(Q. 1Aa.) Let
{
a
n
}
be a sequence of real numbers. If
{
a
n
}
is convergent then prove that its limit
is unique.
[3 marks]
Answer:
Let, if possible, the sequence
{
a
n
}
converge to two different real numbers, say
L
1
and
L
2
, and let
3
δ
=

L
1

L
2

.
[1 mark]
Then for
δ
we have
N
1
, N
2
∈
N
such that

L
i

a
n

< δ
∀
n
≥
N
i
.
[1 mark]
By choosing
N
to be the maximum of
N
1
and
N
2
, we get a contradiction, thus
L
1
=
L
2
.
[1 mark]
(Q. 1Ba.) Let
{
b
n
}
be a sequence of real numbers. If
{
b
n
}
is convergent then prove that its limit
is unique.
[3 marks]
(Q. 1Ca.) Let
{
c
n
}
be a sequence of real numbers. If
{
c
n
}
is convergent then prove that its limit
is unique.
[3 marks]
(Q. 1Da.) Let
{
d
n
}
be a sequence of real numbers. If
{
d
n
}
is convergent then prove that its limit
is unique.
[3 marks]
(Q. 1
·
b.) State the sandwich theorem for sequences.
[3 marks]
Answer:
Let
{
a
n
}
,
{
b
n
}
and
{
c
n
}
be sequences of real numbers such that
[1 mark]
a
n
≤
b
n
≤
c
n
∀
n.
Assume that
{
a
n
}
and
{
c
n
}
are convergent with
lim
n
→∞
a
n
= lim
n
→∞
c
n
=
α
, say.
[1 mark]
Then (the sequence
{
b
n
}
is also convergent and)
lim
n
→∞
b
n
=
α
.
[1 mark]
(Note: The convergence of the middle sequence,
{
b
n
}
, is a part of the conclusion here. If you assume it in the
beginning then your statement is incorrect.)
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(Q. 2A.) Consider the function
f
(
x
) =
(
7

(16)
1
x
7+(16)
1
x
if
x
6
= 0
7
if
x
= 0
.
Prove that there is a point
c
such that
2
< c <
4
and
f
(
c
) =
1
2
.
[6 marks]
Answer:
The function is continuous in the interval
(2
,
4)
, because the exponential function is continuous on the
whole of
R
and
1
/x
is continuous for
x
6
= 0
.
[2 marks]
Consider
g
(
x
) :=
f
(
x
)

1
2
, and note that
g
is also continuous on
(2
,
4)
.
Observe that
g
(2) =
f
(2)

1
2
=
7

4
7 + 4

1
2
<
0
and
g
(4) =
f
(4)

1
2
=
7

2
7 + 2

1
2
>
0
.
[
1 mark each
]
Therefore by the intermediate value theorem, there is some
c
∈
(2
,
4)
such that
g
(
c
) = 0
, or
f
(
c
) =
1
2
.
[2 marks]
(Q. 2B.) Consider the function
f
(
x
) =
(
5

(16)
1
x
5+(16)
1
x
if
x
6
= 0
5
if
x
= 0
.
Prove that there is a point
c
such that
2
< c <
4
and
f
(
c
) =
1
4
.
[6 marks]
(Q. 2C.) Consider the function
f
(
x
) =
(
9

(16)
1
x
9+(16)
1
x
if
x
6
= 0
9
if
x
= 0
.
Prove that there is a point
c
such that
2
< c <
4
and
f
(
c
) =
1
2
.
[6 marks]
(Q. 2D.) Consider the function
f
(
x
) =
(
11

(16)
1
x
11+(16)
1
x
if
x
6
= 0
11
if
x
= 0
.
Prove that there is a point
c
such that
2
< c <
4
and
f
(
c
) =
1
2
.
[6 marks]
(Q. 2A.) Consider the function
f
(
x
) =
(
7

(16)
1
x
7+(16)
1
x
if
x
6
= 0
7
if
x
= 0
.
Prove that there is a point
c
such that
2
< c <
4
and
f
(
c
) =
1
2
.
[6 marks]
Answer:
Note that
f
(
c
) =
1
2
if and only if
7
3
c
= 16
.
[2 marks]
Note that
(
7
3
)
2
<
16
<
(
7
3
)
4
as
2
<
7
3
<
4
.
[2 marks]
The function
x
7→
(
7
3
)
x
is continuous and by intermediate value theorem a required
c
∈
(2
,
4)
exists.
[2 marks]
(Q. 2B.) Consider the function
f
(
x
) =
(
5

(16)
1
x
5+(16)
1
x
if
x
6
= 0
5
if
x
= 0
.
Prove that there is a point
c
such that
2
< c <
4
and
f
(
c
) =
1
4
.
[6 marks]
(Q. 2C.) Consider the function
f
(
x
) =
(
9

(16)
1
x
9+(16)
1
x
if
x
6
= 0
9
if
x
= 0
.
Prove that there is a point
c
such that
2
< c <
4
and
f
(
c
) =
1
2
.
[6 marks]
(Q. 2D.) Consider the function
f
(
x
) =
(
11

(16)
1
x
11+(16)
1
x
if
x
6
= 0
11
if
x
= 0
.
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 Spring '16