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1 | 1818-1821 | 1 = 10β1
0 01 = 10β2
0 001 = 10β3
10βn
f (x)
1
3 333 |
1 | 1819-1822 | 01 = 10β2
0 001 = 10β3
10βn
f (x)
1
3 333 5
10
100 = 102
1000 = 103
10n
We observe that as x gets closer to 0 from the right, the value of f (x) shoots up
higher |
1 | 1820-1823 | 001 = 10β3
10βn
f (x)
1
3 333 5
10
100 = 102
1000 = 103
10n
We observe that as x gets closer to 0 from the right, the value of f (x) shoots up
higher This may be rephrased as: the value of f (x) may be made larger than any given
number by choosing a positive real number very close to 0 |
1 | 1821-1824 | 333 5
10
100 = 102
1000 = 103
10n
We observe that as x gets closer to 0 from the right, the value of f (x) shoots up
higher This may be rephrased as: the value of f (x) may be made larger than any given
number by choosing a positive real number very close to 0 In symbols, we write
lim0
( )
x
f x
β+
= + β
(to be read as: the right hand limit of f (x) at 0 is plus infinity) |
1 | 1822-1825 | 5
10
100 = 102
1000 = 103
10n
We observe that as x gets closer to 0 from the right, the value of f (x) shoots up
higher This may be rephrased as: the value of f (x) may be made larger than any given
number by choosing a positive real number very close to 0 In symbols, we write
lim0
( )
x
f x
β+
= + β
(to be read as: the right hand limit of f (x) at 0 is plus infinity) We wish to emphasise
that + β is NOT a real number and hence the right hand limit of f at 0 does not exist (as
a real number) |
1 | 1823-1826 | This may be rephrased as: the value of f (x) may be made larger than any given
number by choosing a positive real number very close to 0 In symbols, we write
lim0
( )
x
f x
β+
= + β
(to be read as: the right hand limit of f (x) at 0 is plus infinity) We wish to emphasise
that + β is NOT a real number and hence the right hand limit of f at 0 does not exist (as
a real number) Similarly, the left hand limit of f at 0 may be found |
1 | 1824-1827 | In symbols, we write
lim0
( )
x
f x
β+
= + β
(to be read as: the right hand limit of f (x) at 0 is plus infinity) We wish to emphasise
that + β is NOT a real number and hence the right hand limit of f at 0 does not exist (as
a real number) Similarly, the left hand limit of f at 0 may be found The following table is self
explanatory |
1 | 1825-1828 | We wish to emphasise
that + β is NOT a real number and hence the right hand limit of f at 0 does not exist (as
a real number) Similarly, the left hand limit of f at 0 may be found The following table is self
explanatory Table 5 |
1 | 1826-1829 | Similarly, the left hand limit of f at 0 may be found The following table is self
explanatory Table 5 2
x
β 1
β 0 |
1 | 1827-1830 | The following table is self
explanatory Table 5 2
x
β 1
β 0 3
β 0 |
1 | 1828-1831 | Table 5 2
x
β 1
β 0 3
β 0 2
β 10β1
β 10β2
β 10β3
β 10βn
f (x)
β 1
β 3 |
1 | 1829-1832 | 2
x
β 1
β 0 3
β 0 2
β 10β1
β 10β2
β 10β3
β 10βn
f (x)
β 1
β 3 333 |
1 | 1830-1833 | 3
β 0 2
β 10β1
β 10β2
β 10β3
β 10βn
f (x)
β 1
β 3 333 β 5
β 10
β 102
β 103
β 10n
From the Table 5 |
1 | 1831-1834 | 2
β 10β1
β 10β2
β 10β3
β 10βn
f (x)
β 1
β 3 333 β 5
β 10
β 102
β 103
β 10n
From the Table 5 2, we deduce that the
value of f(x) may be made smaller than any
given number by choosing a negative real
number very close to 0 |
1 | 1832-1835 | 333 β 5
β 10
β 102
β 103
β 10n
From the Table 5 2, we deduce that the
value of f(x) may be made smaller than any
given number by choosing a negative real
number very close to 0 In symbols,
we write
lim0
( )
x
f x
ββ
= β β
(to be read as: the left hand limit of f (x) at 0 is
minus infinity) |
1 | 1833-1836 | β 5
β 10
β 102
β 103
β 10n
From the Table 5 2, we deduce that the
value of f(x) may be made smaller than any
given number by choosing a negative real
number very close to 0 In symbols,
we write
lim0
( )
x
f x
ββ
= β β
(to be read as: the left hand limit of f (x) at 0 is
minus infinity) Again, we wish to emphasise
that β β is NOT a real number and hence the
left hand limit of f at 0 does not exist (as a real
number) |
1 | 1834-1837 | 2, we deduce that the
value of f(x) may be made smaller than any
given number by choosing a negative real
number very close to 0 In symbols,
we write
lim0
( )
x
f x
ββ
= β β
(to be read as: the left hand limit of f (x) at 0 is
minus infinity) Again, we wish to emphasise
that β β is NOT a real number and hence the
left hand limit of f at 0 does not exist (as a real
number) The graph of the reciprocal function
given in Fig 5 |
1 | 1835-1838 | In symbols,
we write
lim0
( )
x
f x
ββ
= β β
(to be read as: the left hand limit of f (x) at 0 is
minus infinity) Again, we wish to emphasise
that β β is NOT a real number and hence the
left hand limit of f at 0 does not exist (as a real
number) The graph of the reciprocal function
given in Fig 5 3 is a geometric representation
of the above mentioned facts |
1 | 1836-1839 | Again, we wish to emphasise
that β β is NOT a real number and hence the
left hand limit of f at 0 does not exist (as a real
number) The graph of the reciprocal function
given in Fig 5 3 is a geometric representation
of the above mentioned facts Fig 5 |
1 | 1837-1840 | The graph of the reciprocal function
given in Fig 5 3 is a geometric representation
of the above mentioned facts Fig 5 3
Rationalised 2023-24
MATHEMATICS
110
Example 10 Discuss the continuity of the function f defined by
f (x) =
2, if
1
2, if
1
x
x
x
x
+
β€
ο£±
ο£² β
>
ο£³
Solution The function f is defined at all points of the real line |
1 | 1838-1841 | 3 is a geometric representation
of the above mentioned facts Fig 5 3
Rationalised 2023-24
MATHEMATICS
110
Example 10 Discuss the continuity of the function f defined by
f (x) =
2, if
1
2, if
1
x
x
x
x
+
β€
ο£±
ο£² β
>
ο£³
Solution The function f is defined at all points of the real line Case 1 If c < 1, then f(c) = c + 2 |
1 | 1839-1842 | Fig 5 3
Rationalised 2023-24
MATHEMATICS
110
Example 10 Discuss the continuity of the function f defined by
f (x) =
2, if
1
2, if
1
x
x
x
x
+
β€
ο£±
ο£² β
>
ο£³
Solution The function f is defined at all points of the real line Case 1 If c < 1, then f(c) = c + 2 Therefore, lim
( )
lim(
2)
2
x
c
x
c
f x
x
c
β
=β
+
=
+
Thus, f is continuous at all real numbers less than 1 |
1 | 1840-1843 | 3
Rationalised 2023-24
MATHEMATICS
110
Example 10 Discuss the continuity of the function f defined by
f (x) =
2, if
1
2, if
1
x
x
x
x
+
β€
ο£±
ο£² β
>
ο£³
Solution The function f is defined at all points of the real line Case 1 If c < 1, then f(c) = c + 2 Therefore, lim
( )
lim(
2)
2
x
c
x
c
f x
x
c
β
=β
+
=
+
Thus, f is continuous at all real numbers less than 1 Case 2 If c > 1, then f (c) = c β 2 |
1 | 1841-1844 | Case 1 If c < 1, then f(c) = c + 2 Therefore, lim
( )
lim(
2)
2
x
c
x
c
f x
x
c
β
=β
+
=
+
Thus, f is continuous at all real numbers less than 1 Case 2 If c > 1, then f (c) = c β 2 Therefore,
lim
( )
lim
x
c
x
c
f x
β
=β
(x β 2) = c β 2 = f (c)
Thus, f is continuous at all points x > 1 |
1 | 1842-1845 | Therefore, lim
( )
lim(
2)
2
x
c
x
c
f x
x
c
β
=β
+
=
+
Thus, f is continuous at all real numbers less than 1 Case 2 If c > 1, then f (c) = c β 2 Therefore,
lim
( )
lim
x
c
x
c
f x
β
=β
(x β 2) = c β 2 = f (c)
Thus, f is continuous at all points x > 1 Case 3 If c = 1, then the left hand limit of f at
x = 1 is
β
β
1
1
lim
( )
lim (
2)
1
2
3
x
x
f x
x
β
=β
+
= +
=
The right hand limit of f at x = 1 is
1
1
lim
( )
lim (
2)
1
2
1
x
x
f x
x
+
+
β
=β
β
= β
= β
Since the left and right hand limits of f at x = 1
do not coincide, f is not continuous at x = 1 |
1 | 1843-1846 | Case 2 If c > 1, then f (c) = c β 2 Therefore,
lim
( )
lim
x
c
x
c
f x
β
=β
(x β 2) = c β 2 = f (c)
Thus, f is continuous at all points x > 1 Case 3 If c = 1, then the left hand limit of f at
x = 1 is
β
β
1
1
lim
( )
lim (
2)
1
2
3
x
x
f x
x
β
=β
+
= +
=
The right hand limit of f at x = 1 is
1
1
lim
( )
lim (
2)
1
2
1
x
x
f x
x
+
+
β
=β
β
= β
= β
Since the left and right hand limits of f at x = 1
do not coincide, f is not continuous at x = 1 Hence
x = 1 is the only point of discontinuity of f |
1 | 1844-1847 | Therefore,
lim
( )
lim
x
c
x
c
f x
β
=β
(x β 2) = c β 2 = f (c)
Thus, f is continuous at all points x > 1 Case 3 If c = 1, then the left hand limit of f at
x = 1 is
β
β
1
1
lim
( )
lim (
2)
1
2
3
x
x
f x
x
β
=β
+
= +
=
The right hand limit of f at x = 1 is
1
1
lim
( )
lim (
2)
1
2
1
x
x
f x
x
+
+
β
=β
β
= β
= β
Since the left and right hand limits of f at x = 1
do not coincide, f is not continuous at x = 1 Hence
x = 1 is the only point of discontinuity of f The graph of the function is given in Fig 5 |
1 | 1845-1848 | Case 3 If c = 1, then the left hand limit of f at
x = 1 is
β
β
1
1
lim
( )
lim (
2)
1
2
3
x
x
f x
x
β
=β
+
= +
=
The right hand limit of f at x = 1 is
1
1
lim
( )
lim (
2)
1
2
1
x
x
f x
x
+
+
β
=β
β
= β
= β
Since the left and right hand limits of f at x = 1
do not coincide, f is not continuous at x = 1 Hence
x = 1 is the only point of discontinuity of f The graph of the function is given in Fig 5 4 |
1 | 1846-1849 | Hence
x = 1 is the only point of discontinuity of f The graph of the function is given in Fig 5 4 Example 11 Find all the points of discontinuity of the function f defined by
f (x) =
2, if
1
0,
if
1
2, if
1
x
x
x
x
x
+
<

=
ο£²
 β
>
ο£³
Solution As in the previous example we find that f
is continuous at all real numbers x β 1 |
1 | 1847-1850 | The graph of the function is given in Fig 5 4 Example 11 Find all the points of discontinuity of the function f defined by
f (x) =
2, if
1
0,
if
1
2, if
1
x
x
x
x
x
+
<

=
ο£²
 β
>
ο£³
Solution As in the previous example we find that f
is continuous at all real numbers x β 1 The left
hand limit of f at x = 1 is
β
1
1
lim
( )
lim (
2)
1
2
3
x
x
f x
x
ββ
=β
+
= +
=
The right hand limit of f at x = 1 is
1
1
lim
( )
lim (
2)
1
2
1
x
x
f x
x
+
+
β
=β
β
= β
= β
Since, the left and right hand limits of f at x = 1
do not coincide, f is not continuous at x = 1 |
1 | 1848-1851 | 4 Example 11 Find all the points of discontinuity of the function f defined by
f (x) =
2, if
1
0,
if
1
2, if
1
x
x
x
x
x
+
<

=
ο£²
 β
>
ο£³
Solution As in the previous example we find that f
is continuous at all real numbers x β 1 The left
hand limit of f at x = 1 is
β
1
1
lim
( )
lim (
2)
1
2
3
x
x
f x
x
ββ
=β
+
= +
=
The right hand limit of f at x = 1 is
1
1
lim
( )
lim (
2)
1
2
1
x
x
f x
x
+
+
β
=β
β
= β
= β
Since, the left and right hand limits of f at x = 1
do not coincide, f is not continuous at x = 1 Hence
x = 1 is the only point of discontinuity of f |
1 | 1849-1852 | Example 11 Find all the points of discontinuity of the function f defined by
f (x) =
2, if
1
0,
if
1
2, if
1
x
x
x
x
x
+
<

=
ο£²
 β
>
ο£³
Solution As in the previous example we find that f
is continuous at all real numbers x β 1 The left
hand limit of f at x = 1 is
β
1
1
lim
( )
lim (
2)
1
2
3
x
x
f x
x
ββ
=β
+
= +
=
The right hand limit of f at x = 1 is
1
1
lim
( )
lim (
2)
1
2
1
x
x
f x
x
+
+
β
=β
β
= β
= β
Since, the left and right hand limits of f at x = 1
do not coincide, f is not continuous at x = 1 Hence
x = 1 is the only point of discontinuity of f The
graph of the function is given in the Fig 5 |
1 | 1850-1853 | The left
hand limit of f at x = 1 is
β
1
1
lim
( )
lim (
2)
1
2
3
x
x
f x
x
ββ
=β
+
= +
=
The right hand limit of f at x = 1 is
1
1
lim
( )
lim (
2)
1
2
1
x
x
f x
x
+
+
β
=β
β
= β
= β
Since, the left and right hand limits of f at x = 1
do not coincide, f is not continuous at x = 1 Hence
x = 1 is the only point of discontinuity of f The
graph of the function is given in the Fig 5 5 |
1 | 1851-1854 | Hence
x = 1 is the only point of discontinuity of f The
graph of the function is given in the Fig 5 5 Fig 5 |
1 | 1852-1855 | The
graph of the function is given in the Fig 5 5 Fig 5 4
Fig 5 |
1 | 1853-1856 | 5 Fig 5 4
Fig 5 5
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
111
Example 12 Discuss the continuity of the function defined by
f(x) =
2, if
0
2, if
0
x
x
x
x
+
<
ο£±
ο£²β +
>
ο£³
Solution Observe that the function is defined at all real numbers except at 0 |
1 | 1854-1857 | Fig 5 4
Fig 5 5
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
111
Example 12 Discuss the continuity of the function defined by
f(x) =
2, if
0
2, if
0
x
x
x
x
+
<
ο£±
ο£²β +
>
ο£³
Solution Observe that the function is defined at all real numbers except at 0 Domain
of definition of this function is
D1 βͺ D2 where D1 = {x β R : x < 0} and
D2 = {x β R : x > 0}
Case 1 If c β D1, then lim
( )
lim
x
c
x
c
f x
β
=β
(x + 2)
= c + 2 = f (c) and hence f is continuous in D1 |
1 | 1855-1858 | 4
Fig 5 5
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
111
Example 12 Discuss the continuity of the function defined by
f(x) =
2, if
0
2, if
0
x
x
x
x
+
<
ο£±
ο£²β +
>
ο£³
Solution Observe that the function is defined at all real numbers except at 0 Domain
of definition of this function is
D1 βͺ D2 where D1 = {x β R : x < 0} and
D2 = {x β R : x > 0}
Case 1 If c β D1, then lim
( )
lim
x
c
x
c
f x
β
=β
(x + 2)
= c + 2 = f (c) and hence f is continuous in D1 Case 2 If c β D2, then lim
( )
lim
x
c
x
c
f x
β
=β
(β x + 2)
= β c + 2 = f (c) and hence f is continuous in D2 |
1 | 1856-1859 | 5
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
111
Example 12 Discuss the continuity of the function defined by
f(x) =
2, if
0
2, if
0
x
x
x
x
+
<
ο£±
ο£²β +
>
ο£³
Solution Observe that the function is defined at all real numbers except at 0 Domain
of definition of this function is
D1 βͺ D2 where D1 = {x β R : x < 0} and
D2 = {x β R : x > 0}
Case 1 If c β D1, then lim
( )
lim
x
c
x
c
f x
β
=β
(x + 2)
= c + 2 = f (c) and hence f is continuous in D1 Case 2 If c β D2, then lim
( )
lim
x
c
x
c
f x
β
=β
(β x + 2)
= β c + 2 = f (c) and hence f is continuous in D2 Since f is continuous at all points in the domain of f,
we deduce that f is continuous |
1 | 1857-1860 | Domain
of definition of this function is
D1 βͺ D2 where D1 = {x β R : x < 0} and
D2 = {x β R : x > 0}
Case 1 If c β D1, then lim
( )
lim
x
c
x
c
f x
β
=β
(x + 2)
= c + 2 = f (c) and hence f is continuous in D1 Case 2 If c β D2, then lim
( )
lim
x
c
x
c
f x
β
=β
(β x + 2)
= β c + 2 = f (c) and hence f is continuous in D2 Since f is continuous at all points in the domain of f,
we deduce that f is continuous Graph of this
function is given in the Fig 5 |
1 | 1858-1861 | Case 2 If c β D2, then lim
( )
lim
x
c
x
c
f x
β
=β
(β x + 2)
= β c + 2 = f (c) and hence f is continuous in D2 Since f is continuous at all points in the domain of f,
we deduce that f is continuous Graph of this
function is given in the Fig 5 6 |
1 | 1859-1862 | Since f is continuous at all points in the domain of f,
we deduce that f is continuous Graph of this
function is given in the Fig 5 6 Note that to graph
this function we need to lift the pen from the plane
of the paper, but we need to do that only for those points where the function is not
defined |
1 | 1860-1863 | Graph of this
function is given in the Fig 5 6 Note that to graph
this function we need to lift the pen from the plane
of the paper, but we need to do that only for those points where the function is not
defined Example 13 Discuss the continuity of the function f given by
f (x) =
,2
if
0
, if
0
x
x
x
x
β₯

<

Solution Clearly the function is defined at
every real number |
1 | 1861-1864 | 6 Note that to graph
this function we need to lift the pen from the plane
of the paper, but we need to do that only for those points where the function is not
defined Example 13 Discuss the continuity of the function f given by
f (x) =
,2
if
0
, if
0
x
x
x
x
β₯

<

Solution Clearly the function is defined at
every real number Graph of the function is
given in Fig 5 |
1 | 1862-1865 | Note that to graph
this function we need to lift the pen from the plane
of the paper, but we need to do that only for those points where the function is not
defined Example 13 Discuss the continuity of the function f given by
f (x) =
,2
if
0
, if
0
x
x
x
x
β₯

<

Solution Clearly the function is defined at
every real number Graph of the function is
given in Fig 5 7 |
1 | 1863-1866 | Example 13 Discuss the continuity of the function f given by
f (x) =
,2
if
0
, if
0
x
x
x
x
β₯

<

Solution Clearly the function is defined at
every real number Graph of the function is
given in Fig 5 7 By inspection, it seems prudent
to partition the domain of definition of f into
three disjoint subsets of the real line |
1 | 1864-1867 | Graph of the function is
given in Fig 5 7 By inspection, it seems prudent
to partition the domain of definition of f into
three disjoint subsets of the real line Let
D1 = {x β R : x < 0}, D2 = {0} and
D3 = {x β R : x > 0}
Case 1 At any point in D1, we have f(x) = x2 and it is easy to see that it is continuous
there (see Example 2) |
1 | 1865-1868 | 7 By inspection, it seems prudent
to partition the domain of definition of f into
three disjoint subsets of the real line Let
D1 = {x β R : x < 0}, D2 = {0} and
D3 = {x β R : x > 0}
Case 1 At any point in D1, we have f(x) = x2 and it is easy to see that it is continuous
there (see Example 2) Case 2 At any point in D3, we have f(x) = x and it is easy to see that it is continuous
there (see Example 6) |
1 | 1866-1869 | By inspection, it seems prudent
to partition the domain of definition of f into
three disjoint subsets of the real line Let
D1 = {x β R : x < 0}, D2 = {0} and
D3 = {x β R : x > 0}
Case 1 At any point in D1, we have f(x) = x2 and it is easy to see that it is continuous
there (see Example 2) Case 2 At any point in D3, we have f(x) = x and it is easy to see that it is continuous
there (see Example 6) Fig 5 |
1 | 1867-1870 | Let
D1 = {x β R : x < 0}, D2 = {0} and
D3 = {x β R : x > 0}
Case 1 At any point in D1, we have f(x) = x2 and it is easy to see that it is continuous
there (see Example 2) Case 2 At any point in D3, we have f(x) = x and it is easy to see that it is continuous
there (see Example 6) Fig 5 6
Fig 5 |
1 | 1868-1871 | Case 2 At any point in D3, we have f(x) = x and it is easy to see that it is continuous
there (see Example 6) Fig 5 6
Fig 5 7
Rationalised 2023-24
MATHEMATICS
112
Case 3 Now we analyse the function at x = 0 |
1 | 1869-1872 | Fig 5 6
Fig 5 7
Rationalised 2023-24
MATHEMATICS
112
Case 3 Now we analyse the function at x = 0 The value of the function at 0 is f(0) = 0 |
1 | 1870-1873 | 6
Fig 5 7
Rationalised 2023-24
MATHEMATICS
112
Case 3 Now we analyse the function at x = 0 The value of the function at 0 is f(0) = 0 The left hand limit of f at 0 is
β
2
2
0
0
lim
( )
lim
0
0
x
x
f x
βx
β
=β
=
=
The right hand limit of f at 0 is
0
0
lim
( )
lim
0
x
x
f x
x
+
+
β
=β
=
Thus
lim0
( )
0
x
f x
β
=
= f(0) and hence f is continuous at 0 |
1 | 1871-1874 | 7
Rationalised 2023-24
MATHEMATICS
112
Case 3 Now we analyse the function at x = 0 The value of the function at 0 is f(0) = 0 The left hand limit of f at 0 is
β
2
2
0
0
lim
( )
lim
0
0
x
x
f x
βx
β
=β
=
=
The right hand limit of f at 0 is
0
0
lim
( )
lim
0
x
x
f x
x
+
+
β
=β
=
Thus
lim0
( )
0
x
f x
β
=
= f(0) and hence f is continuous at 0 This means that f is
continuous at every point in its domain and hence, f is a continuous function |
1 | 1872-1875 | The value of the function at 0 is f(0) = 0 The left hand limit of f at 0 is
β
2
2
0
0
lim
( )
lim
0
0
x
x
f x
βx
β
=β
=
=
The right hand limit of f at 0 is
0
0
lim
( )
lim
0
x
x
f x
x
+
+
β
=β
=
Thus
lim0
( )
0
x
f x
β
=
= f(0) and hence f is continuous at 0 This means that f is
continuous at every point in its domain and hence, f is a continuous function Example 14 Show that every polynomial function is continuous |
1 | 1873-1876 | The left hand limit of f at 0 is
β
2
2
0
0
lim
( )
lim
0
0
x
x
f x
βx
β
=β
=
=
The right hand limit of f at 0 is
0
0
lim
( )
lim
0
x
x
f x
x
+
+
β
=β
=
Thus
lim0
( )
0
x
f x
β
=
= f(0) and hence f is continuous at 0 This means that f is
continuous at every point in its domain and hence, f is a continuous function Example 14 Show that every polynomial function is continuous Solution Recall that a function p is a polynomial function if it is defined by
p(x) = a0 + a1 x + |
1 | 1874-1877 | This means that f is
continuous at every point in its domain and hence, f is a continuous function Example 14 Show that every polynomial function is continuous Solution Recall that a function p is a polynomial function if it is defined by
p(x) = a0 + a1 x + + an xn for some natural number n, an β 0 and ai β R |
1 | 1875-1878 | Example 14 Show that every polynomial function is continuous Solution Recall that a function p is a polynomial function if it is defined by
p(x) = a0 + a1 x + + an xn for some natural number n, an β 0 and ai β R Clearly this
function is defined for every real number |
1 | 1876-1879 | Solution Recall that a function p is a polynomial function if it is defined by
p(x) = a0 + a1 x + + an xn for some natural number n, an β 0 and ai β R Clearly this
function is defined for every real number For a fixed real number c, we have
lim
( )
( )
x
c p x
p c
β
=
By definition, p is continuous at c |
1 | 1877-1880 | + an xn for some natural number n, an β 0 and ai β R Clearly this
function is defined for every real number For a fixed real number c, we have
lim
( )
( )
x
c p x
p c
β
=
By definition, p is continuous at c Since c is any real number, p is continuous at
every real number and hence p is a continuous function |
1 | 1878-1881 | Clearly this
function is defined for every real number For a fixed real number c, we have
lim
( )
( )
x
c p x
p c
β
=
By definition, p is continuous at c Since c is any real number, p is continuous at
every real number and hence p is a continuous function Example 15 Find all the points of discontinuity of the greatest integer function defined
by f (x) = [x], where [x] denotes the greatest integer less than or equal to x |
1 | 1879-1882 | For a fixed real number c, we have
lim
( )
( )
x
c p x
p c
β
=
By definition, p is continuous at c Since c is any real number, p is continuous at
every real number and hence p is a continuous function Example 15 Find all the points of discontinuity of the greatest integer function defined
by f (x) = [x], where [x] denotes the greatest integer less than or equal to x Solution First observe that f is defined for all real numbers |
1 | 1880-1883 | Since c is any real number, p is continuous at
every real number and hence p is a continuous function Example 15 Find all the points of discontinuity of the greatest integer function defined
by f (x) = [x], where [x] denotes the greatest integer less than or equal to x Solution First observe that f is defined for all real numbers Graph of the function is
given in Fig 5 |
1 | 1881-1884 | Example 15 Find all the points of discontinuity of the greatest integer function defined
by f (x) = [x], where [x] denotes the greatest integer less than or equal to x Solution First observe that f is defined for all real numbers Graph of the function is
given in Fig 5 8 |
1 | 1882-1885 | Solution First observe that f is defined for all real numbers Graph of the function is
given in Fig 5 8 From the graph it looks like that f is discontinuous at every integral
point |
1 | 1883-1886 | Graph of the function is
given in Fig 5 8 From the graph it looks like that f is discontinuous at every integral
point Below we explore, if this is true |
1 | 1884-1887 | 8 From the graph it looks like that f is discontinuous at every integral
point Below we explore, if this is true Fig 5 |
1 | 1885-1888 | From the graph it looks like that f is discontinuous at every integral
point Below we explore, if this is true Fig 5 8
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
113
Case 1 Let c be a real number which is not equal to any integer |
1 | 1886-1889 | Below we explore, if this is true Fig 5 8
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
113
Case 1 Let c be a real number which is not equal to any integer It is evident from the
graph that for all real numbers close to c the value of the function is equal to [c]; i |
1 | 1887-1890 | Fig 5 8
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
113
Case 1 Let c be a real number which is not equal to any integer It is evident from the
graph that for all real numbers close to c the value of the function is equal to [c]; i e |
1 | 1888-1891 | 8
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
113
Case 1 Let c be a real number which is not equal to any integer It is evident from the
graph that for all real numbers close to c the value of the function is equal to [c]; i e ,
lim
( )
lim [ ]
[ ]
x
c
x
c
f x
x
c
β
=β
= |
1 | 1889-1892 | It is evident from the
graph that for all real numbers close to c the value of the function is equal to [c]; i e ,
lim
( )
lim [ ]
[ ]
x
c
x
c
f x
x
c
β
=β
= Also f(c) = [c] and hence the function is continuous at all real
numbers not equal to integers |
1 | 1890-1893 | e ,
lim
( )
lim [ ]
[ ]
x
c
x
c
f x
x
c
β
=β
= Also f(c) = [c] and hence the function is continuous at all real
numbers not equal to integers Case 2 Let c be an integer |
1 | 1891-1894 | ,
lim
( )
lim [ ]
[ ]
x
c
x
c
f x
x
c
β
=β
= Also f(c) = [c] and hence the function is continuous at all real
numbers not equal to integers Case 2 Let c be an integer Then we can find a sufficiently small real number
r > 0 such that [c β r] = c β 1 whereas [c + r] = c |
1 | 1892-1895 | Also f(c) = [c] and hence the function is continuous at all real
numbers not equal to integers Case 2 Let c be an integer Then we can find a sufficiently small real number
r > 0 such that [c β r] = c β 1 whereas [c + r] = c This, in terms of limits mean that
xlim
βcβ
f (x) = c β 1, lim
x
βc+
f (x) = c
Since these limits cannot be equal to each other for any c, the function is
discontinuous at every integral point |
1 | 1893-1896 | Case 2 Let c be an integer Then we can find a sufficiently small real number
r > 0 such that [c β r] = c β 1 whereas [c + r] = c This, in terms of limits mean that
xlim
βcβ
f (x) = c β 1, lim
x
βc+
f (x) = c
Since these limits cannot be equal to each other for any c, the function is
discontinuous at every integral point 5 |
1 | 1894-1897 | Then we can find a sufficiently small real number
r > 0 such that [c β r] = c β 1 whereas [c + r] = c This, in terms of limits mean that
xlim
βcβ
f (x) = c β 1, lim
x
βc+
f (x) = c
Since these limits cannot be equal to each other for any c, the function is
discontinuous at every integral point 5 2 |
1 | 1895-1898 | This, in terms of limits mean that
xlim
βcβ
f (x) = c β 1, lim
x
βc+
f (x) = c
Since these limits cannot be equal to each other for any c, the function is
discontinuous at every integral point 5 2 1 Algebra of continuous functions
In the previous class, after having understood the concept of limits, we learnt some
algebra of limits |
1 | 1896-1899 | 5 2 1 Algebra of continuous functions
In the previous class, after having understood the concept of limits, we learnt some
algebra of limits Analogously, now we will study some algebra of continuous functions |
1 | 1897-1900 | 2 1 Algebra of continuous functions
In the previous class, after having understood the concept of limits, we learnt some
algebra of limits Analogously, now we will study some algebra of continuous functions Since continuity of a function at a point is entirely dictated by the limit of the function at
that point, it is reasonable to expect results analogous to the case of limits |
1 | 1898-1901 | 1 Algebra of continuous functions
In the previous class, after having understood the concept of limits, we learnt some
algebra of limits Analogously, now we will study some algebra of continuous functions Since continuity of a function at a point is entirely dictated by the limit of the function at
that point, it is reasonable to expect results analogous to the case of limits Theorem 1 Suppose f and g be two real functions continuous at a real number c |
1 | 1899-1902 | Analogously, now we will study some algebra of continuous functions Since continuity of a function at a point is entirely dictated by the limit of the function at
that point, it is reasonable to expect results analogous to the case of limits Theorem 1 Suppose f and g be two real functions continuous at a real number c Then
(1)
f + g is continuous at x = c |
1 | 1900-1903 | Since continuity of a function at a point is entirely dictated by the limit of the function at
that point, it is reasonable to expect results analogous to the case of limits Theorem 1 Suppose f and g be two real functions continuous at a real number c Then
(1)
f + g is continuous at x = c (2)
f β g is continuous at x = c |
1 | 1901-1904 | Theorem 1 Suppose f and g be two real functions continuous at a real number c Then
(1)
f + g is continuous at x = c (2)
f β g is continuous at x = c (3)
f |
1 | 1902-1905 | Then
(1)
f + g is continuous at x = c (2)
f β g is continuous at x = c (3)
f g is continuous at x = c |
1 | 1903-1906 | (2)
f β g is continuous at x = c (3)
f g is continuous at x = c (4)
f
g

ο£Ά

ο£·
ο£
ο£Έ
is continuous at x = c, (provided g(c) β 0) |
1 | 1904-1907 | (3)
f g is continuous at x = c (4)
f
g

ο£Ά

ο£·
ο£
ο£Έ
is continuous at x = c, (provided g(c) β 0) Proof We are investigating continuity of (f + g) at x = c |
1 | 1905-1908 | g is continuous at x = c (4)
f
g

ο£Ά

ο£·
ο£
ο£Έ
is continuous at x = c, (provided g(c) β 0) Proof We are investigating continuity of (f + g) at x = c Clearly it is defined at
x = c |
1 | 1906-1909 | (4)
f
g

ο£Ά

ο£·
ο£
ο£Έ
is continuous at x = c, (provided g(c) β 0) Proof We are investigating continuity of (f + g) at x = c Clearly it is defined at
x = c We have
lim(
)( )
x
c f
g
x
β
+
= lim[ ( )
( )]
x
c f x
g x
β
+
(by definition of f + g)
= lim
( )
lim ( )
x
c
x
c
f x
g x
β
+β
(by the theorem on limits)
= f (c) + g(c)
(as f and g are continuous)
= (f + g) (c)
(by definition of f + g)
Hence, f + g is continuous at x = c |
1 | 1907-1910 | Proof We are investigating continuity of (f + g) at x = c Clearly it is defined at
x = c We have
lim(
)( )
x
c f
g
x
β
+
= lim[ ( )
( )]
x
c f x
g x
β
+
(by definition of f + g)
= lim
( )
lim ( )
x
c
x
c
f x
g x
β
+β
(by the theorem on limits)
= f (c) + g(c)
(as f and g are continuous)
= (f + g) (c)
(by definition of f + g)
Hence, f + g is continuous at x = c Proofs for the remaining parts are similar and left as an exercise to the reader |
1 | 1908-1911 | Clearly it is defined at
x = c We have
lim(
)( )
x
c f
g
x
β
+
= lim[ ( )
( )]
x
c f x
g x
β
+
(by definition of f + g)
= lim
( )
lim ( )
x
c
x
c
f x
g x
β
+β
(by the theorem on limits)
= f (c) + g(c)
(as f and g are continuous)
= (f + g) (c)
(by definition of f + g)
Hence, f + g is continuous at x = c Proofs for the remaining parts are similar and left as an exercise to the reader Rationalised 2023-24
MATHEMATICS
114
Remarks
(i)
As a special case of (3) above, if f is a constant function, i |
1 | 1909-1912 | We have
lim(
)( )
x
c f
g
x
β
+
= lim[ ( )
( )]
x
c f x
g x
β
+
(by definition of f + g)
= lim
( )
lim ( )
x
c
x
c
f x
g x
β
+β
(by the theorem on limits)
= f (c) + g(c)
(as f and g are continuous)
= (f + g) (c)
(by definition of f + g)
Hence, f + g is continuous at x = c Proofs for the remaining parts are similar and left as an exercise to the reader Rationalised 2023-24
MATHEMATICS
114
Remarks
(i)
As a special case of (3) above, if f is a constant function, i e |
1 | 1910-1913 | Proofs for the remaining parts are similar and left as an exercise to the reader Rationalised 2023-24
MATHEMATICS
114
Remarks
(i)
As a special case of (3) above, if f is a constant function, i e , f (x) = Ξ» for some
real number Ξ», then the function (Ξ» |
1 | 1911-1914 | Rationalised 2023-24
MATHEMATICS
114
Remarks
(i)
As a special case of (3) above, if f is a constant function, i e , f (x) = Ξ» for some
real number Ξ», then the function (Ξ» g) defined by (Ξ» |
1 | 1912-1915 | e , f (x) = Ξ» for some
real number Ξ», then the function (Ξ» g) defined by (Ξ» g) (x) = Ξ» |
1 | 1913-1916 | , f (x) = Ξ» for some
real number Ξ», then the function (Ξ» g) defined by (Ξ» g) (x) = Ξ» g(x) is also
continuous |
1 | 1914-1917 | g) defined by (Ξ» g) (x) = Ξ» g(x) is also
continuous In particular if Ξ» = β 1, the continuity of f implies continuity of β f |
1 | 1915-1918 | g) (x) = Ξ» g(x) is also
continuous In particular if Ξ» = β 1, the continuity of f implies continuity of β f (ii)
As a special case of (4) above, if f is the constant function f (x) = Ξ», then the
function g
Ξ» defined by
( )
( )
gx
g x
Ξ»
Ξ»
=
is also continuous wherever g(x) β 0 |
1 | 1916-1919 | g(x) is also
continuous In particular if Ξ» = β 1, the continuity of f implies continuity of β f (ii)
As a special case of (4) above, if f is the constant function f (x) = Ξ», then the
function g
Ξ» defined by
( )
( )
gx
g x
Ξ»
Ξ»
=
is also continuous wherever g(x) β 0 In
particular, the continuity of g implies continuity of 1
g |
1 | 1917-1920 | In particular if Ξ» = β 1, the continuity of f implies continuity of β f (ii)
As a special case of (4) above, if f is the constant function f (x) = Ξ», then the
function g
Ξ» defined by
( )
( )
gx
g x
Ξ»
Ξ»
=
is also continuous wherever g(x) β 0 In
particular, the continuity of g implies continuity of 1
g The above theorem can be exploited to generate many continuous functions |
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