ANSWERS TO SELECTED EXERCISES: HW2
Exercise 6.2.12
Let
f
: [0
,
1]
→
R
be a continuous function. Prove that
lim
n
→∞
Z
1
0
f
(
x
n
)
dx
=
f
(0)
.
Proof
:
First approach.
Let
I
n
=
Z
1
0
f
(
x
n
)
dx
. It is natural to apply the mean
value theorem for the continuous function
f
(
x
n
) on [0
,
1]; we get a sequence
{
c
n
}
such that
Z
1
0
f
(
x
n
)
dx
=
f
(
c
n
n
)
,
for all
n
∈
N
.
It is desirable that
lim
n
→∞
c
n
n
= 0, so that
lim
n
→∞
f
(
c
n
n
) = 0, by continuity of
f
.
However, this cannot be deduced from 0
< c
n
<
1, so this approach fails.
If instead of 0
< c
n
<
1 we had 0
< c
n
≤
1

ε
for some very small (but
fixed!) value of
ε
, then a simple application of the squeeze theorem would yield
lim
n
→∞
c
n
n
= 0. What if we
force
these inequalities to be true? We can make this
happen if we consider
I
n,ε
=
Z
1

ε
0
f
(
x
n
)
dx
instead of
I
n
=
Z
1
0
f
(
x
n
)
dx
. It is
natural then to work as follows:
Second approach.
We split the integral into two parts:
Z
1
0
f
(
x
n
)
dx
=
Z
1

ε
0
f
(
x
n
)
dx
+
Z
1
1

ε
f
(
x
n
)
dx.
We denote the first summand by
I
n,ε
. Applying the mean value theorem, we
get
Z
1

ε
0
f
(
x
n
)
dx
=
f
(
ξ
n
n
)(1

ε
)
,
where 0
< ξ
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 Winter '08
 hitrik
 Topology, lim, Limit of a function, Metric space, Limit of a sequence

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