Problem Set 2
MTH211
Solutions
1
Questions on understanding the lecture
E
What are the domain, the codomain, and the component functions of the
following maps?
(a)
(
) =
+

sin
;
(b)
(
) =
1

2

2
;
(c)
(
) =
3
sin(
)
ln(
)
S
(a) The component functions are
1
(
) =
+
2
(
) =

sin
and
the codomain is
R
2
. Now we need to find the domain, that is, the set where the map
is defined. It’s easy to see that the map is defined for all values of
,
,
, so the
domain is
R
3
.
(b) The component function is just
1
(
) =
1

2

2
and the codomain is
R
. Now
we need to find the domain, that is, the set where the map is defined. Expression
under the square root must be nonnegative, that is,
1

2

2
≥
0
, or the domain
is
2
+
2
≤
1
. It is geometrically the unit disc centered at
(0 0)
in
R
2
.
(c) The component functions are
1
(
) =
3
2
(
) = sin(
)
3
(
) = ln(
)
and
the codomain is
R
3
.
Now we need to find the domain, that is, the set where the
map is defined. Here,
3
and
sin(
)
are always defined, but for
ln(
)
we need
>
0
. In other words, we either must have
>
0
or
<
0
. Thus the domain
of this map is the union of the open I and III quadrants.
:)
1
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E
Given a map
:
R
→
R
, prove that
lim
x
→
a
(
x
) =
l
holds if and only if
lim
x
→
a
(
x
)

l
= 0
does.
S
We first prove the “only if" part. For this, we assume
lim
x
→
a
(
x
) =
l
holds and
we want to show this implies that
lim
x
→
a
(
x
)

l
= 0
. Now it follows from the definition
that
lim
x
→
a
(
x
) =
l
means that for any given
>
0
, there exists a
δ >
0
such that for
any
0
<
x

a
< δ
, we have
(
x
)

l
<
. Note that we have
(
x
)

l
≥
0
so that
(
x
)

l
=

(
x
)

l

0

. Therefore, we see that for any
>
0
, there exists a
δ >
0
such that for any
0
<
x

a
< δ
, we have

(
x
)

l

0

=
(
x
)

l
<
, which implies
that
lim
x
→
a
(
x
)

l
= 0
by definition.
We now prove the “if" part. For this, we assume
lim
x
→
a
(
x
)

l
= 0
holds and we want to
show this implies that
lim
x
→
a
(
x
) =
l
. Now it follows from the definition that
lim
x
→
a
(
x
)

l
= 0
means that for any given
>
0
, there exists a
δ >
0
such that for any
0
<
x

a
< δ
, we
have

(
x
)

l

0

<
. Note that we have
(
x
)

l
≥
0
so that

(
x
)

l

0

=
(
x
)

l
.
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 Spring '13
 Limit, lim, Limit of a function

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