Define the function
(
)
=
{
2
sin
?
(
1
2
)
,
0
,
0
,
=
0.
f(x)=
x
2
sin(
x
2
1
),
0,
x
=0,
x=0.
(a)
Prove that
(
)
f(x) is continuous at
=
0
x=0.
(b)
Determine whether
(
)
f(x) is differentiable at
=
0
x=0. If so, find
(
0
)
f
(0).
(c)
Find an explicit formula for
(
)
f
(x) for
0
x
=0.
(d)
Decide whether
(
)
f
(x) is bounded on any neighborhood of
=
0
x=0.
(e)
Determine whether
(
)
f
(x) is Riemann integrable on
[
1
,
1
]
[1,1].
Complete Solution
(a) Continuity at
=
0
x=0
For
0
x
=0,
?
(
)
?
=
?
2
sin
(
1
/
2
)
?
?
2
?
?f(x)?=?x
2
sin(1/x
2
)??x
2
?
Since:
lim
0
2
=
0
x0
lim
x
2
=0
By the Squeeze Theorem:
lim
0
(
)
=
0
=
(
0
)
x0
lim
f(x)=0=f(0)
f is continuous at
0
0.
(b) Differentiability at
=
0
x=0
Using the definition:
(
0
)
=
lim
0
(
)
(
0
)
=
lim
0
2
sin
(
1
/
2
)
=
lim
0
sin
(
1
/
2
)
f
(0)=
h0
lim
h
f(h)f(0)
=
h0
lim
h
h
2
sin(1/h
2
)
=
h0
lim
hsin(1/h
2
)
Since:
1
sin
(
1
/
2
)
1
?
?
sin
(
1
/
2
)
?
?
1sin(1/h
2
)1?h?hsin(1/h
2
)?h?
Thus:
lim
0
sin
(
1
/
2
)
=
0
h0
lim
hsin(1/h
2
)=0
f is differentiable at
0
0, and
(
0
)
=
0
f
(0)=0.
(c) Formula for
(
)
f
(x) for
0
x
=0
Differentiate:
(
)
=
2
sin
(
1
/
2
)
f(x)=x
2
sin(1/x
2
)
Using product and chain rules:
(
)
=
2
sin
(
1
/
2
)
+
2
cos
(
1
/
2
)
(
2
3
)
f
(x)=2xsin(1/x
2
)+x
2
cos(1/x
2
)(
x
3
2
)
Simplifying:
(
)
=
2
sin
(
1
/
2
)
2
cos
(
1
/
2
)
f
(x)=2xsin(1/x
2
)
x
2
cos(1/x
2
)
(d) Boundedness of
(
)
f
(x) near
0
0
Consider:
(
)
=
2
sin
(
1
/
2
)
2
cos
(
1
/
2
)
f
(x)=2xsin(1/x
2
)
x
2
cos(1/x
2
)
The first term
2
sin
(
1
/
2
)
0
2xsin(1/x
2
)0.
The second term:
?
2
cos
(
1
/
2
)
?
=
2
?
?
x
2
cos(1/x
2
)
=
?x?
2
which diverges to infinity as
0
x0.
(
)
f
(x) is unbounded in every neighborhood of
0
0.
(e) Riemann integrability of
(
)
f
(x) on
[
1
,
1
]
[1,1]
Although
(
)
f
(x) is unbounded near
0
0, it is discontinuous only at a single point
=
0
x=0.
A function that is:
bounded on compact intervals except possibly at finitely many points, and
has a finite number of discontinuities,
is Riemann integrable.
Alternatively, note:
1
1
(
)
=
(
1
)
(
1
)
1
1
f
(x)dx=f(1)f(1)
by the Fundamental Theorem of Calculus, since
f is differentiable everywhere and continuous on
[
1
,
1
]
[1,1].
Thus:
(
)
f
(x) is Riemann integrable on
[
1
,
1
]
[1,1].
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