( )
=c
, we have lim
( )
( )
x
c f x
f c
β
=
and hence, f is continuous
at every point in the domain of f Thus f is a continuous function Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
109
We take this opportunity to explain the concept of infinity |
1 | 1806-1809 | Solution Fix any non zero real number c, we have
1
1
lim
( )
lim
x
c
x
c
f x
x
c
β
=β
=
Also, since for c β 0,
1
f c( )
=c
, we have lim
( )
( )
x
c f x
f c
β
=
and hence, f is continuous
at every point in the domain of f Thus f is a continuous function Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
109
We take this opportunity to explain the concept of infinity This we do by analysing
the function f (x) = 1
x near x = 0 |
1 | 1807-1810 | Thus f is a continuous function Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
109
We take this opportunity to explain the concept of infinity This we do by analysing
the function f (x) = 1
x near x = 0 To carry out this analysis we follow the usual trick of
finding the value of the function at real numbers close to 0 |
1 | 1808-1811 | Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
109
We take this opportunity to explain the concept of infinity This we do by analysing
the function f (x) = 1
x near x = 0 To carry out this analysis we follow the usual trick of
finding the value of the function at real numbers close to 0 Essentially we are trying to
find the right hand limit of f at 0 |
1 | 1809-1812 | This we do by analysing
the function f (x) = 1
x near x = 0 To carry out this analysis we follow the usual trick of
finding the value of the function at real numbers close to 0 Essentially we are trying to
find the right hand limit of f at 0 We tabulate this in the following (Table 5 |
1 | 1810-1813 | To carry out this analysis we follow the usual trick of
finding the value of the function at real numbers close to 0 Essentially we are trying to
find the right hand limit of f at 0 We tabulate this in the following (Table 5 1) |
1 | 1811-1814 | Essentially we are trying to
find the right hand limit of f at 0 We tabulate this in the following (Table 5 1) Table 5 |
1 | 1812-1815 | We tabulate this in the following (Table 5 1) Table 5 1
x
1
0 |
1 | 1813-1816 | 1) Table 5 1
x
1
0 3
0 |
1 | 1814-1817 | Table 5 1
x
1
0 3
0 2
0 |
1 | 1815-1818 | 1
x
1
0 3
0 2
0 1 = 10β1
0 |
1 | 1816-1819 | 3
0 2
0 1 = 10β1
0 01 = 10β2
0 |
1 | 1817-1820 | 2
0 1 = 10β1
0 01 = 10β2
0 001 = 10β3
10βn
f (x)
1
3 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.