Homework answers, week of Sept. 30
#21, p. 54 (Theorem 2.18.) If lim
x
→
a
f
(
x
) =
L
,
L
6
=
0, and if
g
and
k
are such that
g
(
x
)
6
=
0 for 0
<

x

a

<
k
but lim
x
→
a
g
(
x
) =
0, then
lim
x
→
a
f
(
x
)
g
(
x
)
=
∞
.
Discussion.
In order to talk about the limit of
f
(
x
)
/
g
(
x
)
as
x
→
a
, we must
be assured that the ratio exists everywhere in some interval around
a
(except
possibly at
a
itself). This requires that we not divide by zero, so
g
(
x
)
cannot
be zero in some interval around
a
. (An example is a polynomial with a root
at
a
.)
To establish a proof, we start with the deﬁnitions of the various limits and
decide how to tie them together. In particular, since we have the freedom to
choose the various epsilons associated with the limits of
f
and
g
, we show
how to pick them so that the ratio

f
(
x
)
/
g
(
x
)

is as large as desired.
Proof.
We must show that, given
B
>
0, it is possible to ﬁnd
δ
such that
±
±
±
±
f
(
x
)
g
(
x
)
±
±
±
±
>
B
whenever
0
<

x

a

<
δ
,
(1)
or, equivalently,

f
(
x
)

>
B

g
(
x
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 Spring '07
 thieme
 Intermediate Value Theorem, Continuous function, Intermediate Value, Eqs.

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