 1.
 (b) 1. (d) DNE. (f)
. (h) DNE. (j) . (k)
.
 2.
 (b)
. (d) DNE. (e) 0. (f)
.
(g)
. (h)
. (k) 1. (l)
.
 E.
 If we substitute in in the top and the bottom, we get
the case. So we look for a factor of in the top and in
the bottom. Since
and
(use polynomial or synthetic division to divide by ),
we can cancel the from the top and from the bottom. Thus,
.
 3.
 (a) . (d)
. (e)
. (f)
2. (g)
. (h) (l) DNE.
. (k) 0.
 4.
 (a) 1. (b) 0. (d) 0. (e) 0. (h) DNE. (i) 0. (j) DNE (undefined for
).
 5.
 Theorem.
.
Proof (
) If as , then
. Hence
as .
(
) If
as , then
. Therefore
as .
 6.
 For example,
has no limit as , but
.
 8.
 Theorem. If
.
Proof. Since as , we can choose close enough to
that
. Then, given any
, we can choose close
enought to that
. Then, with close
enough to , we have
.
 14.
 (a) 0 (b) 0 (c) 0
(d) Undetermined. For example, if
and
, then
, which depends on .
(e) Undetermined. For example, if
, then
, which depends on .
 16.

, and
. Hence
.
 18.
 (a) bounded, . (b) not bounded. (c) bounded, . (d)
not bounded. (e) bounded,
. (f) bounded, .
 19.
 is not bounded, but
is bounded
for not 0. If
exists and is some nonzero
number, then
is not bounded near 0. Hence if
is bounded for all , then either as or
does not exist.
 20.
 (a)
. (b)
. (c)
. (d) The area of an inscribed polygon
with n sides is
and the area
of the circle is
.