MAT 137Y 20082009 Winter Session, Solutions to Problem Set 4
1
(a) We are given
f
(
x
)
≤
M
and lim
x
→
a
f
(
x
) =
L
exists. Suppose
L
>
M
. Since lim
x
→
a
f
(
x
) =
L
, then for
any
ε
>
0, there exists
δ
>
0 such that 0
<

x

a

<
δ
implies

f
(
x
)

L

<
ε
. In particular, let
ε
=
L

M
>
0. Then there exists
δ
>
0 such that 0
<

x

a

<
δ
implies

f
(
x
)

L

<
L

M
,
which implies
L

(
L

M
)
<
f
(
x
)
<
L
+(
L

M
) =
⇒
M
<
f
(
x
)
<
2
L

M
=
⇒
f
(
x
)
>
M
,
which is a contradiction, since
f
(
x
)
≤
M
for all
x
. Therefore it follows that
L
≤
M
.
(b) The statement is false. Consider the function
f
(
x
) =
(

x
2
,
x
6
=
0
,

1
,
x
=
0
.
Then
f
(
x
)
<
0 for all
x
, but lim
x
→
0
f
(
x
) =
0, so
L
=
M
.
2
(SHE 2.3)
44. Consider the functions
f
(
x
) =
(
0
,
x
<
c
,
1
,
x
≥
c
,
g
(
x
) =
(
1
,
x
<
c
,
0
,
x
≥
c
.
Both limits do not exist at
x
=
c
, but lim
x
→
c
f
(
x
)
g
(
x
) =
0.
46. The statement is false. Since lim
x
→
c
[
f
(
x
) +
g
(
x
)]
and lim
x
→
c
f
(
x
)
exists, then by the properties of
limits,
lim
x
→
c
g
(
x
) =
lim
x
→
c
[
g
(
x
)+
f
(
x
)

f
(
x
)] =
lim
x
→
c
[
f
(
x
)+
g
(
x
)]

lim
x
→
c
f
(
x
)
,
so lim
x
→
c
g
(
x
)
must exist.
48. The statement is false. Let
f
(
x
) =

1. Then lim
x
→
c
f
(
x
) =

1 for any
c
, but
p
f
(
x
)
is not deﬁned
at any
x
=
c
, so lim
x
→
c
p
f
(
x
)
does not exist.
50. The statement is false since it need not be true that lim
x
→
c
f
(
x
)
or lim
x
→
c
g
(
x
)
exists. So, for example,
let
f
(
x
) =
0 and
g
(
x
)
be the Dirichlet function (as deﬁned in SHE 2.2 #47). Then for any
c
,
lim
x
→
c
g
(
x
)
does not exist.
However, if we stipulate that the limits must exist, then the statement is true.
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 lim, Continuous function, x→c, ax bx lim

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