Math 521  Advanced Calculus I
Instructor: J. Metcalfe
Due: April 12, 2010
Assignment 24
1.
Suppose
f
is deﬁned in a neighborhood of
x
, and suppose
f
00
(
x
) exists. Show that
lim
h
→
0
f
(
x
+
h
) +
f
(
x

h
)

2
f
(
x
)
h
2
=
f
00
(
x
)
.
Show by an example that the limit may exist even if
f
00
(
x
) does not.
Let
f
(
x
) =

x
2
for
x
≤
0, and
f
(
x
) =
x
2
for
x
≥
0. Then,
f
0
(
x
) =

x

, and
f
00
does
not exist at
x
= 0. On the other hand, we have, for
x
= 0,
f
(
h
) +
f
(

h
)

2
f
(0) = 0
and thus the limit as
h
→
0 is also 0. This gives the requested counterexample.
Assuming, now, that
f
00
(
x
) exists, we have, by l’Hospital’s rule,
lim
h
→
0
f
(
x
+
h
) +
f
(
x

h
)

2
f
(
x
)
h
2
= lim
h
→
0
f
0
(
x
+
h
)

f
0
(
x

h
)
2
h
=
1
2
±
lim
h
→
0
f
0
(
x
+
h
)

f
0
(
x
)
h
+ lim
h
→
0
f
0
(
x
)

f
0
(
x

h
)
h
²
=
f
00
(
x
)
.
2.
Let
h
(
x
) :=
e

1
/x
2
for
x
6
= 0 and
h
(0) := 0. Show that
h
(
n
)
(0) = 0 for all
n
∈
N
. Conclude that
the remainder term in Taylor’s theorem for
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 Spring '10
 MetCalfe
 Calculus, lim, L’Hospital’s Rule, J. Metcalfe

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